From 2f0da22351d6cd38c0d9f50834dfd73618e94339 Mon Sep 17 00:00:00 2001 From: Jim <0xJ1M@users.noreply.github.com> Date: Fri, 20 Mar 2026 20:58:59 +0000 Subject: [PATCH] Update --- Vol1/README.md | 23080 +++++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 23080 insertions(+) create mode 100644 Vol1/README.md diff --git a/Vol1/README.md b/Vol1/README.md new file mode 100644 index 0000000..f6c5d26 --- /dev/null +++ b/Vol1/README.md @@ -0,0 +1,23080 @@ +# Foundations {#part1} + +### Mathematical logic (To add to as needed) {#SectionOne} + +::: epigraph +There are no facts, only interpretations. + +*Friedrich Nietzsche* +::: + +In this section, we will introduce mathematical logic. This will give us +the tools and basic building blocks to be able to talk about mathematics +formally. What do we mean by 'in a formal way'? Modern mathematics is +built on a bedrock of logic, that is to say, given some statements which +we will take to be true or have already been proven true, what can we +logically deduce must also be true, and what is also false. As an +example, we are familiar with the idea of positive whole numbers, also +called positive integers; we are also familiar with the idea of a +positive whole number being prime when the only other positive whole +numbers that divide it are $1$ and itself, for example, $2$ is prime. +From the facts that the positive whole numbers exist and there is at +least one prime, we can logically deduce there must be infinitely many +primes. We will see the proof of this later. + +In this document we won't be needing the full tools of mathematical +logic, doing so will take us too far afield, instead, we will only cover +the key fundamentals we will need as well as define some terms which +will be used throughout. + +#### Defining a definition + +What is a definition? What does it mean to define something? Definitions +are at the heart of mathematics, without them we wouldn't be able to do +anything at all. A definition is a declaration that gives a formal name +to an object, class of objects, ideas, etc. For example, we can define +prime numbers, such a definition might look something like this. + +::: Def +**Definition 1**. *Definition of a prime number* + +*Consider a positive whole number, we say that this positive whole +number is a prime number if the only other positive whole numbers that +divide it are itself and the number $1$.* +::: + +With this definition whenever we refer to the idea of a prime number, we +know that this prime number must satisfy that it only has two distinct +numbers that divide it, itself and $1$. As we will say throughout this +document, we can use a definition when making logical arguments. +Definitions are the backbone of defining the setup to logical arguments, +if we don't know about the objects we are arguing about then we can't +make any logical deductions, or deduce the truth of mathematical +statements. Now that we know what a definition is, we can start using it +to lay the foundation for the rest of the document. For formality, we +will make, somewhat paradoxically, a formal definition of a definition + +::: definition +**Definition 1**. *Definition* + +*A definition is a statement which gives a formal name to a concept.* +::: + +#### What is truth? + +What is truth? In particular, what is mathematical truth? Loosely +speaking truth and mathematical truth is based on the idea of does the +premise entail this conclusion. That is to say, if we assume that a few +statements are true, then the conclusion we are trying to reach is also +true. This is rather vague at the moment because we haven't defined what +we mean by true. + +##### Logical statements and logical connectives + +We will need a few definitions. + +::: definition +**Definition 2**. *Declarative logical statement* + +*We define a Declarative logical statement to be either true or false. +Here we are using the intuitive definition of true and false.* +::: + +We need to make the definition of declarative logical statements to +define what we mean by true and false, again somewhat paradoxically we +need a definition of true and false to define what we mean when a +declarative logical statement is true. We shall ignore the paradoxical +nature of these definitions. + +::: definition +**Definition 3**. *Assignment of truth* + +*Let $P$ be a declarative logical statement, an assignment of truth is +an interpretation of the statement $P$ that sees $P$ as either true or +false. We write this as $\delta\left(P\right)$.* + +*If this assignment of truth $\delta$ sees $P$ as true we write +$\delta\left(P\right)=1$ and we say that $\delta$ interprets $P$ as +true. If this assignment of truth sees $P$ as false we write +$\delta\left(P\right)=0$ and we say that $\delta$ interprets $P$ as +false.* +::: + +These two definitions will allow us to build the foundations that we +will need. It is first important to note that an assignment of truth is +not an absolute assignment of the truth of a declarative logical +statement. Different assignments of truth, and thus different +interpretations, can give rise to different values of $P$ being true or +false. Now, we have a building blocks to build more complex logical +statements. + +A first natural question is when does one the truth of one logical +statement imply the truth or falseness of another? Thinking about how +this should work gives us a sense that something true should never imply +that something false is true, whereas something false can imply anything +at all. Using this we define the logical implication operator. + +::: definition +**Definition 4**. *Logical implication* + +*Let $P$ and $Q$ be logical statements. We define the logical +implication of the statements $P$ and $Q$, written as $P\Rightarrow Q$, +to have the following logical values* + + *$P$* *$Q$* *$P\Rightarrow Q$* + ------- ------- -------------------- + *1* *1* *1* + *1* *0* *0* + *0* *1* *1* + *0* *0* *1* + + : *The truth table for the logical implication operator.* + +*We read this as $P$ implies $Q$, or if $P$ then $Q$.* +::: + +::: example +**Example 1**. *Let $P =$ "The sky is overcast" and let $Q =$ "The sun +is not visible". We have by the truth table of logical implication that +$P\Rightarrow Q$ is true when* + +1. *$P$ is true and $Q$ is true* + +2. *$P$ is false and $Q$ is true* + +3. *Both $P$ and $Q$ are false.* + +*In words we have $P\Rightarrow Q$ is true in these circumstances* + +1. *If it is true the sky is overcast then the sun is not visible.* + + *That is, if the sky is overcast then the sun is not visible* + +2. *If it is false that the sky is overcast then the sun is not + visible.* + + *That is, if the sky is not overcast then the sun is not visible.* + +3. *If it is false that the sky is overcast then the sun is visible.* + + *That is, if the sky is not overcast then the sun is visible.* + +*In particular case two could be true say when it is nighttime, if it is +nighttime the sun is clearly not visible[^1].* + +*Lets look at these statements the other way, $Q\Rightarrow P$. We have +that is is true when* + +1. *$Q$ is true and $P$ is true* + +2. *$Q$ is false and $P$ is true* + +3. *Both $Q$ and $P$ are false.* + +*In words that is we have $Q\Rightarrow P$ is true in these +circumstances* + +1. *If the sun is not visible then the sky is overcast* + +2. *If the sun is visible then the sky is overcast* + +3. *If the sun is visible then the sky not is overcast* +::: + +There is one definition that arises from logical implication that is +occasionally useful in proving other statements. + +::: definition +**Definition 5**. *Vacuous truth* + +*Let $P$ and $Q$ be statements such that we have $P\Rightarrow Q$. +Suppose that $P$ is false, then by the definition of logical implication +we have that $P\Rightarrow Q$ is true. We say that $P\Rightarrow Q$ is +vacuously true.* +::: + +::: example +**Example 2**. *The statement "All my children are goats" is vacuously +true for someone who doesn't have any children.* +::: + +It is often the case we have theorems in mathematics which are of the +form $P$ implies $Q$ and $Q$ implies $P$, that is two separate logical +sentences can imply each other. This is the logical bi-conditional. + +::: definition +**Definition 6**. *Logical Bi-conditional* + +*Let $P$ and $Q$ be logical statements. We define the logical +Bi-conditional of the statements $P$ and $Q$, written +$P\Leftrightarrow Q$, to have the following logical values* + + *$P$* *$Q$* *$P\Leftrightarrow Q$* + ------- ------- ------------------------ + *1* *1* *1* + *1* *0* *0* + *0* *1* *0* + *0* *0* *1* + + : *The truth table for the logical Bi-conditional operator.* + +*We read this as $P$ if and only $Q$, meaning $P$ implies $Q$ and $Q$ +implies $P$.* +::: + +::: example +**Example 3**. *Let $P =$ "A number is even" and let $Q =$ "It is +divisible by 2". By the truth table of the logical bi-conditional that +$P\Leftrightarrow Q$ is true when* + +1. *Both $P$ and $Q$ are true.* + +2. *Both $P$ and $Q$ are false.* + +*That is in words we have $P\Leftrightarrow Q$ when* + +1. *A number is even if and only if it is divisible by 2* + +2. *A number is not even if and only if it is not divisible by 2* +::: + +Now that we have the logical implication and logical bi-conditional, we +can start defining more complex logical connectives. These are the +logical conjunction, logical disjunction and logical negation + +::: definition +**Definition 7**. *Logical conjunction* + +*Suppose we have two logical statements $P$ and $Q$. We define logical +conjunction, written as $P\wedge Q$, to be true if and only if $P$ and +$Q$ are both true, that is to say the logical conjunction connective has +the following truth table* + + *$P$* *$Q$* *$P\wedge Q$* + ------- ------- --------------- + *1* *1* *1* + *1* *0* *0* + *0* *1* *0* + *0* *0* *0* + + : *The truth table for the logical conjunction operator.* + +*Informally, we call this logical AND rather than logical conjunction.* +::: + +::: example +**Example 4**. *Let $P =$"$x > 2$" and $Q =$"$x < 10$" and suppose that +$P$ and $Q$ are true, then $P\wedge Q$ is true and represents the +expression $2m +\end{equation*}$$ that is $n$ is greater than $m$, where the domain of +discourse $D=\mathbb{N}$ is again the positive whole numbers +$1,2,3,\dots$.* + +*Suppose that $n=2$ and $m=3$, then $P\left(n,m\right)$ is false, if +$n=45$ and $m=7$ then $P\left(n,m\right)$ is true.* +::: + +We see that logical propositions allow us to construct more complex +logical statements and are the building blocks for the more complex +Mathematical statements that we will be using. + +#### Proof + +Logic and truth are two of the corner stones of Mathematics, the third +is proof. Without proof we are unable to verify the truth of any +mathematical statements. So what exactly is a proof? + +::: definition +**Definition 12**. *Mathematical proof* + +*Suppose we have some logical statements which are known or assumed to +be true, and suppose we wish to see if some conclusion if true given +this assumption. We define a Mathematical proof is where we start from +these assumptions and at each step logically deduce additionally true +statements until we have proven the conclusion. In other words a +Mathematical proof can be broken down into a simple question. Do the +assumptions entail this conclusion?* +::: + +This isn't a truly rigours definition of a mathematical proof, and one +can define this rigorously in a course on mathematical logic. To do so +here would be too much of a diversion, instead we will just keep in our +minds that a proof is a series of logical deductions from assumptions to +a conclusion. When we have reached the conclusion we use a special +symbol. We use the symbol $\qed$ at the end of a proof to show that we +are done. + +There are many different types of proof that we will invoke throughout +the rest of this document. + +##### Direct Proof + +The first type of proof we define is direct proof. We define a direct +proof as follows. + +::: definition +**Definition 13**. *Direct Proof* + +*In a direct proof, the conclusion is logically established by using +axioms, definitions and previously proven theorems.* +::: + +We will give an example of direct proof. + +::: example +**Example 8**. *In this example we will breakdown each step of a direct +proof.* + +*Here we will give the definitions we will be using and any assumptions +which we will be using in the prove(i,e previously proven theorem):* + +1. *We say a number is an integer if it is a whole number, such as + $-4,-3,54,8,0,2,7$ and so on.* + +2. *We will also assume that adding and multiplying integers works as + we would have taught in school, for example $5+7=12$, $2*14=28$ + etc.* + +3. *We say that an integer is an even integer if it can be written as + $x=2*m$ where $m$ is any integer.* + +*We now move to the proof.* + +*Suppose we have two such even integers, say $x$ and $y$. We will use +direct proof to show that $x+y$ must also be even.* + +*Proof:* + +*Suppose we have two even integers $x$ and $y$. By the definition of an +even integer we have that $x=2*n$ and $y=2*m$ for some integers $n$ and +$m$. Now consider $xx+y$, we have* + +*$$\begin{equation*} + x+y=2*n+2*m=2*\left(n+m\right) +\end{equation*}$$ Now, $n+m$ is adding two integers together and is an +integer. say $k=n+m$, hence we have that $x+y=2*k$, but by definition of +an even we have that $x+y$ is even. This concludes the proof. $\qed$* +::: + +##### Proof by contradiction + +The second type of proof we define is proof by contradiction. This is a +very powerful tool. + +::: definition +**Definition 14**. *Proof by contradiction* + +*Suppose we have a logical statement $P$ that we wish to find the truth +of. If we suppose that $\neg P$ is true and then assuming $\neg P$ we +can derive another logical statement $Q$ which is known to be false, or +we can derive both $Q$ and $\neg Q$. Then we must have that $\neg P$ is +false and $P$ is true.* +::: + +In other words, proof by contradiction states that if, when making an +assumption, we can derive a false statement, then the assumption itself +must have been invalid. We can justify proof by contradiction using the +following truth table. + + $P$ $\neg P$ $\neg\neg P$ $\neg\neg P\Rightarrow P$ + ----- ---------- -------------- --------------------------- + 1 0 1 1 + 0 1 0 1 + + : The truth table for proof by contradiction. + +::: example +**Example 9**. *Like with the example using direct proof. We will break +down each step of proof by contradiction.* + +*Here we will give the definitions we will be using and any assumptions +which we will be using in the prove(i,e previously proven theorem):* + +1. *We say a number is a rational number if it is the ration of two + integers $a$ and $b$ where $b\neq 0$. Examples of rational numbers + are $\displaystyle \frac{1}{2},\frac{2}{3},-\frac{15}{8}$ and so on. + We say a number is irrational if it is not rational.* + +2. *We say that a rational number $\displaystyle \frac{a}{b}$ is in + simplest form if the only number that divides both $a$ and $b$ is + $1$.* + +3. *Any rational number has a simplest form.* + +4. *We will also assume that adding and multiplying rational numbers + works as we would have taught in school, that is we have for two + rational numbers $\displaystyle \frac{a}{b}$ and + $\displaystyle \frac{c}{d}$ that* + + *$$\begin{equation*} + \frac{a}{b}+\frac{c}{d}= \frac{a*d+b*c}{b*d},\ \frac{a}{b}*\frac{c}{d}=\frac{a*c}{b*d} + \end{equation*}$$* + +5. *We say $\sqrt{2}$ is the number which satisfies + $\sqrt{2}*\sqrt{2}=2$* + +6. *We assume the definition of an even integer from the previous + example* + +7. *If $a*a=a^2$ is an even integer, then so is $a$* + +*We now move to the proof.* + +*We have that $\sqrt{2}$ is an irrational number. This is to say that +$\sqrt{2}$ is not the ratio of two whole numbers $a$ and $b$ where +$\displaystyle \frac{a}{b}$ is in simplest form.* + +*Proof:* + +*Aiming for a proof by contradiction, suppose that $\sqrt{2}$ is a +rational number that is in simplest form. This is to say we have that +$\displaystyle \sqrt{2}=\frac{a}{b}$ for some integers $a,b$. We have by +assumption that $\sqrt{2}$ is the number such that +$\sqrt{2}*\sqrt{2}=2$. Hence we have that* + +*$$\begin{equation*} + \sqrt{2}*\sqrt{2}=\frac{a}{b}*\frac{a}{b}=\frac{a^2}{b^2}=2 +\end{equation*}$$ Where $a^2=a*a$ and $b^2 = b*b$. We can multiply the +above expression by $b^2$ on both sides to get* + +*$$\begin{equation*} + a^2=2*b^2 +\end{equation*}$$* + +*By definition of an even integer we have that $a^2$ is even and so $a$ +must be even, that is $a=2*k$ for some integer $k$. Hence we have that* + +*$$\begin{equation*} + a^2=\left(2*k\right)^2=4*k^2=2*b^2 +\end{equation*}$$ That is $4*k^2=2*b^2$ which implies that $b^2=2*k^2$, +that is $b^2$ is even and so $b$ must be even. This is a contradiction, +as we have that $a$ is even and $b$ is even and so there share a divisor +of $2$, contradicting the fact we assumed that +$\displaystyle\sqrt{2}=\frac{a}{b}$ was in simplest form.* + +*Therefore, $\sqrt{2}$ must be irrational. $\qed$* +::: + +##### Proof by contra-position + +Another type of proof that we define is proof by contra-position, +sometimes called proof by contra-positive. + +::: definition +**Definition 15**. *Proof by contra-position* + +*Suppose we have a logical statement $P$ and we wish to show that $P$ +implies some other statement $Q$. We are able to show that +$P\Rightarrow Q$ if we can show that $\neg Q\Rightarrow \neg P$.* +::: + +Proof by contra-position states that proving a statement of the form +$P\Rightarrow Q$ is the same as showing that $\neg Q\Rightarrow\neg P$. +It is easier to see this from the truth table. + + $P$ $Q$ $\neg P$ $\neg Q$ $P\Rightarrow Q$ $\neg Q\Rightarrow\neg P$ + ----- ----- ---------- ---------- ------------------ --------------------------- + 1 0 0 1 0 0 + 1 1 0 0 1 1 + 0 0 1 1 0 1 + 0 1 1 0 1 1 + + : The truth table for proof by contra-positive. + +Maybe, to make it even clearer, we can use a worded example. Let $P$ +denote the statement "It is raining" and $Q$ denote the statement "I +wear my coat". We have that $P\Rightarrow Q$[^2]. The contra-positive +would be $\neg Q\Rightarrow\neg P$. In words this would be "If I don't +wear my coat" then "It is not raining". + +::: example +**Example 10**. *A more mathematical example can be seen now. We will +let $x$ be an integer and we will show that if $x^2$ is even then $x$ is +even. We will use proof by contra-position. So We will show that if $x$ +is not even then $x^2$ is not even.* + +1. *So, $x$ not being even means $x$ is odd. This means that $x=2n+1$ + for some integer $n$.* + +2. *Now, we have + $x^2=\left(2n+1\right)^2=4n^2+4n+1=2\left(2n^2+2n\right)+1$.* + +3. *Hence, we have shown that $x^2$ is of the form $2k+1$ for some + integer $k$.* + +4. *Therefore $x^2$ is odd.* + +*Concluding the proof by contra-positive.* +::: + +### Sets and mappings {#intro} + +::: epigraph +No one shall expel us from the paradise that Cantor has created for us. + +*David Hilbert* +::: + +#### Sets + +##### Introduction and basic definitions + +We start with the most elementary definition, a Set or less formally, a +collection of 'objects'. This notion of an object is not very rigorous, +what do we mean by an object? Do these objects really exist?[^3] In what +way can one collection of objects differ from another? + +These questions are at the foundation of Mathematics and to justify the +notions and hence tools we need would require a significant detour into +the realm of Mathematical logic. The interested reader would find +so-called Zermelo--Fraenkel set theory to be of interest in formalising +the notion of a set, we will give a brief overview at the end of the +section. To avoid the trip into Mathematical logic, we will instead +define sets with a more 'hands on' approach + +::: definition +**Definition 16**. *Naive definition of a Set* + +*A set is a collection of objects. We list the elements surrounded by +curly brackets $\{$ $\}$.* +::: + +This definition will make sense after we see some examples + +::: example +**Example 11**. *Let $S=\left\{1,2,3,Dogs,Cats,Apples,Pears\right\}$. +Then $S$ is a set.* +::: + +::: example +**Example 12**. *Let +$S=\left\{"Foo", \left\{1,2,3,Dogs,Cats,Apples,Pears\right\}, Apples, Pears\right\}$. +Then $S$ is a set. We note that the set from the previous example is now +in this set.* +::: + +It would be useful to talk about a particular object in some set $S$. +For example we can say that $1$ is in the set from example 2.1. above. +We formalise this idea + +::: definition +**Definition 17**. *Element of a set* + +*An object in a set is called an element of the set.* +::: + +::: definition +**Definition 18**. *Set membership* + +*Let $S$ be a set and let $x$ be an element of the set $S$. We say that +$x$ is a member of the set $S$ and write $x\in S$. If $y$ is some object +which is not in the set $S$ we write that $y\not\in S$.* +::: + +::: example +**Example 13**. *Let $S=\left\{1,2,3,Dogs,Cats,Apples,Pears\right\}$. We +have that $1\in S$ and $Dogs\in S$ but we have that $Blue\not\in S$.* +::: + +The above example shows a few interesting points. Dogs in English is +used when we wish to talk about multiple dogs at once, so it would be +absurd to deny that $Dogs$ could itself be a set, for example +$Dogs=\left\{Lassie, Scooby-Doo, Snoopy, Blue\right\}$. So we have that + +$$\begin{equation*} + S=\left\{1,2,3,\left\{Lassie, Scooby-Doo, Snoopy, Blue\right\},Cats,Apples,Pears\right\} +\end{equation*}$$ + +Does this now mean that $Blue\in S$?. The answer is no, $Blue$ is not +any one of the objects in $S$, however there is an object in $S$ that +does contain $Blue$, namely $Dogs$. This shows that $\in$ only looks at +most one layer deep of $\left\{\dots\right\}$. + +One might wonder if it can ever be the case that a set contains itself, +that is a set like $S=\left\{S\right\}$? Again the answer is no, to see +why we need to define a new way of making sets, where the elements of +the set are conditioned on some statement being true. + +::: example +**Example 14**. *Suppose we want the set of all even integers then we +have* + +*$$\begin{equation*} + S=\left\{x : x\text{ is an even integer}\right\} +\end{equation*}$$ The $:$ symbol stands for such that, so $S$ reads the +elements $x$ such that $x$ is an even integer.* +::: + +Returning to the question of can a set contain itself. Consider the set + +$$\begin{equation*} + S=\left\{R: R\text{ is a set and }R\not\in R\right\} +\end{equation*}$$ That is $S$ is the set of all sets $R$ such that $R$ +is a set and $R$ does not contain itself. Now suppose that $S\in S$. By +definition of $S$ we must conclude that $S\not\in S$. Conversely if +$S\not\in S$ then by definition of $S$ we have that $S\in S$. This is an +issue, and shows the flavour of the issues of allowing a set to contain +itself, so we shall revise our definition to not allow for a set to +contain itself. + +::: definition +**Definition 19**. *Set* + +*A set is a collection of objects such that none of the objects in the +collection is the set itself.* +::: + +##### Subsets and universal quantifiers + +Given a set, we can talk about a smaller collection of the elements of +the set, which we call a subset. + +::: definition +**Definition 20**. *Subset* + +*Let $S$ be a set. If $K$ is also a set such that for every $x\in K$ we +also have that $x\in S$ then we say that $K$ is a subset of $X$, and +write $K\subseteq S$. We say that $K$ is a proper subset of $S$ if we +have that $S\subseteq T$ and $S\neq T$, we denote a proper subset by +$\subset$, hence $\subseteq$ allows for the possibility that $K=S$. We +call $\subseteq$ and $\subset$ the set inclusion operators.* +::: + +Conversely can also define the notion of a super-set, this isn't too +useful for what we are doing but it does sometimes appear in other text +so it worth mentioning it now. + +::: definition +**Definition 21**. *Super-set* + +*Let $S\subseteq T$. We say that $T$ is a super-set of the set $S$ and +we write this as $T\supseteq S$.* +::: + +::: example +**Example 15**. *Let $S=\left\{1,2,3,4,5,6\right\}$ then some subsets of +$S$ are $\left\{1,2\right\}$, $\left\{4\right\}$ and +$\left\{1,2,6\right\}$* +::: + +With the idea of a subset we have our first proposition + +::: {#prop:TwosetsEqualIfContainedInEachOther .proposition} +**Proposition 1**. *Two sets are equal if and only if they are subsets +of each other* + +*Let $X$ and $Y$ be sets. We have that $X=Y$ if and only if +$X\subseteq Y$ and $Y\subseteq X$.* + +*Proof:* + +*This is an if and only if proposition so we have to prove that given +$X=Y$ then $X\subseteq Y$ and $Y\subseteq X$ and then we need to show +that given $X\subseteq Y$ and $Y\subseteq X$, that $X=Y$.* + +*$\left(\Rightarrow\right)$: Suppose that $X=Y$ then we have that $X$ +and $Y$ have the same elements, in particular we have that every +$x\in X$ is also in $Y$ so that $X\subseteq Y$. Likewise +$Y\subseteq X$.* + +*$\left(\Leftarrow\right)$: Suppose that $X\subseteq Y$ and +$Y\subseteq X$. $X\subseteq Y$ means that for every $x\in X$ we have +that $x\in Y$. Likewise $Y\subseteq X$ means that for every $x\in Y$ we +have that $x\in X$. Hence we must have that the elements of $X$ and $Y$ +are the same, that is $X=Y$. $\qed$* +::: + +There is also another property of subsets that is useful. + +::: {#prop:SetInclusionTransitivityProp .proposition} +**Proposition 2**. *Set inclusion transitivity property* + +*Let $R,S$ and $T$ be sets such that $R\subseteq S$ and $S\subseteq T$. +We have that $R\subseteq T$* + +*Proof:* + +*Let $R,S$ and $T$ be sets such that $R\subseteq S$ and $S\subseteq T$. +Suppose that $x\in R$. By assumption we have that $R\subseteq S$ and so +$x\in S$. Likewise by assumption we have that $S\subseteq T$ and so +$x\in T$. Hence $R\subseteq T$.* + +*The result follows. $\qed$* +::: + +A similar result holds if we replace subsets with proper subsets. + +::: {#prop:ProperSetInclusionTransitivityProp .proposition} +**Proposition 3**. *Proper set inclusion transitivity property* + +*Let $R,S$ and $T$ be sets such that $R\subset S$ and $S\subset T$. We +have that $R\subset T$* + +*Proof:* + +*Let $R,S$ and $T$ be sets such that $R\subset S$ and $S\subset T$. +Suppose that $x\in R$. By assumption we have that $R\subset S$ and so +$x\in S$. Likewise by assumption we have that $S\subset T$ and so +$x\in T$. Hence $R\subset T$.* + +*We must show that it is not possible for $R=T$. As $R\subset S$ then by +definition we have that $R\neq S$, likewise as $S\subset T$ then +$S\neq T$. As $R\neq S\neq T$ we conclude that $R\neq T$ and so +$R\subset T$.* + +*The result follows. $\qed$* +::: + +We can also make the following observation. + +::: {#prop:ProperSetSubSetInclusionNotTransitivity .proposition} +**Proposition 4**. *Proper set inclusion and subset inclusion is not +transitive* + +*Let $R,S$ and $T$ be sets such that $R\subseteq S$ and $S\subset T$. We +have that $R\subset T$* + +*Proof:* + +*Let $R,S$ and $T$ be sets such that $R\subseteq S$ and $S\subset T$.* + +*If $R\neq S$ then $R\subset S$ and so proposition +[3](#prop:ProperSetInclusionTransitivityProp){reference-type="ref" +reference="prop:ProperSetInclusionTransitivityProp"} applies. So suppose +that $R=S$ then $R\subseteq S$ and so $\forall x\in R$ we have that +$x\in S$. Now as $S\subset T$ we have that $S\neq T\implies R\neq T$ as +$R=S$.* + +*The result follows. $\qed$* +::: + +We will define what we truly mean by transitivity in the next chapter, +right now it is more important to know that sets satisfy this property +than why this property is named the way it is. As set inclusion is +transitive, so is set equality. + +::: proposition +**Proposition 5**. *Set equality transitivity property* + +*Let $R,S$ and $T$ be sets such that $R=S$ and $S=T$. We have that +$R=T$.* + +*Proof:* + +*Let $R,S$ and $T$ be sets such that $R=S$ and $S=T$. We have that +$R=T$. By equality of sets we have that $R\subseteq S$ and +$S\subseteq R$, likewise we also have that $S\subseteq T$ and +$T\subseteq S$. Now as $R\subseteq S$ and $S\subseteq T$ then we must +have by transitivity of set inclusion that $R\subseteq T$. Moreover as +$T\subseteq S$ and $S\subseteq R$ we again have by transitivity that +$T\subseteq R$. The result follows by equality of sets. $\qed$* +::: + +::: definition +**Definition 22**. *The empty-set* + +*The empty-set is the set that contains no elements. It is denoted by +$\emptyset$.* +::: + +To make our lives a little easier we will introduce some notation + +::: definition +**Definition 23**. *Universal and existential quantifiers* + +*Let $S$ be any set. The universal quantifier $\forall$, meaning for +all, allows us to talk about every element $S$. We can condition the +universal quantifier with a such that ,$:$, in order to pick all the +elements that satisfy a given condition.* + +*The existential quantifier $\exists$ tells us of the existence of an +element in $S$. Just saying an element in a set exists is not +particularly usual and so we normally combine $\exists$ with a +condition.* +::: + +Some examples will help us here. + +::: example +**Example 16**. *Consider the set +$\left\{1,2,3,4,5,\dots\right\}=\mathbb{N}$, we call $\mathbb{N}$ the +natural numbers. Moreover, consider $S=\left\{1,2,3,4,5,6\right\}$* + +1. *We have that $\forall x\in S$ that $x\in\mathbb{N}$, that is every + element of $S$ is also an element of $\mathbb{N}$.* + +2. *We can apply the universal quantifier multiple times in a + statement, for example* + + *$$\begin{equation*} + \forall a\in\mathbb{N},\forall b\in\mathbb{N},\exists c\in\mathbb{N}:a+b=c + \end{equation*}$$* + +3. *Let $a,b\in\mathbb{N}$ that is let $a\in\mathbb{N}$ and let + $b\in\mathbb{N}$. Then we can construct the following set. We say + that $a$ is divisible by $b$ if $\exists c\in\mathbb{N}$ such that + $a=bc$, we write this as $b\mid a$. The set of all such $c$ can be + expressed by* + + *$$\begin{equation*} + C=\left\{c\in\mathbb{N}:a=bc\right\} + \end{equation*}$$* +::: + +The empty set has the interesting property that it is a subset of any +set. + +::: {#prop:EmptySetincontainedineveryset .proposition} +**Proposition 6**. *The empty-set is contained in every set* + +*Let $S$ be any set. Then $\emptyset\subseteq S$* + +*Proof:* + +*We have that $\emptyset\subseteq S$ means that every element of +$\emptyset$ is also contained in $S$. The definition of the empty set +means that there are no elements in $\emptyset$. We can phrase this to +the following statement* + +*$$\begin{equation*} + \forall x: x\in\emptyset\Rightarrow x\in S +\end{equation*}$$ But $x\in\emptyset$ is not true for any $x$ so* + +*$$\begin{equation*} + \forall x: x\in\emptyset\Rightarrow x\in S +\end{equation*}$$* + +*is vacuously true. It hence follows the empty-set is contained in any +set. $\qed$* +::: + +::: {#prop:EmptySetUnique .proposition} +**Proposition 7**. *The empty-set is unique* + +*The empty-set is unique, that is there is only one distinct set which +is the empty-set.* + +*Proof:* + +*Suppose that $\emptyset$ and $\emptyset'$ are two empty sets. By +proposition +[6](#prop:EmptySetincontainedineveryset){reference-type="ref" +reference="prop:EmptySetincontainedineveryset"} we have that +$\emptyset\subseteq\emptyset'$, likewise $\emptyset'\subseteq\emptyset$. +So by proposition +[1](#prop:TwosetsEqualIfContainedInEachOther){reference-type="ref" +reference="prop:TwosetsEqualIfContainedInEachOther"} we have that +$\emptyset=\emptyset'$. Hence the empty-set is unique. $\qed$* +::: + +It would be nice to have more ways to construct sets. Two key ways to do +this are with the union operation and intersection operation. + +::: definition +**Definition 24**. *Union and intersection of sets* + +*Let $S$ and $T$ be any two sets. We define the union of $S$ and $T$, +denoted by $S\cup T$, is the set* + +*$$\begin{equation*} + S\cup T=\left\{x: x\in S\text{ or } x\in T\right\} +\end{equation*}$$* + +*The intersection of $S$ and $T$, denoted by $S\cap T$, is the set* + +*$$\begin{equation*} + S\cap T = \left\{x : x\in S\text{ and } x\in T\right\} +\end{equation*}$$* + +*If we have a finite number of sets, given by $A_1$, $A_2$, $\dots$, +$A_n$ then the union of all of these sets is denoted by* + +*$$\begin{align*} + \bigcup_{i=1}^n A_i +\end{align*}$$* + +*and the intersection is denoted by* + +*$$\begin{align*} + \bigcap_{i=1}^n A_i +\end{align*}$$ Sometimes it is useful to define a union or intersection +of multiple sets given some condition or multiple conditions, usually +when the conditions involve other previously defined sets, this is +denoted as* + +*$$\begin{equation*} + \bigcup_{\substack{\text{Condition 1 for} A \\ \text{Condition 2 for} A\\ \\ \dots}} A +\end{equation*}$$ for the union and for the intersection* + +*$$\begin{equation*} + \bigcap_{\substack{\text{Condition 1 for} A \\ \text{Condition 2 for} A\\ \text{}\dots}} A +\end{equation*}$$* +::: + +::: example +**Example 17**. *Let $S=\left\{1,2,3,4,5,6\right\}$ and let +$T=\left\{2,4,5,6,7,8\right\}$, we have that* + +*$$\begin{align*} + S\cup T &=\left\{1,2,3,4,5,6\right\}\cup \left\{2,4,5,6,7,8\right\}=\left\{1,2,3,4,5,6,2,4,5,6,7,8\right\}=\left\{1,2,3,4,5,6,7,8\right\}\\ + S\cap T &=\left\{1,2,3,4,5,6\right\}\cap \left\{2,4,5,6,7,8\right\}=\left\{1,2,3,4,5,6,2,4,5,6,7,8\right\}=\left\{2,4,5,6\right\}\\ +\end{align*}$$* + +*We note that in the union we have multiple elements, for example we +have two $2$'s. Repeated elements in a set are considered to be the same +element so we don't write them, i.e +$\left\{2,2\right\}=\left\{2\right\}$* +::: + +::: example +**Example 18**. *Let $A_1=\left\{1,2,3\right\}$, +$A_2=\left\{1,2,7,9\right\}$ and $A_3=\left\{1,4,8,12\right\}$. We have +that the union of these sets is given by* + +*$$\begin{align*} + \bigcup_{i=1}^n A_i&=A_1\cup A_2\cup A_3\\ + &=\left\{1,2,3\right\}\cup \left\{1,2,7,9\right\}\cup \left\{1,4,8,12\right\}\\ + &=\left\{1,2,3,4,7,8,9,12\right\} +\end{align*}$$* + +*The intersection of these sets is given by* + +*$$\begin{align*} + \bigcap_{i=1}^n A_i&=A_1\cap A_2\cap A_3\\ + &=\left\{1,2,3\right\}\cap \left\{1,2,7,9\right\}\cap \left\{1,4,8,12\right\}\\ + &=\left\{1\right\} +\end{align*}$$* +::: + +We make one useful definition about intersections + +::: definition +**Definition 25**. *Disjoint sets* + +*Let $X$ and $Y$ be sets. If we have that $X\cap Y =\emptyset$ then we +say that $X$ and $Y$ are disjoint sets.* +::: + +##### Operations on sets + +###### The union, the intersection and set inclusion + +Before we continue we introduce three new ideas that will play a role +throughout the rest of this paper. + +::: definition +**Definition 26**. *Operation* + +*An operation $\circ$ acts on some inputs to produce an output or some +outputs.* +::: + +::: example +**Example 19**. *The union $\cup$ and intersection $\cap$ are examples +of operations. These operators operate on two sets to produce a third.* +::: + +::: definition +**Definition 27**. *Commutative operation* + +*Let $\circ$ be an operation that accepts two inputs, i.e we have +$A\circ B$ for valid inputs $A$ and $B$. We say that $\circ$ is +commutative if and only if $A\circ B = B\circ A$* +::: + +::: example +**Example 20**. *Consider $\mathbb{N}=\left\{1,2,3,4,5,\dots\right\}$. +We are familiar with the idea of addition of positive numbers, say +$1+2=3$. It is clear that the addition operation is commutative for +$\mathbb{N}$, e.g. $1+2=3=2+1$* +::: + +::: definition +**Definition 28**. *Associative operation* + +*Let $\circ$ be an operation that accepts two inputs, i.e we have +$A\circ B$ for valid inputs $A$ and $B$. We say that $\circ$ is +associative if and only if +$\left(A\circ B\right)\circ C = A\circ\left(B\circ C\right)$ where the +operation in the brackets should be computed first.* +::: + +::: example +**Example 21**. *Again consider +$\mathbb{N}=\left\{1,2,3,4,5,\dots\right\}$. The addition operator for +$\mathbb{N}$ is associative, e.g. +$\left(1+2\right)+3=3+3=6=1+5=1+\left(2+3\right)$* +::: + +We note that we have not defined a rigorous notion of addition, to do so +will require us to consider mappings which we do later. + +We have the following proposition about the properties of intersections, +unions and set inclusions. + +::: {#prop:PropertiesOfUnionIntersectionSetinclusion .proposition} +**Proposition 8**. *Properties of intersection, union and set inclusion* + +*Let $A,B,C$ be sets. Then we have that the following properties are +true* + +1. *$A\cap B = B\cap A$* + +2. *$A\cup B = B\cup A$* + +3. *$A\cap B\subseteq A$* + +4. *$A\subseteq A\cup B$* + +5. *$A\subseteq B \Rightarrow A\cap B = A$* + +6. *$A\subseteq B\Rightarrow A\cup B =B$* + +7. *$A\subseteq B \Rightarrow A\cap C \subseteq B\cap C$* + +8. *$A\subseteq B \Rightarrow A\cup C \subseteq B\cup C$* + +9. *$A\cap A = A$* + +10. *$A\cup A =A$* + +11. *$A\cap\left(B\cap C\right)=\left(A\cap B\right)\cap C$* + +12. *$A\cup\left(B\cup C\right)=\left(A\cup B\right)\cup C$* + +13. *$A\cap\left(B\cup C\right)=\left(A\cap B\right)\cup\left(A\cap C\right)$* + +14. *$A\cup\left(B\cap C\right)= \left(A\cup B\right) \cap \left(A\cup C\right)$* + +*Proof:* + +1. *$A\cap B = B\cap A$:* + + *Let $x\in A\cap B$ then $x\in A$ and $x\in B$ by the definition of + the intersection. It is hence clear that $x\in B\cap A$. So we have + $A\cap B\subseteq B\cap A$. Likewise if $x\in B\cap A$ then $x\in B$ + and $x\in A$, so that $x\in A\cap B$. So $B\cap A\subseteq A\cap B$. + It hence follows by proposition + [1](#prop:TwosetsEqualIfContainedInEachOther){reference-type="ref" + reference="prop:TwosetsEqualIfContainedInEachOther"} that + $A\cap B = B\cap A$.* + +2. *$A\cup B = B\cup A$:* + + *Let $x\in A\cup B$ then $x\in A$ or $x\in B$ by the definition of + the union. We hence have that $x\in B\cup A$. So we have + $A\cup B\subseteq B\cup A$. Likewise if $x\in B\cup A$ then $x\in B$ + and $x\in A$, so that $x\in A\cup B$. So $B\cup A\subseteq A\cup B$. + It hence follows by proposition + [1](#prop:TwosetsEqualIfContainedInEachOther){reference-type="ref" + reference="prop:TwosetsEqualIfContainedInEachOther"} that + $A\cup B = B\cup A$.* + +3. *$A\cap B\subseteq A$:* + + *Let $x\in A\cap B$, then by the definition of the intersection + $x\in A$ and $x\in B$. Hence $x\in A\cap B$ means that $x\in A$ so + that $A\cap B\subseteq A$.* + +4. *$A\subseteq A\cup B$:* + + *Let $x\in A$. By the definition of the union of two sets we have + that $y\in A\cup B$ if and only if $y\in A$ or $y\in B$. Hence it + follows that $x\in A\cup B$* + +5. *$A\subseteq B \Rightarrow A\cap B = A$:* + + *Let $A\subseteq B$ and suppose that $x\in A$, then we have that + $x\in B$ as $A\subseteq B$. Hence $x\in A\cap B$. This holds for any + choice of $x\in A$. We conclude that if $A\subseteq B$ then + $A\cap B = A$* + +6. *$A\subseteq B\Rightarrow A\cup B =B$:* + + *Let $A\subseteq B$. Observe that $B\subseteq B$ so that + $A\cup B\subseteq B\cup B= B$, that is to say $A\cup B \subseteq B$. + Now $B\subseteq A\cup B$. Hence $A\cup B = B$.* + +7. *$A\subseteq B \Rightarrow A\cap C \subseteq B\cap C$:* + + *Suppose that $A\subseteq B$ and let $x\in A\cap C$, then by + definition $x\in A$ and $x\in C$. Also we have that as + $A\subseteq B$ that $x\in A$ gives $x\in B$. Hence $x\in B\cap C$. + It follows that $A\cap C\subseteq B\cap C$.* + +8. *$A\subseteq B \Rightarrow A\cup C \subseteq B\cup C$:* + + *Suppose $A\subseteq B$ and let $x\in A\cup C$. We have that + $x\in A$ or $x\in C$. If $x\in A$ then as $A\subseteq B$ we have + that $x\in B$ so that $x\in B\cup C$. If $x\in C$ then clearly + $x\in B\cup C$. Either way we have that $A\cup C\subseteq B\cup C$.* + +9. *$A\cap A = A$:* + + *Let $x\in A$, then by the definition of the intersection we have + that $y\in A\cap A$ if and only if $y\in A$ and $y\in A$, hence + $x\in A\cap A$. So that $A\subseteq A\cap A$. Now If $x\in A\cap A$ + we have by definition of the intersection of two sets that $x\in A$ + and $x\in A$, so the force of deductive logic then drives one to the + conclusion that $x\in A$. So $A\cap A\subseteq A$. Hence + $A\cap A = A$.* + +10. *$A\cup A =A$:* + + *Let $x\in A$, then by the definition of the union of two sets, we + have that $y\in A\cup A$ if and only if $y\in A$ pr $y\in A$, hence + $x\in A\cup A$ so that $A\subseteq A\cup A$. Now suppose that + $x\in A\cup A$, then again by the definition of the union we have + that $x\in A$ so that $A\cup A\subseteq A$. Hence $A=A\cup A$.* + +11. *$A\cap\left(B\cap C\right)=\left(A\cap B\right)\cap C$:* + + *Let $A,B$ and $C$ be sets. Consider $A\cap\left(B\cap C\right)$, we + have that $x\in A\cap\left(B\cap C\right)$ means that $x\in A$ and + $x\in B\cap C$, likewise $x\in B\cap C$ means that $x\in B$ and + $x\in C$. Now as $x\in A$ and $x\in B$ and $x\in C$ so we have that + $x\in A\cap B$ and $x\in C$. Finally we have that + $x\in\left(A\cap B\right)\cap C$ so that + $A\cap\left(B\cap C\right)\subseteq \left(A\cap B\right)\cap C$.* + + *Now consider $\left(A\cap B\right)\cap C$, if + $x\in\left(A\cap B\right)\cap C$ then $x\in A\cap B$ and $x\in C$, + also $x\in A\cap B$ means that $x\in A$ and $x\in B$. As $x\in A$ + and $x\in B$ and $x\in C$ so we have that $x\in A$ and + $x\in B\cap C$ so that $x\in A\cap\left(B\cap C\right)$. Hence + $\left(A\cap B\right)\cap C\subseteq A\cap\left(B\cap C\right)$.* + + *Hence $A\cap\left(B\cap C\right)=\left(A\cap B\right)\cap C$* + +12. *$A\cup\left(B\cup C\right)=\left(A\cup B\right)\cup C$:* + + *Let $A,B$ and $C$ be sets. Consider $A\cup\left(B\cup C\right)$ and + let $x\in A\cup\left(B\cup C\right)$, we have that either $x\in A$ + or $x\in\left(B\cup C\right)$. If $x\in A$ then we have that + $x\in A\cup B$ so that $x\in\left(A\cup B\right)\cup C$. If + $x\in B\cup C$ then either $x\in B$ or $x\in C$. If $x\in B$ then + $X\in A\cup C$ so that $x\in \left(A\cup B\right)\cup C$. Otherwise + $x\in C$ and we have that $x\in \left(A\cup B\right)\cup C$. Hence + we have that + $A\cup\left(B\cup C\right)\subseteq\left(A\cup B\right)\cup C$* + + *Conversely let $x\in\left(A\cup B\right)\cup C$. We have that + either $x\in\left(A\cup B\right)$ or $x\in C$. If + $x\in\left(A\cup B\right)$ then either $x\in A$ or $x\in B$, in + either case we have that $x\in A\cup\left(B\cup C\right)$. If + $x\in C$ then $x\in A\cup\left(B\cup C\right)$. So that + $\left(A\cup B\right)\cup C\subseteq A\cup\left(B\cup C\right)$.* + + *Hence $A\cup\left(B\cup C\right)=\left(A\cup B\right)\cup C$* + +13. *$A\cap\left(B\cup C\right)=\left(A\cap B\right)\cup\left(A\cap C\right)$:* + + *Let $x\in A\cap\left(B\cup C\right)$, then we have that $x\in A$ + and $x\in B\cup C$. We have $x\in B\cup C$ gives us that $x\in B$ or + $x\in C$. If $x\in B$ then $x\in A\cap B$ and so + $x\in\left(A\cap B\right)\cup\left(A\cap C\right)$. Likewise is + $x\in C$ then $x\in A\cap C$ so + $x\in \left(A\cap B\right)\cup\left(A\cap C\right)$. Hence + $A\cap\left(B\cup C\right)\subseteq\left(A\cap B\right)\cup\left(A\cap C\right)$.* + + *For the opposite inclusion, let + $x\in\left(A\cap B\right)\cup\left(A\cap C\right)$ then we have that + either $x\in A\cap B$ or $x\in A\cap C$. If $x\in A\cap B$ then + $x\in A$ and $x\in B$, so we hence have that $x\in B\cup C$ so that + $x\in A\cap\left(B\cup C\right)$. Likewise if we have $x\in A\cap C$ + then $x\in A$ and $x\in C$, so $x\in B\cup C$ and + $x\in A\cap\left(B\cup C\right)$. Hence + $\left(A\cap B\right)\cup\left(A\cap C\right)\subseteq A\cap\left(B\cup C\right)$* + + *So + $A\cap\left(B\cup C\right)=\left(A\cap B\right)\cup\left(A\cap C\right)$.* + +14. *$A\cup\left(B\cap C\right)= \left(A\cup B\right) \cap \left(A\cup C\right)$:* + + *Let $x\in A\cup\left(B\cap C\right)$ then either $x\in A$ or + $x\in B\cap C$. If $x\in A$ then $x\in A\cup B$ and $x\in A\cup C$, + which is to say $x\in\left(A\cup B\right)\cap\left(A\cup C\right)$. + If $x\in B\cap C$ then $x\in B$ and $x\in C$, so it follows that + $x\in A\cup B$ and $x\in A\cup C$ which is to say + $x\in\left(A\cup B\right)\cap\left(A\cup C\right)$. Hence + $A\cup\left(B\cap C\right)\subseteq \left(A\cup B\right) \cap \left(A\cup C\right)$.* + + *Now, suppose that + $x\in\left(A\cup B\right) \cap \left(A\cup C\right)$. We then have + that $x\in A\cup B$ and $x\in A\cup C$. Now $x\in A\cup B$ gives + $x\in A$ or $x\in B$, also $x\in A\cup C$ means that $x\in A$ or + $x\in C$. This gives us two possible outcomes. If $x\in A$ then + $x\in A\cup\left(B\cap C\right)$ so that + $\left(A\cup B\right) \cap \left(A\cup C\right)\subseteq A\cup\left(B\cap C\right)$. + Suppose that $x\not\in A$ then we must have that $x\in B$ and + $x\in C$ as $x\in A\cup B$ and $x\in A\cup C$. Hence $x\in B\cap C$ + so $x\in A\cup\left(B\cap C\right)$. Hence + $\left(A\cup B\right) \cap \left(A\cup C\right)\subseteq A\cup\left(B\cap C\right)$.* + + *So we have that + $A\cup\left(B\cap C\right)= \left(A\cup B\right) \cap \left(A\cup C\right)$.* + +*The proposition now follows. $\qed$* +::: + +::: {#thm:EquivSubsetIntUnion .theorem} +**Theorem 1**. *Equivalence of Subsets with union and intersection* + +*Let $A,B$ be sets. The following are equivalent* + +1. *$A\subseteq B$* + +2. *$A\cap B = A$* + +3. *$A\cup B =B$* + +*Proof:* + +*Suppose $A\subseteq B$. By proposition +[8](#prop:PropertiesOfUnionIntersectionSetinclusion){reference-type="ref" +reference="prop:PropertiesOfUnionIntersectionSetinclusion"} we have +that* + +*$$\begin{equation*} + A=A\cap A \subseteq A\cap B\subseteq A +\end{equation*}$$ Hence $A=A\cap B$.* + +*Now suppose that $A\cap B = A$, then $A\subseteq B$. This shows 1 and 2 +are equivalent.* + +*Suppose $A\subseteq B$. Let $x\in A$ then $x\in B$. Then as $x\in B$ we +have that $x\in A\cup B$ so that $B\subseteq A\cup B$. Suppose that +$x\in A\cup B$, then either $x\in A$ or $x\in B$. If $x\in B$ we are +done and we have that $A\cup B\subseteq B$. If $x\in A$ then as +$A\subseteq B$ we have that $x\in B$ so that $A\cup B\subseteq B$.* + +*Hence $A\cup B = B$.* + +*Now suppose that $A\cup B = B$. Suppose that $x\in A$ then +$x\in A\cup B =B$ so $x\in B$, hence $A\subseteq B$.* + +*This shows the equivalence of 1 and 3.* + +*The equivalence of 2 and 3 now follows. Indeed, suppose that +$A\cap B =A$ then by the equivalence of 1 and 2 we know that +$A\subseteq B$, also by the equivalence of 1 and 3 we know that +$A\cup B = B$. $\qed$* +::: + +###### The complement of a set + +It sometimes becomes useful to talk about the elements that are not in +some set $S$. This only makes sense if $S$ is contained inside some +larger set. + +::: definition +**Definition 29**. *Complement of a set* + +*Let $S$ be a set such that $S\subseteq U$ for some set $U$. We define +the complement of $S$, denoted by $S^C$ as the following set* + +*$$\begin{equation*} + S^C = \left\{x\in U:x\not\in S\right\} +\end{equation*}$$* + +*We can alternatively write $S^C = U\setminus S$, where $\setminus$ is +the set difference operation.* + +*Moreover we can also consider the complement of a set $A$ with respect +to some other set $B$, again occurring inside some larger set $U$ which +is to say $A\subseteq U$ and $B\subseteq U$. We have that* + +*$$\begin{equation*} + A\setminus B = \left\{x\in A : x\not\in B\right\} +\end{equation*}$$* + +*We call* +::: + +::: example +**Example 22**. *Let $U=\left\{1,2,3,4,5,6\right\}$, +$S=\left\{1,2,3,4,6\right\}$ and $T=\left\{2,4,6\right\}$. We have that +$S\subseteq U$ so that* + +*$$\begin{align*} + S^C&=\left\{x\in U:x\not\in S\right\}=\left\{5\right\}\\ + T^C&=\left\{x\in U:x\not\in T\right\}=\left\{1,3,5\right\}\\ +\end{align*}$$* + +*Also* + +*$$\begin{align*} + S\setminus T=\left\{x\in S: x\not\in T\right\}=\left\{1,3\right\}\\ + T\setminus S=\left\{x\in T: x\not\in S\right\}=\emptyset +\end{align*}$$* +::: + +An immediate result follows from the previous definitions of the +complement of a set and set difference. + +::: {#thm:DeMorgan .theorem} +**Theorem 2**. *De-Morgan's laws* + +*Let $A$ and $B$ be subsets of some universal set $U$. We have the +complement laws* + +1. *$\left(A\cap B\right)^C=A^C\cup B^C$* + +2. *$\left(A\cup B\right)^C= A^C\cap B^C$* + +*We also have the set difference laws* + +1. *$U\setminus\left(A\cap B\right)=\left(U\setminus A\right)\cup \left(U\setminus B\right)$* + +2. *$U\setminus\left(A\cup B\right)=\left(U\setminus A\right)\cap \left(U\setminus B\right)$* + +*Proof:* + +*We first prove the complement laws.* + +1. *$\left(A\cap B\right)^C=A^C\cup B^C$:* + + *Let $x\in\left(A\cap B\right)^C$, by the definition of the set + complement we have that $x\not\in \left(A\cap B\right)$. So by the + definition of the intersection and $x$ not being an element of + $A\cap B$ we have that $x\not\in A$ or $x\not\in B$. Suppose that + $x\not\in A$, then by the definition of set complement we have that + $x\in A^C$ so that $x\in A^C\cup B^C$. Likewise if $x\not\in B$ then + $x\in B^C$ so that $x\in A^C\cup B^C$. Hence we have that + $\left(A\cap B\right)^C\subseteq A^C\cup B^C$.* + + *Now suppose $x\in A^C\cup B^C$, then $x\in A^C$ or $x\in B^C$. + Suppose $x\in A^C$ then $x\not\in A$ so that $x\not\in A\cap B$ + hence $x\in\left(A\cap B\right)^C$. Likewise if $x\in B^C$ then + $x\not\in B$ so $x\not\in A\cap B$ so that + $x\in\left(A\cap B\right)^C$. Thus + $A^C\cup B^C\subseteq \left(A\cap B\right)^C$* + + *Hence $\left(A\cap B\right)^C=A^C\cup B^C$.* + +2. *$\left(A\cup B\right)^C= A^C\cap B^C$:* + + *Let $x\in \left(A\cup B\right)^C$, then we have that + $x\not\in A\cup B$ so $x\not\in A$ and $x\not\in B$. This means that + $x\in A^C$ and $x\in B^C$ which is to say $x\in A^C\cap B^C$. So + $\left(A\cup B\right)^C\subseteq A^C\cap B^C$.* + + *Suppose $x\in A^C \cap B^c$ then $x\in A^C$ and $x\in B^C$. + $x\in A^C$ means that $x\not\in A$ and $x\in B^C$ means that + $x\not\in B$, so $x\not\in A$ and $x\not\in B$ hence + $x\not\in A\cup B$. Thus $x\in\left(A\cup B\right)^C$. Hence + $A^C\cap B^C\subseteq\left(A\cup B\right)^C$* + + *Thus $\left(A\cup B\right)^C= A^C\cap B^C$* + +*It is left to prove the set difference laws.* + +1. *$U\setminus\left(A\cap B\right)=\left(U\setminus A\right)\cup \left(U\setminus B\right)$:* + + *Let $X\in U\setminus\left(A\cap B\right)$ then by definition we + have that $x\in U$ and $x\not\in A\cap B$, which is to say that + $x\not\in A$ or $x\not\in B$ with the possibility of being in + neither. If $x\not\in A$ then $x\in \left(U\setminus A\right)$ and + we clearly have + $x\in \left(U\setminus A\right)\cup \left(U\setminus B\right)$. + Likewise if $x\not\in B$ and both cases clearly hold in the case + where $x\not\in A$ and $X\not\in B$. It follows that in every case + that $x\in \left(U\setminus A\right)\cup \left(U\setminus B\right)$. + Hence + $U\setminus\left(A\cap B\right)\subseteq\left(U\setminus A\right)\cup \left(U\setminus B\right)$* + + *Now suppose that + $x\in \left(U\setminus A\right)\cup \left(U\setminus B\right)$ then + by definition we have that $x\in U\setminus A$ or + $x\in U\setminus B$ with the possibility of being in both. If + $x\in U\setminus A$ then $x\in U$ and $X\not\in A$. Hence + $x\not\in A\cap B$, likewise if $X\in Y\setminus B$ then we again + conclude that $X\not\in A\cap B$. However as $x\in U$ then we have + by definition that $x\in U\setminus\left(A\cap B\right)$. We + conclude that + $\left(U\setminus A\right)\cup \left(U\setminus B\right)\subseteq U\setminus\left(A\cap B\right)$* + + *It follows that + $U\setminus\left(A\cap B\right)=\left(U\setminus A\right)\cup \left(U\setminus B\right)$* + +2. *$U\setminus\left(A\cup B\right)=\left(U\setminus A\right)\cap \left(U\setminus B\right)$:* + + *Suppose that $U\setminus\left(A\cup B\right)$ then $x\in U$ and + $x\not\in A\cup B$ so $x\not\in A$ and $x\not\in B$. Clearly then + $x\in U\setminus A$ and $x\in U\setminus B$ so that + $x\in \left(U\setminus A\right)\cap \left(U\setminus B\right)$. So + we have that + $U\setminus\left(A\cup B\right)\subseteq\left(U\setminus A\right)\cap \left(U\setminus B\right)$.* + + *Let $x\in \left(U\setminus A\right)\cap \left(U\setminus B\right)$ + then $x\in U\setminus A$ and $x\in U\setminus B$ which is to say + that $x\in U$ and $x\not\in A$ and $x\not\in B$. Clearly + $x\not\in A$ and $x\not\in B$ gives us that $x\not\in A\cup B$ and + so $x\in U\setminus \left(A\cup B\right)$ by definition. This allows + us to conclude that + $\left(U\setminus A\right)\cap \left(U\setminus B\right)\subseteq U\setminus\left(A\cup B\right)$* + + *Hence + $U\setminus\left(A\cup B\right)=\left(U\setminus A\right)\cap \left(U\setminus B\right)$* + +*This proves the theorem. $\qed$* +::: + +::: {#prop:AdditionComplement .proposition} +**Proposition 9**. *Additional properties of set complements and set +differences* + +*Let $A, B$ and $C$ be a sets such that $A\subseteq U$, $B\subseteq U$ +and $C\subseteq U$. Moreover suppose $U$ is not contained in any other +set. Then we have that* + +1. *$A\cup A^C = U$* + +2. *$A\cap A^C =\emptyset$* + +3. *$\emptyset^C =U$* + +4. *$U^C=\emptyset$* + +5. *If $A\subseteq B$ then $B^C\subseteq A^C$* + +6. *$\left(A^C\right)^C=A$* + +7. *$A\setminus B = A\cap B^C$* + +8. *$\left(A\setminus B\right)^C=A^C\cup B$* + +9. *$A^C\setminus B^C=B\setminus A$* + +10. *$\left(A\setminus B\right)\cap C = \left(A\cap C\right)\setminus\left(B\cap C\right)$* + +11. *$A\setminus\left(B\setminus C\right) = \left(A\cap B\right)\setminus\left(A\cap C\right)$* + +12. *$\left(A\setminus B\right)\cap B=\emptyset$* + +13. *$\left(A\setminus B\right)\cap\left(A\cap B\right)=\emptyset$* + +*Proof:* + +1. *$A\cup A^C = U$:* + + *Let $x\in A\cup A^C$ then $x\in A$ or $x\in A^C$. If $x\in A$ then + as $A\subseteq U$ we have that $x\in U$. If $x\in A^c$ then by the + definition of set complements we have that $x\in A^C$ if and only if + $x\in U$. Hence $A\cup A^C\subseteq U$.* + + *Conversely suppose that $x\in U$. We know that $A\subseteq U$ so if + $x\in A$ we clearly have $x\in A\cup A^C$. So suppose $x\not\in A$ + then by definition of the set complement we have that $x\in A^C$ so + that $x\in A\cup A^C$. Hence $U\subseteq A\cup A^C$.* + + *So $A\cup A^C=U$.* + +2. *$A\cap A^C =\emptyset$:* + + *Let $x\in A\cap A^C$, then $x\in A$ and $x\in A^C$, however + $x\in A^C$ means that $x\not\in A$. This contradicts the fact that + $x\in A$, hence there are no elements $x\in U$ so that $x\in A$ and + $x\in A^C$, this is to say $A\cap A^C= \emptyset$.* + + *Hence $A\cap A^C =\emptyset$.* + +3. *$\emptyset^C =U$:* + + *By the definition of the empty set we have that $\emptyset$ has no + elements. The complement of the empty-set is* + + *$$\begin{equation*} + \emptyset^C=\left\{x\in U:x\not\in\emptyset\right\} + \end{equation*}$$* + + *Hence every $x\in U$ is such that $x\not\in\emptyset$. So + $\emptyset^C\subseteq U$.* + + *Conversely let $x\in U$, then $x\not\in \emptyset$ as $\emptyset$ + has no elements. so $x\in\emptyset^C$ hence + $U\subseteq \emptyset^C$.* + + *It follows that $\emptyset^C=U$.* + +4. *$U^C=\emptyset$:* + + *Let $x\in U^C$, by the definition of set complement we have that* + + *$$\begin{equation*} + U^C=\left\{y\in U:y\not\in U\right\} + \end{equation*}$$* + + *This is clearly empty as no such $y$ can satisfy $y\in U$ and + $y\not\in U$.* + + *Hence $U^C=\emptyset$.* + +5. *If $A\subseteq B$ then $B^C\subseteq A^C$:* + + *Suppose that $A\subseteq B$. We have by proposition + [8](#prop:PropertiesOfUnionIntersectionSetinclusion){reference-type="ref" + reference="prop:PropertiesOfUnionIntersectionSetinclusion"} property + 5 we have that $A\cap B = A$. It follows that + $\left(A\cap B\right)^C = A^C$. Now by De-Morgan's laws we have that + $\left(A\cap B\right)^C= A^C\cup B^C$. Hence $A^C\cup B^C = A^C$. + Finally by theorem + [1](#thm:EquivSubsetIntUnion){reference-type="ref" + reference="thm:EquivSubsetIntUnion"} we know that $X\cup Y = Y$ if + and only if $X\subseteq Y$ for sets $X$ and $Y$. Hence + $B^C\subseteq A^C$.* + +6. *$\left(A^C\right)^C=A$:* + + *Let $x\in \left(A^C\right)^C$. By definition we have that* + + *$$\begin{equation*} + \left(A^C\right)^C=\left\{x\in U : x\not\in A^c\right\} + \end{equation*}$$* + + *Hence $x\in \left(A^C\right)^C$ if and only if $x\not\in A^C$. + However $x\not\in A^C$ means that $x\in A$. Hence + $\left(A^C\right)^C\subseteq A$* + + *Suppose that $x\in A$, then $x\not\in A^C$, moreover by definition + $x\not\in A^C$ if and only if $x\in \left(A^C\right)^C$, hence + $A\subseteq \left(A^C\right)^C$.* + + *Hence $\left(A^C\right)^C=A$* + +7. *$A\setminus B = A\cap B^C$:* + + *Let $x\in A\setminus B$, then by definition we have that + $A\setminus B$ is the set* + + *$$\begin{equation*} + A\setminus B = \left\{y\in A:y\not\in B\right\} + \end{equation*}$$ Hence $x\in A\setminus B$ means that $x\in A$ and + $x\not\in B$. We have that $x\not\in B$ means that $x\in B^C$. So + that $x\in A\cap B^C$. It follows that + $A\setminus B\subseteq A\cap B^C$.* + + *Let $x\in A\cap B^C$, then $x\in A$ and $x\in B^C$. $x\in B^C$ + means that $x\not\in B$, so by definition $x\in A$ and $x\not\in B$ + means that $x\in A\setminus B$. Hence + $A\cap B^C\subseteq A\setminus B$.* + + *Hence $A\setminus B = A\cap B^C$.* + +8. *$\left(A\setminus B\right)^C=A^C\cup B$:* + + *We know that $A\setminus B = A\cap B^C$ by the previous property. + Now by De-Morgan's laws we have that* + + *$$\begin{equation*} + \left(A\setminus B\right)^C=\left(A\cap B^C\right)^C = A^C\cup \left(B^C\right)^C = A^C \cup B + \end{equation*}$$* + +9. *$A^C\setminus B^C=B\setminus A$:* + + *We know that $A^C\setminus B^C = A^C\cap \left(B^C\right)^C$. Now, + $\left(B^C\right)^C=B$ hence + $A^C\cap \left(B^C\right)^C=A^C\cap B = B\cap A^C$. Finally we know + that $B\cap A^C = B\setminus A$ by property 7.* + + *Hence $A^C\setminus B^C=B\setminus A$.* + +10. *$\left(A\setminus B\right)\cap C = \left(A\cap C\right)\setminus\left(B\cap C\right)$:* + +11. *$A\setminus\left(B\setminus C\right) = \left(A\cap B\right)\setminus\left(A\cap C\right)$* + +12. *$\left(A\setminus B\right)\cap B=\emptyset$* + +13. *$\left(A\setminus B\right)\cap\left(A\cap B\right)=\emptyset$* + +*The proposition now follows. $\qed$* +::: + +###### Cartesian Product + +We now look to another method of constructing a set. This method differs +from the union and intersection as it allows us to construct a set where +the elements come in pairs, in particular these pairs are ordered. + +::: definition +**Definition 30**. *Ordered pair* + +*Let $S$ and $T$ be sets. Let $s\in S$ and $t\in T$. We say that the +tuple $\left(s,t\right)$ is an ordered pair of an element in $S$ and an +element in $T$.* +::: + +::: definition +**Definition 31**. *Cartesian product of two sets* + +*Let $S$ and $T$ be sets. We define the Cartesian product of $S$ and +$T$, denoted $S\times T$ to be the set of all ordered pairs of the form +$\left(s,t\right)$ where $s\in S$ and $t\in T$. This is to say that* + +*$$\begin{equation*} + S\times T=\left\{\left(s,t\right):s\in S,t\in T\right\} +\end{equation*}$$* +::: + +::: example +**Example 23**. *Let $S=\left\{1,2,3\right\}$ and +$T=\left\{4,5,6\right\}$. We have that* + +*$$\begin{align*} + S\times T&=\left\{\left(1,4\right),\left(1,5\right),\left(1,6\right),\left(2,4\right),\left(2,5\right),\left(2,6\right),\left(3,4\right),\left(3,5\right),\left(3,5\right)\right\}\\ + T\times S&=\left\{\left(4,1\right),\left(4,2\right),\left(4,3\right),\left(5,1\right),\left(5,2\right),\left(5,3\right),\left(6,1\right),\left(6,2\right),\left(6,3\right)\right\}\\ +\end{align*}$$ This example shows that $S\times T\neq T\times S$ in +general.* +::: + +We can make repeated uses of this idea, we just need to defined an +ordered $n$-tuple. + +::: {#def:orderedNtuple .definition} +**Definition 32**. *Ordered $n$-tuple* + +*Let $S_1,S_2,\dots,S_n$ be sets. Let +$s_1\in S_1,s_2\in S_2,\dots,s_n\in S_n$. We say that +$\left(s_1,s_2,\dots,s_n\right)$ is an ordered $n$-tuple of an elements +in $S_1,S_2,\dots,S_n$.* +::: + +::: {#def:CartProductOfNSet .definition} +**Definition 33**. *Cartesian product of $n$ sets* + +*Let $S_1,S_2,\dots,S_n$ be sets. We define the Cartesian product of +$S_1,S_2,\dots,S_N$, denoted $S_1\times S_2\times\dots\times S_n$ to be +the set of all ordered pairs of the form +$\left(s_1,s_2,\dots,s_n\right)$ where +$s_1\in S_1.s_2\in S_2,\dots s_n\in S_n$. This is to say that* + +*$$\begin{equation*} + S_1\times S_2\times\dots\times S_n=\left\{\left(s_1,s_2,\dots,s_n\right):s_1\in S_1.s_2\in S_2,\dots s_n\in S_n\right\} +\end{equation*}$$* + +*If all the sets are the same we denote this by $S^n$.* +::: + +We make the following observations + +::: {#lem:CartEmpty .lemma} +**Lemma 1**. *Cartesian product is empty if and only if at least one of +the sets in the product is empty* + +*Let $A$ and $B$ be sets. We have that $A\times B=\emptyset$ if and only +if $A=\emptyset$ or $B=\emptyset$.* + +*Proof:* + +*We argue as follows. Suppose that $A\times B\neq \emptyset$ then we +have by definition of a non-empty Cartesian product that +$A\times B\neq \emptyset$ if and only if +$\exists\left(a,b\right)\in A\times B$. Now, by the definition of a +Cartesian product we have that as $\left(a,b\right)\in A\times B$ if and +only if $\exists a\in A$ and $\exists b\in B$, which is to say +$A\neq\emptyset$ and $B\neq\emptyset$.* + +*This proves the result as assuming $A\times B\neq \emptyset$ gives us +$A\neq\emptyset$ and $B\neq\emptyset$. $\qed$* +::: + +::: {#prop:CriterionForComOfCartProd .proposition} +**Proposition 10**. *Criterion for commutativity of the Cartesian +product* + +*Let $A$ and $B$ be sets. We have that $A\times B = B\times A$ only if +at least one of the following holds.* + +1. *$A=B$* + +2. *$A = \emptyset$ or $B=\emptyset$ or $A=B=\emptyset$* + +*Proof:* + +*Let $A$ and $B$ be sets.* + +1. *$A=B$:* + + *Suppose that $A=B$ then without loss of generality[^4] consider* + + *$$\begin{equation*} + A\times B = A\times A = \left\{\left(a,a\right):a\in A\right\} + \end{equation*}$$* + + *Moreover* + + *$$\begin{equation*} + B\times A = A\times A = \left\{\left(a,a\right):a\in A\right\} + \end{equation*}$$* + + *Hence, varying over every $a\in A$ we have that + $A\times B = B\times A$.* + +2. *$A = \emptyset$ or $B=\emptyset$ or $A=B=\emptyset$:* + + *By lemma [1](#lem:CartEmpty){reference-type="ref" + reference="lem:CartEmpty"} we have that if $A=\emptyset$ or + $B=\emptyset$ or $A=B=\emptyset$ then + $A\times B=\emptyset =B\times A$.* + +*The proposition follows. $\qed$* +::: + +We have seen that the Cartesian product is not commutative, but what can +we say about associativity. + +::: example +**Example 24**. *Let $A=\left\{1\right\}$. Consider* + +*$$\begin{align*} + A\times\left(A\times A\right)&=A\times\left\{\left(1,1\right)\right\}=\left\{\left(1,\left(1,1\right)\right)\right\}\\ + \left(A\times A\right)\times A &=\left\{\left(1,1\right)\right\}\times A = \left\{\left(\left(1,1\right),1\right)\right\}\\ +\end{align*}$$* + +*Hence +$A\times\left(A\times A\right)\neq \left(A\times A\right)\times A$. So +in general the Cartesian product is not associative.* +::: + +We have the following criterion for the associativity of the Cartesian +product. + +::: {#prop:CriterionForAssOfCartProd .proposition} +**Proposition 11**. *Criterion for associativity of the Cartesian +product* + +*Let $A,B$ and $C$ be sets. We have that +$A\times\left(B\times C\right)=\left(A\times B\right)\times C$ if and +only if $A=\emptyset$ or $B=\emptyset$ or $C=\emptyset$.* + +*Proof:* + +*Suppose that +$A\times\left(B\times C\right)=\left(A\times B\right)\times C$, we need +to show one of $A,B$ or $C$ is empty.* + +*Consider $A\times\left(B\times C\right)$, we have that* + +*$$\begin{equation*} + A\times\left(B\times C\right)=A\times\left\{\left(b,c\right):b\in B,c\in C\right\}=\left\{\left(a,\left(b,c\right)\right):a\in A, \left(b,c\right)\in B\times C\right\} +\end{equation*}$$* + +*Now consider $\left(A\times B\right)\times C$, we have that* + +*$$\begin{equation*} + \left(A\times B\right)\times C=\left\{\left(a,b\right):a\in A,b\in B\right\}\times C=\left\{\left(\left(a,b\right),c\right):\left(a,b\right)\in A\times B, c\in C\right\} +\end{equation*}$$* + +*Hence for equality we need that $a=\left(a,b\right)$ and +$\left(b,c\right)=c$. However this is not possible as +$\left(a,b\right)\not\in A$ and $\left(b,c\right)\not\in C$. Hence one +of the products must be empty, which implies that one of $A,B$ or $C$ is +empty.* + +*Now suppose that one of $A,B$ or $C$ is empty. Without loss of +generality suppose that $A=\emptyset$, then by lemma +[1](#lem:CartEmpty){reference-type="ref" reference="lem:CartEmpty"} we +know that one of $A\times B=\emptyset$ and +$A\times\left(B\times C\right)=\emptyset$. Also +$\left(A\times B\right)\times C=\emptyset\times C=\emptyset$.* + +*Hence we have that +$\left(A\times B\right)\times C=\emptyset=A\times\left(B\times C\right)$. +is associative. $\qed$* +::: + +It is left to see how the Cartesian product interacts with unions, +intersections and complements. + +::: {#prop:CartProdUnIntComp .proposition} +**Proposition 12**. *Properties of Cartesian products, unions, +intersections and complements* + +*Let $A,B,C$ and $D$ be sets. We have the following properties* + +1. *$\left(A\cap B\right)\times\left(C\cap D\right) =\left(A\times C\right)\cap\left(B\times D\right)$* + +2. *$A\times\left(B\cap C\right)=\left(A\times B\right)\cap \left(A\times C\right)$* + +3. *$\left(A\times B\right)\cap\left(B\times A\right)=\left(A\cap B\right)\times\left(A\cap B\right)$* + +4. *$\left(A\cup B\right)\times\left(C\cup D\right) = \left(A\times C\right)\cup \left(B\times D\right)\cup\left(A\times D\right)\cup\left(B\times C\right)$* + +5. *$A\times\left(B\cup C\right) = \left(A\times B\right)\cup\left(A\times C\right)$* + +6. *$\left(B\cup C\right)\times A = \left(B\times A\right)\cup\left(C\times A\right)$* + +7. *If $A\subseteq B$ and $C\subseteq D$ then + $A\times C\subseteq B\times D$. Moreover if $A\neq\emptyset$ and + $C\neq\emptyset$ then* + + *$$\begin{equation*} + A\times C\subseteq B\times T \iff A\subseteq B\text{ and } C\subseteq D + \end{equation*}$$* + +8. *If $A\subseteq B$ then $A\times C\subseteq B\times C$* + +9. *If $C\subseteq D$ then $A\times C\subseteq A\times D$* + +10. *$A\times\left(B\setminus C\right)=\left(A\times B\right)\setminus\left(A\times C\right)$* + +11. *$\left(A\setminus B\right)\times C = \left(A\times C\right)\setminus\left( B\times C\right)$* + +12. *$\left(A\times B\right)\setminus\left(C\times D\right)=\left(A\times\left(B\setminus D\right)\right)\cup\left(\left(A\setminus B\right)\times C\right)$* + +13. *Suppose $A\subseteq C$ and $B\subseteq D$ and consider + $C\setminus A$ and $T\setminus B$. We have* + + *$$\begin{align*} + \left(C\setminus A\right)\times D &= \left(C\times D\right)\setminus\left(A\times D\right)\\ + C\times\left(D\setminus B\right) &=\left(C\times D\right)\setminus \left(C\times B\right) + \end{align*}$$* + +*Proof:* + +1. *$\left(A\cap B\right)\times\left(C\cap D\right) =\left(A\times C\right)\cap\left(B\times D\right)$:* + + *Let + $\left(x,y\right)\in\left(A\cap B\right)\times\left(C\cap D\right)$, + then by definition of the Cartesian product we have that + $\left(x,y\right)\in\left(A\cap B\right)\times\left(C\cap D\right)$ + if and if only $x\in A$ and $x\in B$ and $y\in C$ and $y\in D$. + $x\in A$ and $x\in B$ and $y\in C$ and $y\in D$ means that + $\left(x,y\right)\in A\times C$ and $\left(x,y\right)\in B\times D$, + finally this happens if and only if + $\left(x,y\right)\in \left(A\times C\right)\cap\left(B\times D\right)$.* + +2. *$A\times\left(B\cap C\right)=\left(A\times B\right)\cap \left(A\times C\right)$:* + + *We know that $A\cap A=A$. By the previous property we have that* + + *$$\begin{equation*} + A\times\left(C\cap D\right)=\left(A\cap A\right)\times\left(B\cap C\right)=\left(A\times B\right)\cap \left(A\times C\right) + \end{equation*}$$* + +3. *$\left(A\times B\right)\cap\left(B\times A\right)=\left(A\cap B\right)\times\left(A\cap B\right)$:* + + *By property 1 we have* + + *$$\begin{equation*} + \left(A\times B\right)\cap\left(B\times A\right)=\left(A\cap B\right)\times \left(B\cap A\right) = \left(A\cap B\right)\times \left(A\cap B\right) + \end{equation*}$$* + +4. *$\left(A\cup B\right)\times\left(C\cup D\right) = \left(A\times C\right)\cup \left(B\times D\right)\cup\left(A\times D\right)\cup\left(B\times C\right)$:* + + *Let + $\left(x,y\right)\in \left(A\cup B\right)\times\left(C\cup D\right)$, + then by definition of Cartesian product and the union of sets we + have that + $\left(x,y\right)\in \left(A\cup B\right)\times\left(C\cup D\right)$ + if and only if $x\in A$ or $x\in B$ and $y\in C$ or $y\in D$.* + + *$x\in A$ or $x\in B$ and $y\in C$ or $y\in D$ will occur if and + only if ($x\in A$ or $x\in B$ and $y\in C$) or ($x\in A$ or $x\in B$ + and $y\in D$).* + + *($x\in A$ or $x\in B$ and $y\in C$) or ($x\in A$ or $x\in B$ and + $y\in D$) occurs if and only if ($x\in A$ and $y\in C$) or ($x\in B$ + and $y\in C$) or ($x\in A$ and $y\in D$) or ($x\in B$ and + $y\in D$).* + + *By the definition of the Cartesian product we have that ($x\in A$ + and $y\in C$) or ($x\in B$ and $y\in C$) or ($x\in A$ and $y\in D$) + or ($x\in B$ and $y\in D$) if and only if + $\left(x,y\right)\in A\times C$ or $\left(x,y\right)\in A\times D$ + or$\left(x,y\right)\in B\times C$ or + $\left(x,y\right)\in B\times D$. Hence by the definition of the + union of two sets, $\left(x,y\right)\in A\times C$ or + $\left(x,y\right)\in A\times D$ or$\left(x,y\right)\in B\times C$ or + $\left(x,y\right)\in B\times D$ occurs if and only if + $\left(x,y\right)\in \left(A\times C\right)\cup \left(B\times D\right)\cup\left(A\times D\right)\cup\left(B\times C\right)$.* + +5. *$A\times\left(B\cup C\right) = \left(A\times B\right)\cup\left(A\times C\right)$:* + + *We know $A=A\cup A$ and so by the previous property we have that* + + *$$\begin{align*} + A\times\left(B\cup C\right)&=\left(A\cup A\right)\times\left(B\cup C\right)\\ + &=\left(A\times B\right)\cup \left(A\times C\right)\cup\left(A\times C\right)\cup\left(A\times B\right)\\ + &=\left(A\times B\right)\cup\left(A\times C\right) + \end{align*}$$* + +6. *$\left(B\cup C\right)\times A = \left(B\times A\right)\cup\left(C\times A\right)$:* + + *Again $A=A\cup A$ and so by property 4 we have* + + *$$\begin{align*} + \left(B\cup C\right)\times A&=\left(B\cup C\right)\times\left(A\cup A\right)\\ + &=\left(B\times A\right)\cup \left(B\times A\right)\cup\left(C\times A\right)\cup\left(C\times A\right)\\ + &=\left(B\times A\right)\cup\left(C\times A\right) + \end{align*}$$* + +7. *If $A\subseteq B$ and $C\subseteq D$ then + $A\times C\subseteq B\times D$. Moreover if $A\neq\emptyset$ and + $C\neq\emptyset$ then* + + *$$\begin{equation*} + A\times C\subseteq B\times T \iff A\subseteq B\text{ and } C\subseteq D + \end{equation*}$$:* + + *Let $A\subseteq B$ and $C\subseteq D$. If $A=\emptyset$ or + $C=\emptyset$ then by lemma [1](#lem:CartEmpty){reference-type="ref" + reference="lem:CartEmpty"} we have $A\times C=\emptyset$ and by + proposition + [6](#prop:EmptySetincontainedineveryset){reference-type="ref" + reference="prop:EmptySetincontainedineveryset"} we have + $A\times C=\emptyset \subseteq B\subseteq D$.* + + *So suppose that $A\neq\emptyset$ and $C\neq\emptyset$ then lemma + [1](#lem:CartEmpty){reference-type="ref" reference="lem:CartEmpty"} + gives $A\times C\neq\emptyset$. Then we have that + $\left(x,y\right)\in A\times C$ if and if only $x\in A$ and + $y\in C$. We have $A\subseteq B$ so $x\in B$ and $C\subseteq D$ so + $y\in D$, hence $\left(x,y\right)\in B\times D$. Hence + $A\times C\subseteq B\times D$.* + + *It is left to prove that if $A\neq\emptyset$ and $C\neq\emptyset$ + and $A\times C\subseteq B\times D$, then $A\subseteq B$ and + $C\subseteq D$. Suppose $A\times C\subseteq B\times D$. If + $A=\emptyset$ then $A\times C=\emptyset$ by lemma + [1](#lem:CartEmpty){reference-type="ref" reference="lem:CartEmpty"} + and $A\times C=\emptyset\subseteq B\times D$ irrespective of $C$, so + $C$ need not be a subset of $D$. Likewise if $C=\emptyset$ then + $A\times C=\emptyset\subseteq B\times D$ irrespective of $A$ so $A$ + need not be a subset of $B$.* + + *So suppose that $A\neq\emptyset$ and $C\neq\emptyset$ then + $\exists x\in A$ and $\exists y\in C$ such that + $\left(x,y\right)\in A\times C$, we have that + $A\times C\subseteq B\times T$ and so + $\left(X,y\right)\in B\times D$ so $x\in B$ and $y\in D$.* + + *Hence for $A\neq\emptyset$ and $C\not\emptyset$, we have that + $A\subseteq B$ and $C\subseteq D$ gives + $A\times C\subseteq B\times D$ and $A\times C\subseteq B\times D$ + gives $A\subseteq B$ and $C\subseteq D$. Hence we have* + + *$$\begin{equation*} + A\times C\subseteq B\times D\iff A\subseteq B\text{ and } C\subseteq D + \end{equation*}$$* + +8. *If $A\subseteq B$ then $A\times C\subseteq B\times C$:* + + *Let $A$ be such that $A\subseteq B$. We have for any set $C$ that + $C\subseteq C$, hence by the previous property we know that* + + *$$\begin{equation*} + A\subseteq B\text{ and } C\subseteq C\Rightarrow A\times C\subseteq B\times C + \end{equation*}$$* + +9. *If $C\subseteq D$ then $A\times C\subseteq A\times D$:* + + *Let $C$ be such that $C\subseteq D$. We have that $A\subseteq A$ + and so by property 7 we have that* + + *$$\begin{equation*} + A\subseteq A\text{ and } C\subseteq D\Rightarrow A\times C\subseteq A\times D + \end{equation*}$$* + +10. *$A\times\left(B\setminus C\right)=\left(A\times B\right)\setminus\left(A\times C\right)$:* + + *Let $\left(x,y\right)\in A\times\left(B\setminus C\right)$ then we + have that $\left(x,y\right)\in A\times\left(B\setminus C\right)$ if + and only if $x\in A$ and $y\in B\setminus C$. $y\in B\setminus C$ + means that $y\in B$ and $y\not\in C$. Thus, $x\in A$ and $y\in B$ + and $y\not\in C$ happens if and only if + $\left(x,y\right)\in A\times B$ and + $\left(x,y\right)\not\in A\times C$. Hence by definition of the + difference of two sets we have that $\left(x,y\right)\in A\times B$ + and $\left(x,y\right)\not\in A\times C$ if and only if + $\left(x,y\right)\in \left(A\times B\right)\setminus\left(A\times C\right)$.* + +11. *$\left(A\setminus B\right)\times C = \left(A\times C\right)\setminus\left( B\times C\right)$:* + + *Let $\left(x,y\right)\in \left(A\setminus B\right)\times C$ then we + have that $\left(x,y\right)\in \left(A\setminus B\right)\times C$ if + and only if $x\in A\setminus B$ and $y\in C$, moreover + $x\in A\setminus B$ means that $x\in A$ and $x\not\in B$. Hence + $x\in A$ and $x\not\in B$ and $y\in C$ occurs if and only if + $\left(x,y\right)\in A\times C$ and + $\left(x,y\right)\not\in B\times C$. Hence by definition we have + that $\left(x,y\right)\in A\times C$ and + $\left(x,y\right)\not\in B\times C$ if and only if + $\left(x,y\right)\in\left(A\times C\right)\setminus\left( B\times C\right)$.* + +12. *$\left(A\times B\right)\setminus\left(C\times D\right)=\left(A\times\left(B\setminus D\right)\right)\cup\left(\left(A\setminus B\right)\times C\right)$:* + + *Let + $\left(x,y\right)\in \left(A\times B\right)\setminus\left(C\times D\right)$, + then we have that $\left(x,y\right)\in A\times B$ and + $\left(x,y\right)\not\in C\times D$, which happens if and only if + $x\in A$ and $y\in B$ and $x\not\in C$ and $y\not\in D$. Now, + $x\in A$ and $y\in B$ and $x\not\in C$ and $y\not\in D$ means that + either $x\in A$ and $y\in B$ and $x\not\in C$ or $x\in A$ and + $y\in B$ and $y\not\in D$. In the first case, $x\in A$ and $y\in B$ + and $x\not\in C$, we have that $x\in A\setminus C$ and $y\in B$, in + the second case, $x\in A$ and $y\in B$ and $y\not\in D$ we have + $x\in A$ and $y\in B\setminus D$.* + + *$x\in A$ and $y\in B$ and $x\not\in C$ or $x\in A$ and $y\in B$ and + $y\not\in D$ occurs if and only if $x\in A\setminus C$ and $y\in B$ + or $x\in A$ and $y\in B\setminus D$. Now by the definition of the + Cartesian product we have that $x\in A\setminus C$ and $y\in B$ + gives us that + $\left(x,y\right)\in \left(A\setminus C\right)\times B$ and $x\in A$ + and $y\in B\setminus D$ gives us + $\left(x,y\right)\in A\times \left(C\setminus D\right)$.* + + *Hence $x\in A\setminus C$ and $y\in B$ or $x\in A$ and + $y\in B\setminus D$ occurs if and only if + $\left(x,y\right)\in \left(A\setminus C\right)\times B$ or + $\left(x,y\right)\in A\times \left(C\setminus D\right)$, from which + we deduce that + $\left(x,y\right)\in \left(A\setminus C\right)\times B$ or + $\left(x,y\right)\in A\times \left(C\setminus D\right)$ if and only + if + $\left(x,y\right)\in \left(A\setminus C\right)\times B\cup A\times \left(C\setminus D\right)$.* + +13. *Suppose $A\subseteq C$ and $B\subseteq D$ and consider + $C\setminus A$ and $T\setminus B$. We have* + + *$$\begin{align*} + \left(C\setminus A\right)\times D &= \left(C\times D\right)\setminus\left(A\times D\right)\\ + C\times\left(D\setminus B\right) &=\left(C\times D\right)\setminus \left(C\times B\right) + \end{align*}$$* + + *Recall that + $C\setminus A=\left\{x: x\in C\text{ and } x\not\in A\right\}$. Now + we have by property 11. that* + + *$$\begin{equation*} + \left(C\setminus A\right)\times D= \left(C\times D\right)\setminus \left(A\times D\right) + \end{equation*}$$* + + *Likewise, by property 10. we have that* + + *$$\begin{equation*} + C\times\left(D\setminus B\right)= \left(C\times D\right)\setminus \left(C\times B\right) + \end{equation*}$$* + +*Hence the result has been shown. $\qed$* +::: + +###### Power Set + +We make one final definition of an elementary operation for sets. + +::: definition +**Definition 34**. *Power set* + +*Let $S$ be a set. We define the power set of the set $S$, denoted +$P\left(S\right)$ to be the set which contains all of the possible +subsets of $S$.* +::: + +::: example +**Example 25**. *Let $S=\left\{1,2,3\right\}$ then we have that* + +*$$\begin{equation*} + P\left(S\right)=\left\{\emptyset,\left\{1\right\},\left\{2\right\},\left\{3\right\},\left\{1,2\right\},\left\{1,3\right\},\left\{2,3\right\},S\right\} +\end{equation*}$$* +::: + +##### Set Partitions + +Recall the idea of disjoint sets, that is if $X$ and $Y$ are sets then +$X$ and $Y$ are disjoint if $X\cap Y=\emptyset$. This is saying that $X$ +and $Y$ have no elements in common. Now suppose we have a set $S$ such +that $X\cup Y=S$ but $X\cap Y=\emptyset$. Then $S$ is made of two +distinct pieces. Of course there is nothing special about $S$ being made +of only two pieces, and could be made of many many pieces. We capture +this idea in the next definition. + +::: definition +**Definition 35**. *Partition of a set* + +*Let $S$ be a set and define $\mathbb{S}$ to be the set of subsets of +$S$. We say that $\mathbb{S}$ is a partition of $S$ if the following +hold.* + +1. *$\forall S_1,S_2\in\mathbb{S}$ we have $S_1\cap S_2=\emptyset$ + whenever $S_1\neq S_2$* + +2. *Taking the union of every $T\in\mathbb{S}$ gives us $S$ that is* + + *$$\begin{equation*} + S=\bigcup_{T\in\mathbb{S}} T + \end{equation*}$$* + +3. *$\forall T\in\mathbb{S}$ we have that $T\neq\emptyset$.* + +*If the number of sets in $\mathbb{S}$ is finite with say $n$ elements +then we call $\mathbb{S}$ an $n$-component partition* +::: + +::: example +**Example 26**. *Let $S=\left\{1,2,3,4\right\}$ and let +$S_1=\left\{2,4\right\}$ and $S_2=\left\{1,3\right\}$. Then $S_1$ and +$S_2$ partition $S$. Interestingly we have that $S_1^C=S_2$ and +$S_2^C = S_1$, so the complements of these sets still forms a partition* + +*If instead we have $S_3 = \left\{1\right\}$ and +$S_4=\left\{2,3,4\right\}$ then we also have a partition where the +complements are also a partition. Now if $S_5=\left\{2\right\}$, +$S_6=\left\{1,3\right\}$ and $S_7=\left\{4\right\}$ then $S_5,S_6$ and +$S_7$ is a partition of $S$.* +::: + +The fact in the first two examples we had two sets partitioning $S$ +where the complements also partitioned $S$ is not a coincidence. + +::: proposition +**Proposition 13**. *Complements of 2-component partition is partition* + +*Let $S$ be a set such that $A\subseteq S$ and $B\subseteq S$ is a +$2$-component partition for $S$. We have that $A$ and $B$ partition $S$ +if and only if $A^C$ and $B^C$ partition $S$.* + +*Proof:* + +*$\left(\Rightarrow\right):$ Suppose that $A\subseteq S$ and +$B\subseteq S$ partition $S$. By definition we have that* + +1. *$A\cap B = \emptyset$* + +2. *$A\cup B = S$* + +3. *$A\neg\emptyset$ and $B\neq \emptyset$* + +*We need to show that $A^C$ and $B^C$ is a partition that is* + +1. *$A^C\cap B^C = \emptyset$* + +2. *$A^C\cup B^C = S$* + +3. *$A^C\neq\emptyset$ and $B^C\neq \emptyset$* + + + +1. *$A^C\cap B^C = \emptyset$:* + + *As $A\cup B = S$ we have on taking the complement of both sides + that* + + *$$\begin{align*} + A\cup B &= S\\ + \left(A\cup B\right)^C &= S^C\\ + A^C\cap B^C &= \emptyset + \end{align*}$$* + + *So $A^C\cap B^C = \emptyset$.* + +2. *$A^C\cup B^C = S$:* + + *Likewise as $A\cap B = \emptyset$ then on taking the complement of + both sides we have that* + + *$$\begin{align*} + A\cap B &= \emptyset\\ + \left(A\cap B\right)^C &= \emptyset^C\\ + A^C\cup B^C &= S + \end{align*}$$* + + *So $A^C\cup B^C = S$.* + +3. *$A^C\neq\emptyset$ and $B^C\neq \emptyset$:* + + *Suppose that $A^C = \emptyset$ then by taking the complement of + both sides we have that $A=S$ which implies $B=\emptyset$, which is + a contradiction as $A$ and $B$ partition $S$. Likewise if we suppose + that $B^C=\emptyset$ we will have to conclude that $A=\emptyset$ + which will be a contradiction. It thus follows that neither $A^C$ or + $B^C$ can be empty.* + + *Hence $A^C\neq\emptyset$ and $B^C\neq \emptyset$.* + +*It follows that $A^C$ and $B^C$ is a partition of $S$* + +*$\left(\Leftarrow\right)$: Suppose that $A^C$ and $B^C$ is a partition +of $S$. We have that $A^C\subseteq S$ and $B^C\subset S$. By the +previous part we have that $\left(A^C\right)^C$ and $\left(B^C\right)^C$ +is a partition of $S$. However $\left(A^C\right)^C=A$ and +$\left(B^C\right)^C=B$. Thus $A$ and $B$ is a partition of $S$* + +*The result now follows. $\qed$* +::: + +There are some additional results we can state about partitions that +relate to the operations we can do on sets. We will require the +following lemma. + +::: lemma +**Lemma 2**. *Set difference and intersection are disjoint sets* + +*Let $S$ and $T$ be two sets. We have that $S\setminus T$ and $S\cap T$ +are disjoint sets, which is to say that* + +*$$\begin{equation*} + \left(S\setminus T\right)\cap \left(S\cap T\right)=\emptyset +\end{equation*}$$* + +*Proof:* + +*Suppose that $x\in \left(S\setminus T\right)\cap \left(S\cap T\right)$ +then by definition $x\in S\setminus T$ and $x\in S\cap T$. As +$x\in S\setminus T$ then we have that $x\in S$ and $x\not\in T$, +likewise as $x\in S\cap T$ then $x\in S$ and $x\in T$. It is clear that +no such $x$ can exist hence +$\left(S\setminus T\right)\cap \left(S\cap T\right)=\emptyset$.* +::: + +##### A brief look at Zermelo--Fraenkel set theory {#subsubSec:ZFCAxioms} + +At the start of this section we introduced the idea of Zermelo--Fraenkel +set theory. This is the complete formalisation of set theory and the +true bedrock of mathematics. The Zermelo--Fraenkel set theory axioms, +hence now referred to as ZF, are given as follows. + +::: definition +**Definition 36**. *Zermelo--Fraenkel set theory axioms* + +*The Zermelo-Fraenkel set theory axioms are the following.* + +1. *The axiom of extensionality:* + + *The axiom of extensionality asserts that two sets are equal if and + only if they contain the same elements.* + +2. *The axiom of the empty-set:* + + *The axiom of the empty-set asserts that there exists a set which + contains no elements* + +3. *The axiom of pairing:* + + *The axiom of pairing asserts that given any set $A$ and any set + $B$, there is a set $C$ such that, given any set $D$, $D$ is a + member of $C$ if and only if $D$ is equal to $A$ or $D$ is equal to + $B$. This is to say, given two sets, there is a set whose members + are exactly the two given sets.* + +4. *The axiom of specification:* + + *The axiom of specification asserts that we can construct a set + which satisfies a given condition, so long as this condition is not + inherently contradictory.* + +5. *The axiom of unions:* + + *The axiom of unions asserts that we can perform the union of two + sets $A$ and $B$* + +6. *The axiom of powers:* + + *The axiom of powers asserts that for any set $S$ we can construct a + set $P\left(S\right)$ whose elements are all the possible subsets of + $S$.* + +7. *The axiom of infinity:* + + *The axiom of infinity asserts that there is at least one infinite + set $A$, that is at least one set with infinitely many elements. + That is we have a set $A$ such that the $\emptyset\in A$ and if + $x\in A$ then the set $x\cup\left\{x\right\}$ is also in $A$.* + +8. *The axiom of replacement:* + + *We will need the next section to fully understand this axiom, + however informally asserts that for some set $S$, and form another + set by replacing the elements of $S$ by other sets according to any + definite rule.* + +9. *The axiom of foundation:* + + *The axiom of foundation asserts that for every non-empty set $S$, + there exists an element $x\in S$ such that $x$ and $S$ are disjoint. + This also asserts that no set can contain itself.* +::: + +There is also one axiom which we have left off. This is the +controversial axiom of choice. + +::: definition +**Definition 37**. *The axiom of choice* + +*Let $S$ be a set of non-empty sets. The axiom of choice asserts that +there is a way to pick an element of each of the sets in $S$.* +::: + +With the axiom of choice we have the following + +::: definition +**Definition 38**. *ZFC axioms* + +*The axioms of ZF along with the axiom of choice gives us the ZFC +axioms* +::: + +We can already see that our "hands-on" approach to set theory has +somewhat indirectly captured the essence of the ZF axioms. We can use +the ZF axiom to prove in a truly rigours way what we did with out +"hands-on" approach. Although an interesting field of study itself, we +will not really need to use the ZF axioms, although occasionally we may +rely on choice. + +There is one other thing that needs bringing up, ZFC has one more +component, the axioms alone are not enough to prove anything. We need +the notion of inclusion, that is being an element of a set. That is we +include the symbol $\in$ along with the axioms, where $\in$ takes on the +meaning we defined earlier. With this we can in theory use ZFC to start +proving and building up mathematics from the bedrock. + +#### Mappings + +##### Introduction and basic definitions + +Now that we have the of a set what can we use it for? Many areas of +mathematics can be broken down into the theory of sets, in particular +how we can get from one set to another. Without this idea we wouldn't be +able to get very far at all. As an example, you may have seen, in a +calculus course for example, the idea of a function $f\left(x\right)$, +say $f\left(x\right)=x^2$ where $x$ can be any number we choose. Say +$x=2$ then $f\left(2\right)=4$. You may have also seen functions where +we are not allowed to use any number we wish for example, if we take +$f\left(x\right)=\sqrt{x}$ then we are only allowed positive numbers if +we want a to find an answer using the numbers we are familiar with, such +as $1, 88.125, \pi,\sqrt{2}$ etc. This set we will denote by +$\mathbb{R}$. The alert reader may now see how sets will come into play, +to define in a rigours way the ideas of $f\left(x\right)=x^2$ and other +such functions, we need to consider what are the allowable inputs which +once done will give us the possible outputs. That is if we have a set +whose elements are inputs and we define some form of function, which we +will now call a map, then we will get another set whose elements are +what inputs will be 'mapped' to. + +::: definition +**Definition 39**. *Mapping* + +*Let $X$ and $Y$ be sets. Suppose we have some rule or description, +which we will denote by $f$, by which for each $x\in X$ there is some +element $f\left(x\right)\in Y$. We say that the rule (description) is a +mapping or map or function from $X$ to $Y$. We denote a mapping with the +following notation* + +*$$\begin{align*} + f:X&\mathlarger{\mathlarger{\rightarrow}}Y\\ + x&\mapsto f\left(x\right) +\end{align*}$$* + +*where the first line tells us what sets the mapping is between, and the +bottom line tells us where each element $x\in X$ gets mapped to* +::: + +::: definition +**Definition 40**. *Domain* + +*Let $f:X\mathlarger{\mathlarger{\rightarrow}}Y$ be a mapping between +two sets $X$ and $Y$. We say that the set $X$ is the domain of the +mapping $f$. The domain contains the elements which the map can act on. +We can write this as* + +*$$\begin{equation*} + \mathop{\mathrm{Dom}}\left(f\right)=X +\end{equation*}$$* +::: + +::: definition +**Definition 41**. *Co-Domain* + +*Let $f:X\mathlarger{\mathlarger{\rightarrow}}Y$ be a mapping between +two sets $X$ and $Y$. We say that the set $Y$ is the Co-domain of the +mapping $f$. The co-domain contains the possible elements that the map +can send elements of $X$ to. We can write this as* + +*$$\begin{equation*} + \mathop{\mathrm{Cdm}}\left(f\right)=Y +\end{equation*}$$* +::: + +We have some examples of mappings. + +::: example +**Example 27**. *Let $X=\left\{1,2,3\right\}$ and let $Y=X$. Define the +map* + +*$$\begin{align*} + f:X&\mathlarger{\mathlarger{\rightarrow}}Y\\ + x&\mapsto f\left(x\right)=x +\end{align*}$$* + +*To see what $f$ does we will take each element of $X$ one at a time. +Starting with $1$ we have that $1\mapsto f\left(1\right)=1$, for $2$ we +have $2\mapsto f\left(2\right)=2$ and finally +$3\mapsto f\left(3\right)=3$. Hence the map $f$ takes an element of $X$ +and leaves it alone. A map which takes every element of its domain and +leaves it alone is called an identity map, or if you prefer the do +nothing at all map.* +::: + +::: {#exmp:Mapping 1 .example} +**Example 28**. *Let $X=Y=\mathbb{N}$. Let $f$ be the map given by* + +*$$\begin{align*} + f:X&\mathlarger{\mathlarger{\rightarrow}}Y\\ + x&\mapsto f\left(x\right)=2x +\end{align*}$$* + +*It is clear to see that every element in the domain gets doubled, i.e +$f\left(1\right)=2$, $f\left(2\right)=4$, $f\left(3\right)=6$ and so +on.* +::: + +A map does not need to be given by an explicit mathematical formulae + +::: example +**Example 29**. *Let +$A=\text{The set of all humans currently alive on planet earth}$, from +which it should be clear to see that $\text{You}\in A$ [^5]. Let +$B=\left\{0,1\right\}$. Let $f$ be the mapping given by* + +*$$\begin{align*} + f:A&\mathlarger{\mathlarger{\rightarrow}}B\\ + a&\mapsto f\left(a\right)= \begin{cases} + 1,\ \text{If } a \text{ has hair on their head}\\ + 0,\ \text{If } a \text{ does not have hair on their head}\\ + \end{cases} +\end{align*}$$* + +*Then $f$ is a map which indicates if a given person has hair on their +head or not.* +::: + +The above definition of a mapping can be made more general + +::: definition +**Definition 42**. *Piecewise mapping* + +*Let $f:X\rightarrow Y$ be a mapping. We say that $f$ is a piecewise +mapping if we need multiple rules or descriptions to fully describe $f$. +That we wish to define the mapping using different rules based on the +input. If for each of this input ranges we define a mapping +$g_1,g_2,g_3,\dots$ then we can write the piecewise function as follows* + +*$$\begin{align*} + f:X&\rightarrow Y\\ + x&\mapsto f\left(x\right)=\begin{cases} + g_1\left(x\right),\ \text{Condition for }g_1\\ + g_2\left(x\right),\ \text{Condition for }g_2\\ + g_3\left(x\right),\ \text{Condition for }g_3\\ + \dots + \end{cases} +\end{align*}$$* +::: + +::: example +**Example 30**. *Let $f:\mathbb{N}\rightarrow\mathbb{N}$ be defined by* + +*$$\begin{align*} + f:\mathbb{N}&\rightarrow\mathbb{N}\\ + x &\mapsto f\left(x\right) = \begin{cases} + 2x,\ \text{If } $x <5$\\ + 5x,\ \text{Otherwise} + \end{cases} +\end{align*}$$* + +*We have that $f\left(1\right)=2$, $f\left(2\right)=4$ and so on up to +$f\left(4\right)=8$, then $f\left(5\right)=25$ and so on.* +::: + +We make one more useful definition that will be useful throughout the +rest of the text, + +::: definition +**Definition 43**. *Closure of a mapping* + +*Let $X$ be a set. If we have a mapping such that $f:X^n\rightarrow X$. +We say the mapping has closure on the set $X$, or we say that $f$ is a +closed mapping.* +::: + +##### The image and pre-image + +We now define a more technical notion of how a mapping $f$ maps an +element in the domain to the co-domain. + +::: definition +**Definition 44**. *Image of an element* + +*Let $f:X\mathlarger{\mathlarger{\rightarrow}}Y$ be a mapping of between +two sets $X$ and $Y$, and let $x\in X$ be an element of the domain. We +say that $f\left(x\right)\in Y$ is the image of the element $x$.* +::: + +Which in turn allows us to define a subset of the co-domain for which +every element $x\in X$ gets mapped to + +::: {#def:ImageMapping .definition} +**Definition 45**. *Image of a mapping* + +*Let $f:X\mathlarger{\mathlarger{\rightarrow}}Y$ be a mapping of between +two sets $X$ and $Y$. We define the set* + +*$$\begin{equation*} + \mathop{\mathrm{Image}}\left(f\right)=f\left(X\right)=\left\{f\left(x\right):x\in X\right\}\subseteq Y +\end{equation*}$$ To be the image of the domain, sometimes called the +range of $f$. That is the image is the set of all possible outputs of +the mapping $f$ with the domain $X$.* + +*Moreover, suppose that $A\subseteq X$ then we define the image of the +subset $A$ to be* + +*$$\begin{equation*} + f\left(A\right)=\left\{f\left(x\right):x\in A\right\}\subseteq f\left(X\right)\subseteq Y +\end{equation*}$$* + +*That is we can consider the image of subsets of $X$.* +::: + +::: example +**Example 31**. *Consider the mapping in example +[28](#exmp:Mapping 1){reference-type="ref" reference="exmp:Mapping 1"}, +we have that $X=Y=\mathbb{N}$ and is $f$ the map* + +*$$\begin{align*} + f:X&\mathlarger{\mathlarger{\rightarrow}}Y\\ + x&\mapsto f\left(x\right)=2x +\end{align*}$$* + +*then we have that +$\mathop{\mathrm{Image}}\left(f\right)=f\left(\mathbb{N}\right)=\left\{2x:x\in\mathbb{N}\right\}$* +::: + +::: example +**Example 32**. *Let $f$ be an arbitrary mapping such that +$f:\emptyset\mathlarger{\mathlarger{\rightarrow}}Y$ for some set $Y$. +What is $\mathop{\mathrm{Image}}\left(f\right)$?. We have by the +definition of a mapping [45](#def:ImageMapping){reference-type="ref" +reference="def:ImageMapping"}, we have that* + +*$$\begin{equation*} + \mathop{\mathrm{Image}}\left(f\right)=\left\{f\left(x\right):x\in\emptyset\right\} +\end{equation*}$$* + +*However, we know that the empty set has no elements, so there are no +elements that $f$ can send anything to, so +$\mathop{\mathrm{Image}}\left(f\right)=\emptyset$.* +::: + +Likewise we can define how a mapping is mapped to from the domain to the +co-domain. This is called the pre-image. + +::: definition +**Definition 46**. *Pre-image of an element* + +*Let $f:X\mathlarger{\mathlarger{\rightarrow}}Y$ be a mapping of between +two sets $X$ and $Y$, and let $y\in Y$ be an element of the co-domain. +If $f\left(x\right)=y$ then we say that $f\left(x\right)\in X$ is the +pre-image of the element $y$ and we denote this $f^{-1}\left(y\right)$.* +::: + +Which in turn allows us to define a subset of the domain for which every +element $y\in Y$ gets mapped to + +::: {#def:PreImageMapping .definition} +**Definition 47**. *Pre-image of a mapping* + +*Let $f:X\mathlarger{\mathlarger{\rightarrow}}Y$ be a mapping of between +two sets $X$ and $Y$. We define the set* + +*$$\begin{equation*} + \mathop{\mathrm{PreImage}}\left(f\right)=f^{-1}\left(Y\right)=\left\{x\in X:f\left(x\right)\in Y\right\}\subseteq X +\end{equation*}$$ To be the pre-image of the co-domain. That is the +pre-image is the set of all possible inputs that give the given +outputs.* + +*Moreover, suppose that $B\subseteq Y$ then we define the pre-image of +the subset $B$ to be* + +*$$\begin{equation*} + f^{-1}\left(B\right)=\left\{x\in X:f\left(x\right)\in B\right\}\subseteq f^{-1}\left(Y\right)\subseteq X +\end{equation*}$$* +::: + +::: example +**Example 33**. *Consider the mapping +$f:\mathbb{N}\rightarrow\mathbb{N}$ given by* + +*$$\begin{align*} + f:\mathbb{N}&\rightarrow\mathbb{N}\\ + x&\mapsto f\left(x\right)=\frac{x}{2} +\end{align*}$$* + +*We have that $\frac{x}{2}$ is defined in the naturals only when $x$ is +an even number, hence the pre-image must consist of the even numbers.* + +*$$\begin{equation*} + \mathop{\mathrm{PreImage}}\left(f\right)=f^{-1}\left(\mathbb{N}\right)=\left\{x\in\mathbb{N}:\frac{x}{2}\in\mathbb{N}\right\}=\left\{0,2,4,6,8\dots\right\} +\end{equation*}$$* +::: + +::: example +**Example 34**. *Consider the mapping +$f:\mathbb{N}\rightarrow\mathbb{N}$ given by* + +*$$\begin{align*} + f:\mathbb{N}&\rightarrow\mathbb{N}\\ + x&\mapsto f\left(x\right)=x^2 +\end{align*}$$* + +*We have that the pre-image is given by* + +*$$\begin{equation*} + \mathop{\mathrm{PreImage}}\left(f\right)=\left\{x\in\mathbb{N}:x^2\in\mathbb{N}\right\}=\left\{0,1,2,3,4\dots\right\}=\mathbb{N} +\end{equation*}$$* +::: + +With these definitions we can make the following observations + +::: {#prop:PropertyImagePreImage .proposition} +**Proposition 14**. *Properties of the image and pre-image* + +*Let $f:X\rightarrow Y$ be a mapping and let $A\subseteq X$ and +$B\subseteq Y$. We have that the following properties hold for the image +and pre-image* + +1. *$f\left(X\right)\subseteq Y$* + +2. *$f\left(f^{-1}\left(Y\right)\right)=f\left(X\right)$* + +3. *$f\left(f^{-1}\left(B\right)\right)\subseteq B$* + +4. *$f\left(f^{-1}\left(B\right)\right)=B\cap f\left(X\right)$* + +5. *$f\left(f^{-1}\left(f\left(A\right)\right)\right)=f\left(A\right)$* + +6. *$f\left(A\right)=\emptyset\iff A=\emptyset$* + +7. *$B\subseteq f\left(A\right)\iff\exists C\subseteq A: f\left(C\right)=B$* + +8. *$f\left(X\setminus A\right)\subseteq f\left(A\right)\iff f\left(A\right)=f\left(X\right)$* + +9. *$f\left(X\right)\setminus f\left(A\right)\subseteq f\left(X\setminus A\right)$* + +10. *$f\left(A\cup f^{-1}\left(B\right)\right)\subseteq f\left(A\right)\cup B$* + +11. *$f\left(A\cap f^{-1}\left(B\right)\right)= f\left(A\right)\cap B$* + +*Likewise the following properties hold for the pre-image* + +1. *$f^{-1}\left(Y\right)=X$* + +2. *$f^{-1}\left(f\left(X\right)\right)=X$* + +3. *$A\subseteq f^{-1}\left(f\left(A\right)\right)$* + +4. *Suppose that instead of the mapping $f:X\rightarrow Y$ we consider + a new mapping based on $f$, which we we call $\Bar{f}$. We define + $\Bar{f}$ to be the mapping* + + *$$\begin{align*} + \Bar{f}:A&\mathlarger{\mathlarger{\rightarrow}}Y\\ + x&\mapsto \Bar{f}\left(x\right)=f\left(x\right) + \end{align*}$$* + + *that is $\Bar{f}$ maps every element of $a\in A$ to what + $f\left(a\right)$ does. With this new mapping we have the following + property* + + *$$\begin{equation*} + \left(\Bar{f}\right)^{-1}\left(B\right)=A\cap f^{-1}\left(B\right) + \end{equation*}$$* + +5. *$f^{-1}\left(f\left(f^{-1}\left(B\right)\right)\right)=f^{-1}\left(B\right)$* + +6. *$f^{-1}\left(B\right)=\emptyset\iff B\subseteq Y\setminus f\left(X\right)$* + +7. *$A\subseteq f^{-1}\left(B\right)\iff f\left(A\right)\subseteq B$* + +8. *$f^{-1}\left(Y\setminus B\right)\subseteq f^{-1}\left(B\right)\iff f^{-1}\left(B\right)=X$* + +9. *$f^{-1}\left(Y\setminus B\right)= X\setminus f^{-1}\left(B\right)$* + +10. *$A\cup f^{-1}\left(B\right)\subseteq f^{-1}\left(f\left(A\right)\cup B\right)$* + +11. *$A\cap f^{-1}\left(B\right)\subseteq f^{-1}\left(f\left(A\right)\cap B\right)$* + +*Proof:* + +*We start with the properties of the image.* + +1. *$f\left(X\right)\subseteq Y$:* + + *This holds by definition of the image.* + +2. *$f\left(f^{-1}\left(Y\right)\right)=f\left(X\right)$:* + + *Let $x\in f\left(f^{-1}\left(Y\right)\right)$ and recall the + definition of the image and pre-image.* + + *$$\begin{align*} + f\left(A\right)&=\left\{f\left(x\right):x\in A\right\}\subseteq f\left(X\right)\subseteq Y\\ + f^{-1}\left(B\right)&=\left\{x\in X:f\left(x\right)\in B\right\}\subseteq f^{-1}\left(Y\right)\subseteq X + \end{align*}$$* + + *We have that* + + *$$\begin{equation*} + f\left(f^{-1}\left(Y\right)\right)=\left\{f\left(y\right):y\in f^{-1}\left(Y\right)\right\} + \end{equation*}$$* + + *Hence $x\in f\left(f^{-1}\left(Y\right)\right)$ means that + $x=f\left(y\right)$ for some $y\in f^{-1}\left(Y\right)$, + additionally we conclude that $y\in X$. Moreover by the definition + of the pre-image we have that $f^{-1}\left(Y\right)\subseteq X$. It + thus follows that $x\in f\left(X\right)$ and so + $f\left(f^{-1}\left(Y\right)\right)\subseteq f\left(X\right)$.* + + *Now suppose that $x\in f\left(X\right)$, that is + $x=f\left(x'\right)$ for some $x'\in X$. Now by definition of the + pre-image as $x'\in X$ with $f\left(x'\right)\in Y$ we have that + $x'\in f^{-1}\left(Y\right)$. Hence by definition of the set + $f\left(f^{-1}\left(Y\right)\right)$ we must conclude that + $f\left(x'\right)\in f\left(f^{-1}\left(Y\right)\right)$, which is + to say $x\in f\left(f^{-1}\left(Y\right)\right)$. Hence + $f\left(X\right)\subseteq f\left(f^{-1}\left(Y\right)\right)$.* + + *It follows that + $f\left(f^{-1}\left(Y\right)\right)=f\left(X\right)$.* + +3. *$f\left(f^{-1}\left(B\right)\right)\subseteq B$:* + + *Suppose that $x\in f\left(f^{-1}\left(B\right)\right)$ where + $B\subseteq Y$. We hence have that $x=f\left(b\right)$ for some + $b\in f^{-1}\left(B\right)$, hence $b\in X$ giving us + $f\left(b\right)\in B$ and so + $f\left(f^{-1}\left(B\right)\right)\subseteq B$.* + +4. *$f\left(f^{-1}\left(B\right)\right)=B\cap f\left(X\right)$:* + + *Let $x\in f\left(f^{-1}\left(B\right)\right)$ then by property 3 we + have that $x\in B$. Additionally as + $x\in f\left(f^{-1}\left(B\right)\right)$ and $B\subseteq Y$ then + $f\left(f^{-1}\left(B\right)\right)\subseteq f\left(f^{-1}\left(Y\right)\right)$ + and so $x\in f\left(f^{-1}\left(Y\right)\right)$. Now by property 2 + we have that $f\left(f^{-1}\left(Y\right)\right)=f\left(X\right)$ + thus $x\in f\left(X\right)$ and so $x\in B\cap f\left(X\right)$. It + follows that + $f\left(f^{-1}\left(B\right)\right)\subseteq B\cap f\left(X\right)$.* + + *Now suppose that $x\in B\cap f\left(X\right)$. By definition of + $f\left(X\right)$ we have $x\in f\left(X\right)$ gives us that + $x=f\left(x'\right)$ where $x'\in X$, moreover we also have that + $x\in B$. Now we have the set $f\left(f^{-1}\left(B\right)\right)$ + is given by* + + *$$\begin{equation*} + f\left(f^{-1}\left(B\right)\right)=\left\{f\left(b\right):b\in f^{-1}\left(B\right)\right\} + \end{equation*}$$* + + *We have that $x=f\left(x'\right)$ and so + $x'\in f^{-1}\left(B\right)$, hence clearly by definition of the + image we have that $x\in f\left(f^{-1}\left(B\right)\right)$. It + follows that + $B\cap f\left(X\right)\subseteq f\left(f^{-1}\left(B\right)\right)$.* + + *Hence the result + $f\left(f^{-1}\left(B\right)\right)=B\cap f\left(X\right)$.* + +5. *$f\left(f^{-1}\left(f\left(A\right)\right)\right)=f\left(A\right)$:* + + *By property $4$ we have that* + + *$$\begin{equation*} + f\left(f^{-1}\left(f\left(A\right)\right)\right)=f\left(A\right)\cap f\left(X\right) + \end{equation*}$$ as $f\left(A\right)\subseteq Y$. Finally + $f\left(A\right)\cap f\left(X\right)=f\left(A\right)$ as + $f\left(A\right)\subseteq f\left(X\right)$. The result follows.* + +6. *$f\left(A\right)=\emptyset\iff A=\emptyset$:* + + *$\left(\Leftarrow\right)$: Suppose that + $f\left(A\right)=\emptyset$. By definition of the image we have + that* + + *$$\begin{equation*} + f\left(A\right)=\left\{f\left(x\right):x\in A\right\} + \end{equation*}$$ By set equality we must have that + $f\left(A\right)=\left\{f\left(x\right):x\in A\right\}=\emptyset$. + Hence there can be no elements $f\left(x\right)$ where $x\in A$ + which can only occur if $A=\emptyset$ for if not then + $f\left(A\right)$ has at least one element for some $x'\in A$, + contradicting the fact that $f\left(A\right)=\emptyset$. It follows + that $A=\emptyset$.* + + *$\left(\Rightarrow\right)$: Suppose that $A=\emptyset$, we have + that the image of the empty set is given by* + + *$$\begin{equation*} + f\left(A\right)=f\left(\emptyset\right)=\left\{f\left(x\right):x\in \emptyset\right\}=\emptyset + \end{equation*}$$* + + *It follows that $f\left(A\right)=\emptyset$.* + +7. *$B\subseteq f\left(A\right)\iff\exists C\subseteq A: f\left(C\right)=B$:* + + *$\left(\Rightarrow\right)$: Suppose that + $B\subseteq f\left(A\right)$. We show that + $\exists C\subseteq A: f\left(C\right)=B$. So, suppose that $x\in B$ + then we have that $x\in f\left(A\right)$ by assumption. By + definition of the image we have that* + + *$$\begin{equation*} + f\left(A\right)=\left\{f\left(x\right):x\in A\right\} + \end{equation*}$$* + + *Hence we have $x\in f\left(A\right)$ gives us that + $x=f\left(x'\right)$ for some $x'\in A$. We define the required set + $C$ as follows.* + + *$$\begin{equation*} + C = \bigcup_{\substack{x'\in A \\ f\left(x'\right)\in B}} x' + \end{equation*}$$ That is $C$ is defined to be those elements + $x'\in A$ such that $f\left(x'\right)\in B$ which is a subset of + $f\left(A\right)$. Clearly $C\subseteq A$ as each $x'\in C$ is by + construction an element of $A$. Additionally we also have + $f\left(C\right)=B$ by construction of $C$.* + + *$\left(\Leftarrow\right)$: Suppose that + $\exists C\subseteq A: f\left(C\right)=B$. As $f\left(C\right)=B$ we + have by the definition of the image that* + + *$$\begin{equation*} + f\left(C\right)=\left\{f\left(x\right):x\in C\right\} + \end{equation*}$$ that is $x\in f\left(C\right)$ gives + $x=f\left(c\right)$ for some $c\in C$ and additionally $x\in B$ by + assumption. Now $C\subseteq A$ so $c\in A$. Hence + $x\in f\left(A\right)$, hence we must conclude that + $B\subseteq f\left(A\right)$, possibly being equal if $C=A$.* + + *The result follows.* + +8. *$f\left(X\setminus A\right)\subseteq f\left(A\right)\iff f\left(A\right)=f\left(X\right)$:* + + *$\left(\Rightarrow\right)$: Suppose that + $f\left(X\setminus A\right)\subseteq f\left(A\right)$ and recall the + definition of the complement of sets. We have that* + + *$$\begin{equation*} + X\setminus A = \left\{x\in X: x\not\in A\right\} + \end{equation*}$$ Now, $A\subseteq X$ by hypothesis of the + proposition. So if $x\in f\left(X\setminus A\right)$ then by + definition of the image we have that* + + *$$\begin{equation*} + f\left(X\setminus A\right)=\left\{f\left(x\right): x\in X\setminus A\right\}=\left\{f\left(x\right):x\in X\text{ and } x\not\in A\right\} + \end{equation*}$$ but then if $x\not\in A$ then + $x\not\in f\left(A\right)$. However if $A=X$ then we have that + $X\setminus A = \emptyset$ from which it follows by property 6 that + $f\left(X\setminus A\right)=\emptyset$ and so as the empty set is a + subset of any set we conclude that + $\emptyset\subseteq f\left(A\right)$, that is we must have + $f\left(A\right)=f\left(X\right)$.* + + *$\left(\Leftarrow\right):$ Suppose that + $f\left(A\right)=f\left(X\right)$, by definition of the image we + have that* + + *$$\begin{equation*} + f\left(A\right)=\left\{f\left(a\right):a\in A\right\}=\left\{f\left(x\right):x\in X\right)=f\left(X\right) + \end{equation*}$$* + + *Now consider $f\left(X\setminus A\right)$ this set is given by* + + *$$\begin{equation*} + f\left(X\setminus A\right)=\left\{f\left(x\right): x\in X\setminus A\right\}=\left\{f\left(x\right):x\in X\text{ and } x\not\in A\right\} + \end{equation*}$$ But as all such $x\in A$ must also be $x\in X$ by + assumption we conclude that $f\left(X\setminus A\right)=\emptyset$ + and the empty set is clearly contained in any other set. Hence + $f\left(X\setminus A\right)\subseteq f\left(A\right)$. The result + has now been shown.* + +9. *$f\left(X\right)\setminus f\left(A\right)\subseteq f\left(X\setminus A\right)$:* + + *Let $x\in f\left(X\right)\setminus f\left(A\right)$. By definition + we have that* + + *$$\begin{equation*} + f\left(X\right)\setminus f\left(A\right)=\left\{x\in f\left(X\right):x\not\in f\left(A\right)\right\} + \end{equation*}$$* + + *Hence $x\in f\left(X\right)\setminus f\left(A\right)$ gives us that + $x\in f\left(X\right)$ and $x\not\in f\left(A\right)$. That is + $\exists y\in X$ with $y\nexists A$ such that $x=f\left(y\right)$, + this is $y\in X\setminus A$. Hence it follows that + $x\in f\left(X\setminus A\right)$. That is + $f\left(X\right)\setminus f\left(A\right)\subseteq f\left(X\setminus A\right)$.* + +10. *$f\left(A\cup f^{-1}\left(B\right)\right)\subseteq f\left(A\right)\cup B$:* + + *Let $x\in f\left(A\cup f^{-1}\left(B\right)\right)$. This is our + first usage of the pre-image of a set so we recall the definition, + we have that* + + *$$\begin{equation*} + f^{-1}\left(B\right)=\left\{x\in X:f\left(x\right)\in B\right)\subseteq X + \end{equation*}$$* + + *Hence the image $f\left(A\cup f^{-1}\left(B\right)\right)$ is given + by* + + *$$\begin{align*} + f\left(A\cup f^{-1}\left(B\right)\right)&=\left\{f\left(y\right):y\in A\cup f^{-1}\left(B\right)\right\}\\ + &=\left\{f\left(y\right):y\in A\text{ or } y\in f^{-1}\left(B\right)\right\}\\ + &=\left\{f\left(y\right):y\in A\text{ or } y\in X : f\left(y\right)\in B\right\} + \end{align*}$$* + + *Now, $x\in f\left(A\cup f^{-1}\left(B\right)\right)$ gives us that + either $\exists y\in A$ with $x=f\left(y\right)$ or $\exists y\in X$ + with $f\left(y\right)\in B$. In the first case where + $\exists y\in A$ with $x=f\left(y\right)$ then by definition of the + image we have that $x\in f\left(A\right)$ and so is clearly in the + union $f\left(A\right)\cup B$. Now for the second case we have that + $x\in B$ as $y\in X$ such that $x=f\left(y\right)\in B$, likewise it + is in the union $f\left(A\right)\cup B$.* + + *Hence $x\in f\left(A\right)\cup B$ and we have that + $f\left(A\cup f^{-1}\left(B\right)\right)\subseteq f\left(A\right)\cup B$. + Hence the result.* + +11. *$f\left(A\cap f^{-1}\left(B\right)\right)= f\left(A\right)\cap B$:* + + *Let $x\in f\left(A\cap f^{-1}\left(B\right)\right)$, the image of + $A\cap f^{-1}\left(B\right)$ is given by* + + *$$\begin{align*} + f\left(A\cap f^{-1}\left(B\right)\right)&=\left\{f\left(y\right):y\in A\cap f^{-1}\left(B\right)\right\}\\ + &=\left\{f\left(y\right):y\in A\text{ and } y\in f^{-1}\left(B\right)\right\}\\ + &=\left\{f\left(y\right):y\in A\text{ and } y\in X : f\left(y\right)\in B\right\}\\ + \end{align*}$$* + + *Now $x\in f\left(A\cap f^{-1}\left(B\right)\right)$ gives us that + $\exists y\in A$ with $x=f\left(y\right)$ and $\exists y\in X$ with + $f\left(y\right)\in B$. Hence we clearly have that + $x\in f\left(A\right)$ and $x\in B$ and so is in the intersection + $f\left(A\right)\cap B$. Hence we have that + $f\left(A\cap f^{-1}\left(B\right)\right)\subseteq f\left(A\right)\cap B$.* + + *Now suppose that $x\in f\left(A\right)\cap B$. We have that + $x\in f\left(A\right)$ and $x\in B$, from the first of these having + $x\in f\left(A\right)$ means that $\exists y\in A$ such that + $x=f\left(y\right)$. Now as $x\in B$ means there is some $y'\in X$ + with $x=f\left(y'\right)$. However as $f\left(A\right)\cap B$ then + we must have that $f\left(y'\right)\in f\left(A\right)$ hence + $y'\in A$. Hence both $y$ and $y'$ are in the set + $A\cap f^{-1}\left(B\right)$ and so we have + $x\in f\left(A\cap f^{-1}\left(B\right)\right)$ and therefore + $f\left(A\right)\cap B\subseteq f\left(A\cap f^{-1}\left(B\right)\right)$.* + + *The result + $f\left(A\cap f^{-1}\left(B\right)\right)= f\left(A\right)\cap B$ + follows.* + +*We now turn our attention to the results for the pre-image.* + +1. *$f^{-1}\left(Y\right)=X$:* + + *By definition of the pre-image we have that* + + *$$\begin{equation*} + f^{-1}\left(Y\right)=\left\{x\in X:f\left(x\right)\in Y\right\}\subseteq X + \end{equation*}$$* + + *Clearly $f^{-1}\left(Y\right)\subseteq X$ by definition. Now if + $x\in X$ then we must also clearly have $f\left(x\right)\in Y$ and + so $X\subseteq f^{-1}\left(Y\right)$. Hence + $f^{-1}\left(Y\right)=X$.* + +2. *$f^{-1}\left(f\left(X\right)\right)=X$:* + + *Let $y\in f^{-1}\left(f\left(X\right)\right)$, we have that the set + $f^{-1}\left(f\left(X\right)\right)$ is given by* + + *$$\begin{equation*} + f^{-1}\left(f\left(X\right)\right)=\left\{x\in X: f\left(x\right)\in f\left(X\right)\right\}\\ + \end{equation*}$$* + + *It is hence clear that for any + $x\in f^{-1}\left(f\left(X\right)\right)$ we have clearly have + $x\in X$, that is $f^{-1}\left(f\left(X\right)\right)\subseteq X$. + Likewise if $x\in X$ then clearly $x\in f\left(X\right)$ and so by + the definition of $f^{-1}\left(f\left(X\right)\right)$ we have that + $x\in f^{-1}\left(f\left(X\right)\right)$. That is + $X\subseteq f^{-1}\left(f\left(X\right)\right)$. The result + follows.* + +3. *$A\subseteq f^{-1}\left(f\left(A\right)\right)$:* + + *Suppose that $x\in A\subseteq X$. By property $2$. of the pre-image + we have that $f^{-1}\left(f\left(X\right)\right)=X$. Hence + $x\in A\subseteq f^{-1}\left(f\left(X\right)\right)=X$ giving the + result.* + +4. *Suppose that instead of the mapping $f:X\rightarrow Y$ we consider + a new mapping based on $f$, which we we call $\Bar{f}$. We define + $\Bar{f}$ to be the mapping* + + *$$\begin{align*} + \Bar{f}:A&\mathlarger{\mathlarger{\rightarrow}}Y\\ + x&\mapsto \Bar{f}\left(x\right)=f\left(x\right) + \end{align*}$$* + + *that is $\Bar{f}$ maps every element of $a\in A$ to what + $f\left(a\right)$ does. With this new mapping we have the following + property* + + *$$\begin{equation*} + \left(\Bar{f}\right)^{-1}\left(B\right)=A\cap f^{-1}\left(B\right): + \end{equation*}$$* + + *Let $x\in \left(\Bar{f}\right)^{-1}\left(B\right)$. We have that + $\left(\Bar{f}\right)^{-1}\left(B\right)$ is given by* + + *$$\begin{equation*} + \left(\Bar{f}\right)^{-1}\left(B\right)=\left\{x\in A:\Bar{f}\left(x\right)\in B\right\} + \end{equation*}$$* + + *So $x\in \left(\Bar{f}\right)^{-1}\left(B\right)$ gives that + $x\in A$, moreover as $\Bar{f}\left(x\right)\in B$ and $\Bar{f}$ + maps every $x\in A$ to $f\left(x\right)$ then + $\Bar{f}\left(x\right)=f\left(x\right)\in B$. It follows that + $x\in f^{-1}\left(B\right)$ and so + $x\in A\cap f^{-1}\left(B\right)$. Thus + $\left(\Bar{f}\right)^{-1}\left(B\right)\subseteq A\cap f^{-1}\left(B\right)$.* + + *Now, suppose that $x\in A\cap f^{-1}\left(B\right)$, by definition + of $\Bar{f}$ we have that $\Bar{f}\left(x\right)$. Now + $x\in f^{-1}\left(B\right)$ means that $f\left(x\right)\in B$, now + as $\Bar{f}\left(x\right)$ maps any $x\in A$ to $f\left(x\right)$ we + have that $\Bar{f}\left(x\right)=f\left(x\right)$ and so + $x\in \left(\Bar{f}\right)^{-1}\left(B\right)$* + + *Hence + $\left(\Bar{f}\right)^{-1}\left(B\right)=A\cap f^{-1}\left(B\right)$* + +5. *$f^{-1}\left(f\left(f^{-1}\left(B\right)\right)\right)=f^{-1}\left(B\right)$:* + + *This follows by property 2. $f^{-1}\left(f\left(X\right)\right)=X$. + Indeed we have* + + *$$\begin{equation*} + f^{-1}\left(f\left(f^{-1}\left(B\right)\right)\right)=f^{-1}\left(B\right) + \end{equation*}$$* + +6. *$f^{-1}\left(B\right)=\emptyset\iff B\subseteq Y\setminus f\left(X\right)$:* + + *$\left(\Rightarrow\right):$ Suppose + $f^{-1}\left(B\right)=\emptyset$, by definition of the pre-image we + have* + + *$$\begin{equation*} + f^{-1}\left(B\right)=\left\{x\in X:f\left(x\right)\in B\right\}=\emptyset + \end{equation*}$$* + + *Hence the pre-image being empty means that there are no elements + $x\in X$ with $f\left(x\right)\in B$. Now the set + $Y\setminus f\left(X\right)$ is given* + + *$$\begin{equation*} + Y\setminus f\left(X\right)=\left\{y\in Y: y\not\in f\left(X\right)\right\} + \end{equation*}$$* + + *Thus as there are no $x\in X$ with $f\left(x\right)\in B$, then + $Y\setminus f\left(x\right)$ will not remove any + $f\left(x\right)\in B$, that is + $B\subseteq Y\setminus f\left(X\right)$.* + + *$\left(\Leftarrow\right):$ Suppose that + $B\subseteq Y\setminus f\left(X\right)$. We Have that + $Y\setminus f\left(X\right)$ is precisely the set of $y\in Y$ with + $y\not\in f\left(X\right)$, therefore the set + $B\subseteq Y\setminus f\left(X\right)$ means that if + $f\left(b\right)\in B$ then we have have that + $b\not\in f\left(X\right)$ and hence $b\not\in X$. This holds for + any $f\left(b\right)\in B$ and hence we must have that the pre-image + of $B$ is empty. This is to say $f^{-1}\left(B\right)=\emptyset$.* + +7. *$A\subseteq f^{-1}\left(B\right)\iff f\left(A\right)\subseteq B$:* + + *$\left(\Rightarrow\right):$ Suppose that + $A\subseteq f^{-1}\left(B\right)$. Recall the definition of the + image* + + *$$\begin{equation*} + f\left(A\right)=\left\{f\left(x\right):x\in A\right\} + \end{equation*}$$* + + *Now for some $a\in A$ we have that $a\in f^{-1}\left(B\right)$ and + so there is some $x\in X$ such that $f\left(x\right)\in B$, in + particular $a=x$ and so $x\in A$ which gives + $f\left(A\right)\subseteq B$.* + + *$\left(\Leftarrow\right):$ Now, suppose that + $f\left(A\right)\subseteq B$ we have that for some + $y\in f\left(A\right)$ that $y\in B$ and in particular by definition + there is some $x\in A$ such that + $f\left(x\right)=y\in f\left(A\right)$. Hence as $A\subseteq X$ we + have that $x\in X$ and so by definition of the pre-image we have + that $x\in f^{-1}\left(B\right)$. This is to say we conclude that + $A\subseteq f^{-1}\left(B\right)$.* + +8. *$f^{-1}\left(Y\setminus B\right)\subseteq f^{-1}\left(B\right)\iff f^{-1}\left(B\right)=X$:* + + *Suppose that + $f^{-1}\left(Y\setminus B\right)\subseteq f^{-1}\left(B\right)$. We + have that pre-image of $Y\setminus B$ is given by* + + *$$\begin{equation*} + f^{-1}\left(Y\setminus B\right)=\left\{x\in X: f\left(x\right) \in Y\setminus B\right\}=\left\{x\in X: f\left(x\right)\in Y \text{ and } f\left(x\right)\not\in B\right\} + \end{equation*}$$* + + *Hence by definition $y\in f^{-1}\left(Y\setminus B\right)$ gives us + that $y=x$ for some $x\in X$ with $f\left(x\right)\in Y$ and + $f\left(x\right)\not\in B$, but then we can't have + $y\in f^{-1}\left(B\right)$ by the definition of the pre-image on + $B$. Hence we conclude that + $f^{-1}\left(Y\setminus B\right)\subseteq f^{-1}\left(B\right)$ + holds if and only if $Y\setminus B = \emptyset$ from which $B= Y$ + and so by property $1$. we have that $f^{-1}\left(B\right)= X$.* + +9. *$f^{-1}\left(Y\setminus B\right)= X\setminus f^{-1}\left(B\right)$:* + + *Suppose that $x\in f^{-1}\left(Y\setminus B\right)$ then by + definition we have that $f\left(x\right)\in y$ and + $f\left(x\right)\not\in B$ for some $x\in X$, but this is clearly + the definition of $X\setminus f^{-1}\left(B\right)$ and so + $x\in X\setminus f^{-1}\left(B\right)$.* + + *Conversely if $x\in X\setminus f^{-1}\left(B\right)$ then + $f\left(x\right)\not\in B$ but by definition of $f$ we have that + $f\left(x\right)\in Y$ and so + $x\in f^{-1}\left(Y\setminus B\right)$.* + + *It follows that + $f^{-1}\left(Y\setminus B\right)= X\setminus f^{-1}\left(B\right)$.* + +10. *$A\cup f^{-1}\left(B\right)\subseteq f^{-1}\left(f\left(A\right)\cup B\right)$:* + + *Let $x\in A\cup f^{-1}\left(B\right)$. We have that either $x\in A$ + or $x\in f^{-1}\left(B\right)$. If $x\in A$ then + $f\left(x\right)\in f\left(A\right)$ and so + $f\left(x\right)\in f\left(A\right)\cup B$, the result follows on + taking the pre-image as* + + *$$\begin{equation*} + f^{-1}\left(f\left(A\right)\cup B\right)=\left\{x\in X: f\left(x\right)\in f\left(A\right)\cup B\right\} + \end{equation*}$$ This is to say that + $x\in f^{-1}\left(f\left(A\right)\cup B\right)=\left\{x\in X: f\left(x\right)\in f\left(A\right)\cup B\right\}$.* + + *Now if $x\in f^{-1}\left(B\right)$ then we have by definition that + $f\left(x\right)\in B$ and by a similar argument to above we + conclude that $f\left(x\right)\in f\left(A\right)\cup B$ so that on + taking the pre-image we conclude that + $x\in f^{-1}\left(f\left(A\right)\cup B\right)=\left\{x\in X: f\left(x\right)\in f\left(A\right)\cup B\right\}$.* + + *Hence it follows that + $A\cup f^{-1}\left(B\right)\subseteq f^{-1}\left(f\left(A\right)\cup B\right)$.* + +11. *$A\cap f^{-1}\left(B\right)\subseteq f^{-1}\left(f\left(A\right)\cap B\right)$:* + + *Suppose that $x\in A\cap f^{-1}\left(B\right)$ then $x\in A$ and + $x\in f^{-1}\left(B\right)$ and so $f\left(x\right)\in B$. As + $x\in A$ then $f\left(x\right)\in f\left(A\right)$ and hence as + $f\left(x\right)\in f\left(A\right)$ and $f\left(x\right)\in B$ then + $f\left(x\right)\in f\left(A\right)\cap B$. The result follows on + taking the pre-image.* + + *Hence + $A\cap f^{-1}\left(B\right)\subseteq f^{-1}\left(f\left(A\right)\cap B\right)$* + +*The proposition now follows. $\qed$* +::: + +##### Injective, surjective and bijective mappings + +Armed with the examples we have seen we can make a few comments about +mappings. Consider example [28](#exmp:Mapping 1){reference-type="ref" +reference="exmp:Mapping 1"} where we have that $X=Y=\mathbb{N}$ and is +$f$ the map + +$$\begin{align*} + f:X&\mathlarger{\mathlarger{\rightarrow}}Y\\ + x&\mapsto f\left(x\right)=2x +\end{align*}$$ + +We have that for every $x,y\in X$ with $f\left(x\right)=f\left(y\right)$ +that $x=y$, which is to say if the image of two different elements +agree, then the elements are in-fact the same. This is clear to see, +suppose that $x,y\in X$ with $f\left(x\right)=f\left(y\right)$, then we +have that + +$$\begin{align*} + f\left(x\right)&=f\left(y\right)\\ + 2x&=2y\\ + x&=y +\end{align*}$$ + +Another way of expressing this idea is that two distinct elements in the +domain will have distinct images, we say a mapping with this property is +an injective mapping. Now, if we consider +$\mathop{\mathrm{Image}}\left(f\right)\subseteq Y$ and consider the map + +$$\begin{align*} + g:X&\mathlarger{\mathlarger{\rightarrow}}\mathop{\mathrm{Image}}\left(f\right)\\ + x&\mapsto g\left(x\right)=2x +\end{align*}$$ Then, for every +$y\in\mathop{\mathrm{Image}}\left(f\right)$, we have that there exists +some element $x\in X$ such that $y=g\left(x\right)$. Again, we can show +this. Let $y\in\mathop{\mathrm{Image}}\left(f\right)$, then we need to +show that $\exists x\in X$ such that $g\left(x\right)=y$. Now + +$$\begin{align*} + y&=g\left(x\right)\\ + y&=2x\\ + \frac{y}{2}&=x +\end{align*}$$ + +We hence will need to take $\displaystyle x=\frac{y}{2}$, however we +first then to verify that $\displaystyle x=\frac{y}{2}\in X$. We note +that $y\in\mathop{\mathrm{Image}}\left(f\right)$ means that $y=2k$ for +some $k\in\mathbb{N}$, so + +$$\begin{align*} + x&=\frac{y}{2}\\ + x&=\frac{2k}{2}\\ + x&=k +\end{align*}$$ + +as $x\in X=\mathbb{N}$ and $k\in\mathbb{N}$ then we can rest safe in the +knowledge that our choice for $x$ indeed works. As a sanity check we +have that + +$$\begin{equation*} + g\left(x\right)=2x=2\frac{y}{2}=y +\end{equation*}$$ + +This choice of $x$ works for any choice of $y$. Another way to express +this idea is that every element in the image of the mapping is the image +of some element in the domain, we say a mapping with this property is a +surjective mapping. + +It is worth noting that the mapping $g$ is both injective and +surjective, this makes $g$ a special type of mapping. If we take an +element in the domain $x$ and consider its image +$g\left(x\right)\in\mathop{\mathrm{Image}}\left(f\right)$, then as $g$ +is injective we know that $g\left(x\right)$ is a distinct element in +$\mathop{\mathrm{Image}}\left(f\right)$. Moreover, as $g$ is surjective +then there is an element in the domain, say $a$ with the property that +$g\left(a\right)=g\left(x\right)$, but as $g$ is injective then we know +that $a=x$. This means that we can go between elements of the domain and +elements of the image in a distinct way, a mapping with this property is +called a bijective mapping and the domain and image are said to be in +bijection with each other. + +We formalise these ideas now to a mapping between any two sets. + +::: definition +**Definition 48**. *Injective, surjective and bijective maps* + +*Let $f:X\mathlarger{\mathlarger{\rightarrow}}Y$ be a mapping between +two sets $X$ and $Y$.* + +1. *We say that $f$ is an injective mapping, sometimes called a + one-to-one mapping, if* + + *$$\begin{equation*} + \forall x,y\in X,\ f\left(x\right)=f\left(y\right) \Rightarrow x=y + \end{equation*}$$* + + *That is we have that $f\left(x\right)=f\left(y\right)$ for + $x,y\in X$ then $x=y$. If we know that $f$ is injective we can write + the mapping as* + + *$$\begin{equation*} + f:X\mathlarger{\mathlarger{\hookrightarrow}}Y + \end{equation*}$$* + + *which is read as $f$ is an injective mapping from $X$ to $Y$.* + +2. *We say that $f$ is a surjective mapping, sometimes called a onto + mapping, if* + + *$$\begin{equation*} + \forall y\in Y,\exists x\in X: y=f\left(x\right) + \end{equation*}$$* + + *That is we have that for each $y\in Y$, there exists some $x\in X$ + such that $f\left(x\right)=y$. If we know that $f$ is a surjective + then we can write the mapping as* + + *$$\begin{equation*} + f:X\mathlarger{\mathlarger{\twoheadrightarrow}}Y + \end{equation*}$$* + + *which is read as $f$ is a surjective mapping from $X$ to $Y$* + +3. *We say that $f$ is a bijective mapping, sometimes called a + one-to-one and unto mapping, if $f$ is both injective and + surjective. If we know that $f$ is a bijection then we can write the + mapping as* + + *$$\begin{equation*} + f:X% + \mathlarger{\mathlarger{\hookrightarrow}}\mathrel{\mspace{-27.5mu}}\mathlarger{\mathlarger{\rightarrow}} + Y + \end{equation*}$$* + + *which is read as $f$ is a bijective mapping from $X$ to $Y$.* +::: + +We will look for additional examples of each type of mapping. + +::: example +**Example 35**. *Let +$f:\mathbb{N}\mathlarger{\mathlarger{\rightarrow}}\mathbb{N}$ where +$f\left(x\right)=x$. We will prove that $f$ is a bijective mapping.* + +*Proof:* + +*To show $f$ is bijective we show that $f$ is injective and surjective. +To see that $f$ is an injection, suppose that +$f\left(x\right)=f\left(y\right)$ where $x,y\in N$, the domain. then we +have that* + +*$$\begin{align*} + f\left(x\right)&=f\left(y\right)\\ + x&=y +\end{align*}$$ This shows $f$ is injective as this holds for any choice +of $x,y\in \mathbb{N}$. To see that $f$ is surjective consider +$y\in\mathbb{N}$, the co-domain, we show there exists an +$x\in\mathbb{N}$, the domain, so that $f\left(x\right)=y$. We have* + +*$$\begin{align*} + y&=f\left(x\right)\\ + y&=x +\end{align*}$$ so we take $x=y$. This works for every $y\in\mathbb{N}$, +the co-domain, so $f$ is surjective.* + +*As $f$ is both injective and surjective it is by definition a bijective +map, that is $f:\mathbb{N}% + \mathlarger{\mathlarger{\hookrightarrow}}\mathrel{\mspace{-27.5mu}}\mathlarger{\mathlarger{\rightarrow}} +\mathbb{N}$. $\qed$* +::: + +::: example +**Example 36**. *Let +$f:\mathbb{N}\mathlarger{\mathlarger{\rightarrow}}\mathbb{N}$ where* + +*$$\begin{equation*} + f\left(x\right)=\begin{cases} + x,\ \text{If } x \text{ is odd}\\ + \frac{x}{2},\ \text{If } x \text{ is even}\\ + \end{cases} +\end{equation*}$$* + +*Is $f$ injective? To see if it is we would need to show that +$f\left(x\right)=f\left(y\right)$ with $x,y\in \mathbb{N}$ means that +$x=y$. It becomes clear that there are $x,y\in\mathbb{N}$ where this +does not hold, for example $f\left(1\right)=1$ and $f\left(2\right)=1$ +so $f\left(1\right)=f\left(2\right)$ but $1\neq 2$. This shows that $f$ +is not injective. Is $f$ surjective? To see if it is we would need to +show that $\forall y\in\mathbb{N},\exists x\in\mathbb{N}$ such that +$y=f\left(x\right)$. Note that for every even input $x=2k$ we have that +$\displaystyle f\left(x\right)=\frac{2k}{2}=k$. So for any +$y\in\mathbb{N}$ if we take $x=2y$ then every $y\in\mathbb{N}$ gets +mapped to to by $2y$. So $f$ is surjective.* + +*As $f$ was not injective we have that $f$ is not a bijection, so we +have +$f:\mathbb{N}\mathlarger{\mathlarger{\twoheadrightarrow}}\mathbb{N}$.* +::: + +::: example +**Example 37**. *Let $X=\left\{1,2\right\}$ and $Y=\left\{3,4,5\right\}$ +and define the map $f:X\mathlarger{\mathlarger{\rightarrow}}Y$ by* + +*$$\begin{equation*} + f\left(1\right)=3,\ f\left(2\right)=4 +\end{equation*}$$* + +*Then it is clear that $f$ is injective, as each input is mapped to a +distinct output. More formally suppose that +$f\left(x\right)=f\left(y\right)$ where $x,y\in X$. We have that by the +definition of the mapping $f\left(1\right)=3,\ f\left(2\right)=4$. In +the first case we have $f\left(x\right)=f\left(y\right)=3$ and so +$x=y=1$, likewise in the second case we have that +$f\left(x\right)=f\left(y\right)=4$ and so $x=y=2$. This proves +injectivity.* + +*To see that $f$ is not surjective, consider the image +$\mathop{\mathrm{Image}}\left(f\right)=\left\{f\left(x\right):x\in X\right\}=\left\{3,4\right\}\neq Y$. +So $\exists y\in Y$ such that $\not\exists x\in X$ with +$y=f\left(x\right)$.* + +*It hence follows that $f$ is not bijective, that is +$f:\left\{1,2\right\}\mathlarger{\mathlarger{\hookrightarrow}}\left\{3,4,5\right\}$.* +::: + +::: example +**Example 38**. *Let $X=\left\{1,2,3\right\}$ and $Y=\left\{4,5\right\}$ +and define the map $f:X\mathlarger{\mathlarger{\rightarrow}}Y$ by* + +*$$\begin{equation*} + f\left(1\right)=4,\ f\left(2\right)=4,\ f\left(3\right)=5 +\end{equation*}$$* + +*We have that $f$ is not injective as +$f\left(1\right)=f\left(2\right)=4$ but $1\neq 2$. However we have that +$f$ is surjective as the image of $f$ is +$\mathop{\mathrm{Image}}\left(f\right)=\left\{f\left(x\right):x\in X\right\}=\left\{4,5\right\}=Y$.* + +*By definition $f$ is not bijective, hence +$f:\left\{1,2,3\right\}\mathlarger{\mathlarger{\twoheadrightarrow}}\left\{4,5\right\}$.* +::: + +We note that we can always construct a mapping $g$ from +$f:X\rightarrow Y$ such that +$g:X\mathlarger{\mathlarger{\rightarrow}}\mathop{\mathrm{Image}}\left(f\right)$ +is a surjection. + +::: {#prob:RestOfCodomainToImageIsSurjective .proposition} +**Proposition 15**. *The restriction of a mappings co-domain to its +image is a surjective mapping* + +*Let $f:X\mathlarger{\mathlarger{\rightarrow}}Y$ be a mapping and +consider +$\mathop{\mathrm{Image}}\left(f\right)=\left\{f\left(x\right):x\in X\right\}$. +Consider the following mapping* + +*$$\begin{align*} + g:X&\mathlarger{\mathlarger{\rightarrow}}\mathop{\mathrm{Image}}\left(f\right)\\ + x&\mapsto f\left(x\right) +\end{align*}$$* + +*Then $g$ is a surjective map.* + +*Proof:* + +*Let $f:X\mathlarger{\mathlarger{\rightarrow}}Y$ and consider +$\mathop{\mathrm{Image}}\left(f\right)=\left\{f\left(x\right):x\in X\right\}$. +By the definition of the image of a mapping +[45](#def:ImageMapping){reference-type="ref" +reference="def:ImageMapping"} we have that +$\mathop{\mathrm{Image}}\left(f\right)\subseteq Y$. Moreover, by the +definition of the image of a map we have that +$y\in\mathop{\mathrm{Image}}\left(f\right)$ if and only if +$\exists x\in X$ such that $y=f\left(x\right)$. This will hold for all +$y\in\mathop{\mathrm{Image}}\left(f\right)$ so $g$ is a surjection. +$\qed$.* +::: + +In the proof we used the idea of restricting the co-domain of the +function so that it was the image +$\mathop{\mathrm{Image}}\left(f\right)$ rather than $Y$, while leaving +the domain $X$ unchanged. In actuality we didn't restrict the co-domain +at all but instead only considered those elements of the co-domain that +actually get mapped to. It should be clear that the image +$\mathop{\mathrm{Image}}\left(f\right)$, the elements that actually get +mapped to, only depends on the allowable inputs for the function, that +is only depend on the domain $X$. In many fields of mathematics it is +sometimes desirable to restrict the domain $X$ that is being worked with +to a smaller subset of the domain $A\subseteq X$. As a quick example of +why this is useful, and which we will see later, is for inverse +mappings. For now the key idea of an inverse map is to be able to create +a bijection between a mapping and its domain and co-domain to enable us +to unambiguously go between the two. Why is this useful? + +For an example, suppose that you wanted to go on holiday abroad then +you'll need to convert your currency to the currency that is in use +where you go to. Suppose that you use gold coins where as the contry you +vist only uses silver coins. The exchange rate from gold coins to silver +coins is given by the following mapping $E\left(x\right) = Ax^2$ where +the domain is the set of all the numbers that we are familiar with, that +is $\mathbb{R}$, and $A$ is some positive number which is greater than +0. + +Suppose we wish to convert $50$ gold coins into the new currency, then +we will have $E\left(50\right)=A*50^2=2500A$ silver coins. Finally +suppose that after our holiday we have some silver coins left over that +we wish to convert back to gold coins, say $2500A-y$ where $00$ for every $x\in\mathbb{N}$. Observe also +that $f$ is an injective map, indeed let $x,y\in\mathbb{N}$ and suppose +$f\left(x\right)=f\left(y\right)$ then* + +*$$\begin{align*} + f\left(x\right)&=f\left(y\right)\\ + x+1&=y+1\\ + x&=y +\end{align*}$$* + +*It is also worth noting that $g$ is not injective as we have +$g\left(1\right)=0=g\left(0\right)$ but $1\neq 0$. We note that $f$ is +the right inverse of $g$ as the calculation above shows.* +::: + +::: example +**Example 45**. *Let $X=\mathbb{R}$ and define +$Y=\mathbb{R}^+=\left\{x\in\mathbb{R}:x\geq 0\right\}$, the set of +familiar numbers. Let +$f:\mathbb{R}\mathlarger{\mathlarger{\rightarrow}}\mathbb{R}^+$ be +define by $f\left(x\right)=x^2$. We can define two possible right +inverses of $f$. The first is given by +$g_1:\mathbb{R}^+\mathlarger{\mathlarger{\rightarrow}}\mathbb{R}$ where +$g_1\left(x\right)=\sqrt{x}$. Indeed* + +*$$\begin{equation*} + f\circ g_1\left(x\right)=f\left(g_1\left(x\right)\right)=f\left(\sqrt{x}\right)=\left(\sqrt{x}\right)^2=x=\mathop{\mathrm{id}}_{\mathbb{R}}\left(x\right) +\end{equation*}$$ The second, as you may have guessed, is given by +$g_2:\mathbb{R}^+\mathlarger{\mathlarger{\rightarrow}}\mathbb{R}$ where +$g_1\left(x\right)=-\sqrt{x}$ where likewise we have* + +*$$\begin{equation*} + f\circ g_2\left(x\right)=f\left(g_2\left(x\right)\right)=f\left(-\sqrt{x}\right)=\left(-\sqrt{x}\right)^2=x=\mathop{\mathrm{id}}_{\mathbb{R}}\left(x\right) +\end{equation*}$$* + +*We note that $f$ is surjective. Let $y\in\mathbb{R}^+$ then +$f\left(x\right)=y\Rightarrow x^2=y\Rightarrow x=\pm\sqrt{y}\in\mathbb{R}$, +hence every output of $f$ is mapped to by some input. It is clear that +$f$ is not injective as $f\left(2\right)=4=f\left(-2\right)$.* + +*Does $f$ have a left inverse?. By the definition of a left inverse we +will need to find some +$g:\mathbb{R}^+\mathlarger{\mathlarger{\rightarrow}}\mathbb{R}$ such +that $g\circ f=id_{\mathbb{R}}$. So for each input of $f$, $g$ will have +to send $f\left(x\right)$ back to $x$, hence we might require that $f$ +be injective, for if not then $\exists x,y\in\mathbb{R}$ such that +$f\left(x\right)=f\left(y\right)$ with $x\neq y$ and we have the problem +where $g$ could send $f\left(x\right)$ back to either $x$ or $y$, and if +it is sent back to $y$ then we don't have the identity mapping!* + +*Now, $f$ is not injective as we have seen that +$f\left(2\right)=4=f\left(-2\right)$, so if there where a left inverse +$g$ it wouldn't know where to send $4$ back to, it could have been +either $2$ or $-2$.* +::: + +::: example +**Example 46**. *Let $X=\mathbb{R}$ and +$Y=\mathbb{R}\setminus\left\{0\right\}=\left\{x\in\mathbb{R}:x\neq 0\right\}$. +You may have seen the function $e^x$ before, we shall consider this +mapping, that is the mapping +$f:\mathbb{R}\mathlarger{\mathlarger{\rightarrow}}\mathbb{R}\setminus\left\{0\right\}$ +given by $f\left(x\right)=e^x=\exp\left(x\right)$. We can define a left +inverse to $f$ by +$g:\mathbb{R}\setminus\left\{0\right\}\mathlarger{\mathlarger{\rightarrow}}\mathbb{R}$ +where $g\left(x\right)=\ln\left(x\right)$, where $\ln\left(x\right)$ is +the natural logarithm, the logarithm to the base $e$. We will discuss +logarithms in more detail later but for now we can think of +$\ln\left(x\right)=y$ as asking the question $e^y=x$, that is value of +$y$ do we need to raise $e$ to to get $x$. This $g$ is indeed a left +inverse of $f$ as* + +*$$\begin{equation*} + g\circ f\left(x\right)=g\left(f\left(x\right)\right)=g\left(e^x\right)=\ln\left(e^x\right)=x=\mathop{\mathrm{id}}_{\mathbb{R}} +\end{equation*}$$* + +*Like in the previous example, we can ask the question does $f$ have a +right inverse? By definition for $f$ to have a right inverse, there +needs to be a mapping +$g:\mathbb{R}\setminus\left\{0\right\}\mathlarger{\mathlarger{\rightarrow}}\mathbb{R}$ +such that +$f\circ g=\mathop{\mathrm{id}}_\mathbb{R}\setminus\left\{0\right\}$. So +for each $g\left(y\right)$ with +$y\in\mathbb{R}\setminus\left\{0\right\}$ we have that $f$ will send +$g\left(y\right)$ back to $y$. This will happen if every output of $f$ +has some input that generates it, that is $f$ is a surjection. If this +not the case then there is some element +$y\in\mathbb{R}\setminus\left\{0\right\}$ that is not mapped to by +$f\left(x\right)$ for some $x\in\mathbb{R}$.* + +*For example we have that $\not\exists x\in\mathbb{R}$ such that +$e^x=-1$ for example. So $f$ is not surjective in this case we are not +able to define a right inverse that makes sense.* +::: + +We can generalise the last two examples to the next two propositions. + +::: {#prop:LeftInverseIffInjective .proposition} +**Proposition 27**. *Condition for the existence of a left inverse* + +*Let $f:X\mathlarger{\mathlarger{\rightarrow}}Y$ be a mapping with +$X\neq\emptyset$. We have that $f$ has a left inverse +$g:Y\mathlarger{\mathlarger{\rightarrow}}X$ such that +$g\circ f=\mathop{\mathrm{id}}_X$ if and only if $f$ is an injective +mapping.* + +*Proof:* + +*$\left(\Rightarrow\right)$: Suppose that $f$ has a left inverse +$g:Y\mathlarger{\mathlarger{\rightarrow}}X$ such that +$g\circ f=\mathop{\mathrm{id}}_X$. We know by proposition +[26](#prop:IdentityMapProperties){reference-type="ref" +reference="prop:IdentityMapProperties"} that $\mathop{\mathrm{id}}_X$ is +an injective mapping, moreover we know by proposition +[21](#prop:CompositeMapInectSurjectProp){reference-type="ref" +reference="prop:CompositeMapInectSurjectProp"} that if a composite map +$g\circ f$ is injective then so is $f$. Hence as +$g\circ f = \mathop{\mathrm{id}}_X$ and $\mathop{\mathrm{id}}_X$ is +injective, we conclude that $f$ is an injective map.* + +*$\left(\Leftarrow\right)$: Suppose that $f$ is an injective map, then +$\forall x,y\in X$ we have that +$f\left(x\right)=f\left(y\right)\Rightarrow x=y$. Let $x\in X$, we need +to construct a map which acts as a left inverse to $f$.* + +*Consider the following map +$\mathrel{h\restriction_{\mathop{\mathrm{Image}}\left(f\right)}}:\mathop{\mathrm{Image}}\left(f\right)\mathlarger{\mathlarger{\rightarrow}}X$, +where we send $y\in\mathop{\mathrm{Image}}\left(f\right)$ back to the +element that it was mapped from. Now, define $g$ as follows* + +*$$\begin{align*} + g:Y&\mathlarger{\mathlarger{\rightarrow}}X\\ + y&\mapsto g\left(y\right)=\begin{cases} + x,\ \text{If } y\in Y\setminus\mathop{\mathrm{Image}}\left(f\right)\\ + h\left(y\right),\ \text{If } y\in\mathop{\mathrm{Image}}\left(f\right) + \end{cases} +\end{align*}$$* + +*We note that if $\mathop{\mathrm{Image}}\left(f\right) = Y$ then we do +not need to consider the first case +$x,\ \text{If } y\in Y\setminus\mathop{\mathrm{Image}}\left(f\right)$, +however if $\mathop{\mathrm{Image}}\left(f\right) \subset Y$ then there +exists at least one $x$ for this case.* + +*Now with this $g$ we have that* + +*$$\begin{equation*} + g\circ f\left(x\right)=g\left(f\left(x\right)\right)=h\left(f\left(x\right)\right)=x=\mathop{\mathrm{id}}_X +\end{equation*}$$* + +*Hence $g$ is indeed a left inverse of $f$.* + +*The proposition now follows. $\qed$* +::: + +::: {#prop:RightInverseIffSurjective .proposition} +**Proposition 28**. *Condition for the existence of a right inverse* + +*Let $f:X\mathlarger{\mathlarger{\rightarrow}}Y$ be a mapping with +$X\neq\emptyset$. We have that $f$ has a right inverse +$g:Y\mathlarger{\mathlarger{\rightarrow}}X$ such that +$f\circ g=\mathop{\mathrm{id}}_Y$ if and only if $f$ is a surjective +mapping.* + +*Proof:* + +*$\left(\Rightarrow\right)$: Suppose that $f$ has a right inverse +$g:Y\mathlarger{\mathlarger{\rightarrow}}X$ such that +$f\circ g=\mathop{\mathrm{id}}_Y$. We know by proposition +[26](#prop:IdentityMapProperties){reference-type="ref" +reference="prop:IdentityMapProperties"} that $\mathop{\mathrm{id}}_X$ is +a surjective mapping, moreover we know by proposition +[21](#prop:CompositeMapInectSurjectProp){reference-type="ref" +reference="prop:CompositeMapInectSurjectProp"} that if a composite map +$f\circ g$ is surjective then so is $f$. Hence as +$f\circ g = \mathop{\mathrm{id}}_Y$ and $\mathop{\mathrm{id}}_Y$ is +surjective, we conclude that $f$ is a surjective map.* + +*$\left(\Leftarrow\right)$: Suppose that $f$ is a surjective map, then +$\forall y\in Y,\exists x\in X: f\left(x\right)=y$. We need to construct +a $g:Y\mathlarger{\mathlarger{\rightarrow}}X$ such that +$f\circ g=\mathop{\mathrm{id}}_Y$. As $f$ is surjective we have that +$\forall y\in Y,\exists x\in X: f\left(x\right)=y$, in particular we +know that there maybe more than one such $x$ so that +$f\left(x\right)=y$, if this is the case we pick for that $y$ one of the +possible choices of $x$. Hence we can define $g\left(y\right)=x$ for +every $y\in Y$ then we have that +$f\circ g\left(y\right)=f\left(g\left(y\right)\right)=f\left(x\right)=y=\mathop{\mathrm{id}}_Y$* + +*The proposition now follows. $\qed$* +::: + +These two propositions give the following immediate results + +::: {#LeftInverseOfInjectionIsSurjective .proposition} +**Proposition 29**. *Left inverse of injective mapping is a surjection* + +*Let $f:X\rightarrow Y$ be an injection with left inverse +$g:Y\rightarrow X$. We have that $g$ is a surjection.* + +*Proof let $f$ and $g$ be as stated. Then by definition of a left +inverse we have that $g\circ f =\mathop{\mathrm{id}}_X$. Moreover we +have the identity mapping $\mathop{\mathrm{id}}_X$ is an injection as it +is bijective. We then have by proposition +[21](#prop:CompositeMapInectSurjectProp){reference-type="ref" +reference="prop:CompositeMapInectSurjectProp"} that $g$ is a surjection. +$\qed$* +::: + +::: {#RightInverseOfSurjecctionisInection .proposition} +**Proposition 30**. *Right inverse of surjective mapping is an +injection* + +*Let $f:X\rightarrow Y$ be a surjection with right inverse +$g:Y\rightarrow X$. We have that $g$ is an injection.* + +*Proof let $f$ and $g$ be as stated. Then by definition of a right +inverse we have that $f\circ g =\mathop{\mathrm{id}}_Y$. Moreover we +have the identity mapping $\mathop{\mathrm{id}}_Y$ is a surjection as it +is bijective. We then have by proposition +[21](#prop:CompositeMapInectSurjectProp){reference-type="ref" +reference="prop:CompositeMapInectSurjectProp"} that $g$ is an injection. +$\qed$* +::: + +The ideas of a left and right inverse will allow us to construct the +idea of a so-called two-sided inverse, that is an inverse which is both +a left inverse and a right inverse. this will allow us to consider when +a mappings can be inverted without regards to how we compose the +mappings. However there is one final result about left and right inverse +that will be required in order to pave the way. + +::: {#prop:BijectionHasLeftRightInverse .proposition} +**Proposition 31**. *Bijection has a left and right inverse* + +*Let $f:X\rightarrow Y$ be a bijective mapping. We have that there +exists a left inverse $g:Y\rightarrow X$ and there exists a right +inverse $h:Y\rightarrow X$ such that* + +*$$\begin{align*} + g\circ f &= \mathop{\mathrm{id}}_X\\ + f\circ h&=\mathop{\mathrm{id}}_Y +\end{align*}$$* + +*Proof:* + +*Let $f:X\rightarrow Y$ be a bijection. We have that as $f$ is a +bijection then we know that $f$ is both injective and surjective. Now by +proposition [27](#prop:LeftInverseIffInjective){reference-type="ref" +reference="prop:LeftInverseIffInjective"} that a left inverse exists if +and only if $f$ is an injective mapping. Likewise by proposition +[28](#prop:RightInverseIffSurjective){reference-type="ref" +reference="prop:RightInverseIffSurjective"} we have that a right inverse +exists if and only if $f$ is a surjective mapping. Hence we have the +existence of a left and right inverse. As required. $\qed$* +::: + +::: {#prop:LeftRightInverseImpliesBijection .proposition} +**Proposition 32**. *The existence of a left and right inverse implies a +bijection* + +*Let $f:X\mathlarger{\mathlarger{\rightarrow}}Y$ be a mapping such that +$\exists g_1:Y\mathlarger{\mathlarger{\rightarrow}}X$ such that +$g_1\circ f=\mathop{\mathrm{id}}_X$ and +$\exists g_2:Y\mathlarger{\mathlarger{\rightarrow}}X$ such that +$f\circ g_2=\mathop{\mathrm{id}}_Y$. We have that $f$ is a bijection.* + +*Proof:* + +*Let $f:X\mathlarger{\mathlarger{\rightarrow}}Y$ be a mapping such that +$\exists g_1:Y\mathlarger{\mathlarger{\rightarrow}}X$ such that +$g_1\circ f=\mathop{\mathrm{id}}_X$ and +$\exists g_2:Y\mathlarger{\mathlarger{\rightarrow}}X$ such that +$f\circ g_2=\mathop{\mathrm{id}}_Y$. We have by proposition +[27](#prop:LeftInverseIffInjective){reference-type="ref" +reference="prop:LeftInverseIffInjective"} that as $g_1$ is a left +inverse of $f$ then $f$ must be injective. Likewise by proposition +[28](#prop:RightInverseIffSurjective){reference-type="ref" +reference="prop:RightInverseIffSurjective"} that as $g_2$ is a right +inverse of $f$ then $f$ must be surjective. It hence follows by +definition that $f$ is a bijective mapping. $\qed$* +::: + +These propositions are useful in proving the following. + +::: proposition +**Proposition 33**. *Bijection if and only if left and right inverses +exist* + +*Let $f:X\mathlarger{\mathlarger{\rightarrow}}Y$ be a mapping. We have +that $f$ is bijective if and only if +$\exists g_1:Y\mathlarger{\mathlarger{\rightarrow}}X$ such that +$g_1\circ f=\mathop{\mathrm{id}}_X$ and +$\exists g_2:Y\mathlarger{\mathlarger{\rightarrow}}X$ such that +$f\circ g_2=\mathop{\mathrm{id}}_Y$.* + +*Proof:* + +*$\left(\Rightarrow\right)$: Let $f: X\rightarrow Y$ be a bijective +mapping. We have by proposition +[31](#prop:BijectionHasLeftRightInverse){reference-type="ref" +reference="prop:BijectionHasLeftRightInverse"} we have that $f$ being a +bijection gives the existence of a left and right inverse.* + +*$\left(\Leftarrow\right)$: Suppose we have a mapping $f:X\rightarrow Y$ +such that $\exists g_1:Y\mathlarger{\mathlarger{\rightarrow}}X$ such +that $g_1\circ f=\mathop{\mathrm{id}}_X$ and +$\exists g_2:Y\mathlarger{\mathlarger{\rightarrow}}X$ such that +$f\circ g_2=\mathop{\mathrm{id}}_Y$. Then $f$ has both a left inverse +and a right inverse, hence by proposition +[32](#prop:LeftRightInverseImpliesBijection){reference-type="ref" +reference="prop:LeftRightInverseImpliesBijection"} we have that $f$ is a +bijection.* + +*The result is shown. $\qed$* +::: + +We have seen that if $f:X\rightarrow Y$ is a bijection then $f$ has both +a left and a right inverse, likewise if these two inverses exist then we +have that $f$ is a bijection. This property is key to defining what we +mean by the inverse to a bijective mapping. + +::: definition +**Definition 54**. *Inverse* + +*Let $f:X\mathlarger{\mathlarger{\rightarrow}}Y$ be a mapping. We say +that the mapping $g:Y\mathlarger{\mathlarger{\rightarrow}}X$ is an +inverse[^6] of $f$ if we have that $g$ is both a left inverse and a +right inverse for $f$. This is to say, $g$ is an inverse of $f$ if we +have that* + +*$$\begin{align*} + g\circ f&=\mathop{\mathrm{id}}_X\\ + f\circ g &=\mathop{\mathrm{id}}_Y +\end{align*}$$* + +*We sometimes use the notation $f^{-1}$ to denote the inverse.* +::: + +::: example +**Example 47**. *Let +$f:\mathbb{R}^+\mathlarger{\mathlarger{\rightarrow}}\mathbb{R}^+$ be +such that $f\left(x\right)=x^2$, then we have that +$g:\mathbb{R}^+\mathlarger{\mathlarger{\rightarrow}}\mathbb{R}^+$ with +$g\left(x\right)=\sqrt{x}$ is an inverse of $f$. Indeed* + +*$$\begin{align*} + g\circ f\left(x\right)&=g\left(f\left(x\right)\right)=g\left(x^2\right)=\sqrt{x^2}=x=\mathop{\mathrm{id}}_{\mathbb{R}^+}\\ + f\circ g\left(x\right)&=f\left(g\left(x\right)\right)=f\left(\sqrt{x}\right)=\left(\sqrt{x}\right)^2=x=\mathop{\mathrm{id}}_{\mathbb{R}^+}\\ +\end{align*}$$* +::: + +::: example +**Example 48**. *The identity mapping +$\mathop{\mathrm{id}}_X:X\mathlarger{\mathlarger{\rightarrow}}X$ with +$\mathop{\mathrm{id}}_X\left(x\right)=x,\ \forall x\in X$ is its own +inverse, indeed* + +*$$\begin{equation*} + \mathop{\mathrm{id}}_X\circ\mathop{\mathrm{id}}_X=\mathop{\mathrm{id}}_X\left(\mathop{\mathrm{id}}_X\left(x\right)\right)=\mathop{\mathrm{id}}_X\left(x\right)=x=\mathop{\mathrm{id}}_X +\end{equation*}$$* +::: + +::: example +**Example 49**. *Let +$f:\left\{1,2\right\}\mathlarger{\mathlarger{\rightarrow}}\left\{a,b\right\}$ +be such that $f\left(1\right)=a$ and $f\left(2\right)=b$. We have that +$g:\left\{a,b\right\}\mathlarger{\mathlarger{\rightarrow}}\left\{1,2\right\}$ +with $g\left(a\right)=1$ and $g\left(b\right)=2$ is an inverse to $f$. +Indeed we have that* + +*$$\begin{align*} + f\left(g\left(a\right)\right)&=f\left(1\right)=a\\ + f\left(g\left(b\right)\right)&=f\left(2\right)=b\\ +\end{align*}$$* + +*It also follows that $g$ is an inverse to $f$, indeed* + +*$$\begin{align*} + g\left(f\left(1\right)\right)&=g\left(a\right)=1\\ + g\left(f\left(1\right)\right)&=g\left(b\right)=2\\ +\end{align*}$$* +::: + +::: example +**Example 50**. *Let +$f:\mathbb{R}\mathlarger{\mathlarger{\rightarrow}}\mathbb{R}^+$ be given +by $f\left(x\right)=e^x$. We have that +$g:\mathbb{R}^+\mathlarger{\mathlarger{\rightarrow}}\mathbb{R}$ where +$g\left(x\right)=\ln\left(x\right)$ is an inverse of $f$.* +::: + +We shall prove that the composition of a mapping and its inverse gives +the identity mapping. Firstly, we will need to show the following +propositions. + +::: {#prop:MappingInjectiveSurjectiveIFFInverseIsMapping .proposition} +**Proposition 34**. *Mapping is injective and surjective if and only if +the inverse is a mapping* + +*Let $f:X\rightarrow Y$ be a mapping. We have that $f$ is a bijection if +and only if $f^{-1}$, the inverse of $f$, is a mapping.* + +*Proof:* + +*$\left(\Rightarrow\right):$ Let $f:X\rightarrow Y$ be a bijection, then +$f$ is both surjective and injective. Let $y\in Y$, then as $f$ is +surjective we have that $\exists x\in X$ such that $f\left(x\right)=y$, +moreover by injectivity of $f$ we have that there is only one such $x$ +which does this. Define $g:Y\rightarrow X$ by* + +*$$\begin{equation*} + g\left(y\right)=x +\end{equation*}$$* + +*As $y\in Y$ is an arbitrary element, it follows that* + +*$$\begin{equation*} + \forall y\in Y:\exists x\in X : g\left(y\right)=x +\end{equation*}$$ such that $x$ is unique for a given $y$. That is $g$ +is a mapping. Now by the definition of $g$ we have that* + +*$$\begin{equation*} + \forall y\in Y: f\left(g\left(y\right)\right)=y +\end{equation*}$$ Now, let $x\in X$ and let* + +*$$\begin{equation*} + x'=g\left(f\left(x\right)\right) +\end{equation*}$$ then* + +*$$\begin{equation*} + f\left(x'\right)=f\left(g\left(f\left(x\right)\right)\right)=f\left(x\right) +\end{equation*}$$ by the above. However, $f$ is an injection so we have +that $x'=x$ and thus $x=g\left(f\left(x\right)\right)$.* + +*It follows that $f$ and $g$ are inverse mappings of each other.* + +*$\left(\Leftarrow\right):$ Suppose that $f:X\rightarrow Y$ is a +mapping, moreover suppose that $f^{-1}:Y\rightarrow X$ is also a mapping +which is the inverse of $f$. We show that $f$ must be a bijection.* + +1. *$f$ is injective:* + + *Let $x,y\in X$ and suppose that $f\left(x\right)=f\left(y\right)$* + + *$$\begin{align*} + f\left(x\right)&=f\left(y\right)\\ + f^{-1}\left(f\left(x\right)\right)&=f^{-1}\left(f\left(y\right)\right)\\ + \Rightarrow x&=y,\ \text{As } f^{-1} \text{ is the inverse of f} + \end{align*}$$ Hence we have that $f$ is injective.* + +2. *$f$ is surjective:* + + *Suppose that $y\in Y$. We then have that* + + *$$\begin{align*} + y&\in Y\\ + \Rightarrow f^{-1}\left(y\right)&\in X,\ \text{As } f^{-1} \text{ is the inverse of f}\\ + \Rightarrow f\left(^{-1}\left(y\right)\right)&=y,\ \text{By definition of an inverse mapping}\\ + \Rightarrow \exists x\in X: f\left(x\right)&= y,\ \text{Where } x=f^{-1}\left(y\right) + \end{align*}$$ Hence we have that $f$ is surjective.* + +*As $f$ is both injective and surjective it is a bijection. $\qed$* +::: + +We can now show that the inverse of a bijective mapping is also a +bijective mapping. + +::: {#prop:InverseBijectionIsBijection .proposition} +**Proposition 35**. *Inverse of a bijective mapping is a bijective +mapping* + +*Let $f:X\rightarrow Y$ be a bijective mapping. We have that +$f^{-1}:Y\rightarrow X$, the inverse of $f$, is also a bijection.* + +*Proof:* + +*Let $f:X\rightarrow Y$ be a bijective mapping. By definition of being a +bijection we have that $f$ is both injective and surjective. By +proposition +[34](#prop:MappingInjectiveSurjectiveIFFInverseIsMapping){reference-type="ref" +reference="prop:MappingInjectiveSurjectiveIFFInverseIsMapping"} we have +that $f^{-1}$ is a mapping. Now it is clear that the inverse of the +inverse is the original mapping that is.* + +*$$\begin{equation*} + \left(f^{-1}\right)^{-1}=f +\end{equation*}$$* + +*Now, $f$ is a bijection and thus is a mapping. But as $f$ is a mapping +we have that by proposition +[34](#prop:MappingInjectiveSurjectiveIFFInverseIsMapping){reference-type="ref" +reference="prop:MappingInjectiveSurjectiveIFFInverseIsMapping"} we have +that $f^{-1}$ is a bijection. As required. $\qed$* +::: + +We can now see that the composition of a bijective mapping with its +inverse must be the identity map. + +::: {#prop:BijectionWithInverseIsIdentity .proposition} +**Proposition 36**. *Composition of bijective mapping with the inverses +is the identity mapping* + +*Let $f:X\rightarrow Y$ be a bijective mapping, and let +$f^{-1}:Y\rightarrow X$ be the inverse mapping of $f$. We have that* + +*$$\begin{align*} + f\circ f^{-1} &=\mathop{\mathrm{id}}_Y\\ + f^{-1}\circ f &= \mathop{\mathrm{id}}_X +\end{align*}$$* + +*Proof:* + +*Let $f:X\rightarrow Y$ be a bijective mapping, with inverse given by +$f^{-1}:Y\rightarrow X$. As $f$ is bijective we have that by proposition +[35](#prop:InverseBijectionIsBijection){reference-type="ref" +reference="prop:InverseBijectionIsBijection"} we have that $f^{-1}$ is a +bijection. Let $x\in X$, then we have that* + +*$$\begin{equation*} + \exists y\in Y: f\left(x\right)=y \Rightarrow f^{-1}\left(y\right)=x +\end{equation*}$$* + +*Hence, we have that* + +*$$\begin{align*} + f^{-1}\circ f\left(x\right)&=f^{-1}\left(f\left(x\right)\right),\ \text{By function composition}\\ + &=f^{-1}\left(y\right),\ \text{By above}\\ + &=x,\ \text{By above}\\ + &=\mathop{\mathrm{id}}_X\left(x\right),\ \text{By the definition of the identity map of } X +\end{align*}$$* + +*We have that the domain of $f^{-1}\circ f$ is clearly $X$, likewise the +co-domain is $X$, which is the same as $\mathop{\mathrm{id}}_X$. +Moreover $\forall x\in X$ we have +$f^{-1}\circ f\left(x\right)=x=\mathop{\mathrm{id}}_X\left(x\right)$. So +the mappings are equal.* + +*Likewise, let $y\in Y$, then we have that* + +*$$\begin{equation*} + \exists x\in X: f^{-1}\left(y\right)=x \Rightarrow f\left(x\right)=y +\end{equation*}$$* + +*Hence, we have that* + +*$$\begin{align*} + f\circ f^{-1}\left(y\right)&=f\left(f^{-1}\left(y\right)\right),\ \text{By function composition}\\ + &=f\left(x\right),\ \text{By above}\\ + &=y,\ \text{By above}\\ + &=\mathop{\mathrm{id}}_Y\left(y\right),\ \text{By the definition of the identity map of } Y +\end{align*}$$* + +*We have that the domain of $f\circ f^{-1}$ is clearly $Y$, likewise +the co-domain is $Y$, which is the same as $\mathop{\mathrm{id}}_Y$. +Moreover $\forall y\in Y$ we have +$f\circ f^{-1}\left(y\right)=y=\mathop{\mathrm{id}}_Y\left(y\right)$. +So the mappings are equal.* + +*In both cases the composition yields the required identity mappings, as +required. $\qed$* +::: + +### The Natural numbers + +::: epigraph +The natural numbers are the work of God. All the rest is the work of +mankind. + +*Leopold Kronecker (Paraphrased)* +::: + +#### Constructing the Natural numbers + +We now have enough tools and core theory to start building up from the +foundations of mathematics. We do this using the ZFC axioms, although +perhaps not with the complete rigour we should be using. We touched on +these briefly in section +[2.1.5](#subsubSec:ZFCAxioms){reference-type="ref" +reference="subsubSec:ZFCAxioms"}. We will state them again. + +1. The axiom of extensionality: + + The axiom of extensionality asserts that two sets are equal if and + only if they contain the same elements. + +2. The axiom of the empty-set: + + The axiom of the empty-set asserts that there exists a set which + contains no elements + +3. The axiom of pairing: + + The axiom of pairing asserts that given any set $A$ and any set $B$, + there is a set $C$ such that, given any set $D$, $D$ is a member of + $C$ if and only if $D$ is equal to $A$ or $D$ is equal to $B$. This + is to say, given two sets, there is a set whose members are exactly + the two given sets. + +4. The axiom of specification: + + The axiom of specification asserts that we can construct a set which + satisfies a given condition, so long as this condition is not + inherently contradictory. + +5. The axiom of unions: + + The axiom of unions asserts that we can perform the union of two + sets $A$ and $B$ + +6. The axiom of powers: + + The axiom of powers asserts that for any set $S$ we can construct a + set $P\left(S\right)$ whose elements are all the possible subsets of + $S$. + +7. The axiom of infinity: + + The axiom of infinity asserts that there is at least one infinite + set $A$, that is at least one set with infinitely many elements. + That is we have a set $A$ such that the $\emptyset\in A$ and if + $x\in A$ then the set $x\cup\left\{x\right\}$ is also in $A$. + +8. The axiom of replacement: + + We will need the next section to fully understand this axiom, + however informally asserts that for some set $S$, and form another + set by replacing the elements of $S$ by other sets according to any + definite rule. + +9. The axiom of foundation: + + The axiom of foundation asserts that for every non-empty set $S$, + there exists an element $x\in S$ such that $x$ and $S$ are disjoint. + This also asserts that no set can contain itself. + +We also recall that we include the symbol $\in$ in the ZFC axioms, which +allows us to talk about element inclusions in sets. In other words, ZFC +defines a set of axioms that allow us to talk about sets and elements of +sets. Next, we have that, formally speaking, ZFC is allowed to make +statements about mappings. Finally, we will ZFC has the power to prove +the results in the previous two sections we made on sets and mappings, +so we will assume these as well. We will use this as the building blocks +for building the natural numbers. How can we do this from the ZFC +axioms? + +As it stands right now ZFC only gives us the existence of the empty set, +and there is at least a set which contains infinitely many elements. We +start with the empty set, a set which contains no elements, we can use +the ZFC axioms to build a new set which contains the empty set. + +Our ultimate goal is to identify each natural number with the number of +elements in some corresponding set. Hence naturally the empty set +containing no elements would be identified with the number $0$, and so +on. The question is given that we only have the empty set, how can we +build a new set? We can use the axiom of powers. This states that we can +take any set $S$ and construct a new set $P\left(S\right)$ whose +elements are the possible subsets of $S$. Applying this to the +empty-set, a set which contains no elements and thus has no subsets +except for itself, must give us +$P\left(\emptyset\right)=\left\{\emptyset\right\}$. This is sufficient +for what we need to do. + +So, we have two sets, $\emptyset$ and $\left\{\emptyset\right\}$. We +shall identify $\emptyset$ with $0$ and $\left\{\emptyset\right\}$ with +$1$. + +::: {#def:Zero .definition} +**Definition 55**. *Zero* + +*We define the number zero to be $\emptyset$. That is, we say Zero is a +set that contains no elements.* +::: + +::: {#def:One .definition} +**Definition 56**. *One* + +*We define the number zero to be $\left\{\emptyset\right\}$. That is, we +say One is the set whose only element is $\emptyset$.* +::: + +How do we define any more numbers? We can use the axiom of unions. This +raises the question why not use the axiom of powers again? If we apply +the axiom of powers to $\left\{\emptyset\right\}$ we get the set + +$$\begin{equation*} + P\left(\left\{\emptyset\right\}\right)=\left\{\emptyset,\left\{\emptyset\right\}\right\} +\end{equation*}$$ If we assume we already know what the natural numbers +are, we could identify this with the number $2$. However, a repeated +application of the axiom of powers would give us + +$$\begin{equation*} + P\left(\left\{\emptyset,\left\{\emptyset\right\}\right\}\right)=\left\{\emptyset,\left\{\emptyset\right\},\left\{\left\{\emptyset\right\}\right\},\left\{\emptyset,\left\{\emptyset\right\}\right\}\right\} +\end{equation*}$$ Which we would identify with the number $4$. Another +application would give us a set that we would identify with the number +$8$. Clearly, we are skipping numbers such as $3,5,7,9$ etc. We can't +get additional numbers that aren't powers of $2$. Instead, we can define +an operation that will allow us to construct each number one at a time. + +This operation uses the axiom of unions, and starts of with the numbers +$0$ and $1$, which we recall are the sets $\emptyset$, and +$\left\{\emptyset\right\}$ respectively. Applying the axiom of unions to +these two sets gives us + +$$\begin{equation*} + \emptyset\cup\left\{\emptyset\right\}=\left\{\emptyset,\left\{\emptyset\right\}\right\} +\end{equation*}$$ This is in agreement with +$P\left(\left\{\emptyset\right\}\right)$, so we can identify this with +the number $2$. Now, the axiom of pairing allows us to create a set that +contains as elements any two sets that have already been created. +Applying this to $\left\{\emptyset,\left\{\emptyset\right\}\right\}$ +with itself allows us to create the set +$\left\{\left\{\emptyset,\left\{\emptyset\right\}\right\}\right\}$. +Hence we can now apply this operation again on the set +$\left\{\emptyset,\left\{\emptyset\right\}\right\}$ to get + +$$\begin{equation*} + \left\{\emptyset,\left\{\emptyset\right\}\right\}\cup\left\{\left\{\emptyset,\left\{\emptyset\right\}\right\}\right\}=\left\{\emptyset,\left\{\emptyset\right\},\left\{\left\{\emptyset,\left\{\emptyset\right\}\right\}\right\}\right\} +\end{equation*}$$ A set of $3$ elements so we identify this with the +number $3$. We can keep doing this to build the Natural numbers. Lets +make some definitions + +::: definition +**Definition 57**. *The successor operation* + +*Let $x$ be a set. We define the successor operation, denoted by $S$ to +be given by* + +*$$\begin{equation} + S\left(x\right)= x\cup\left\{x\right\} +\end{equation}$$* +::: + +We call this the successor function, as it is clear in the context of +the Natural numbers that $S\left(n\right)=n+1$, but we shall prove this +later. + +This definition allows us to essentially make any finite number. This +leads us to our first potential definition for the Natural numbers. We +first need to define the idea of recursion. + +We have the following proposition + +::: {#prop:EqualSuccOp .proposition} +**Proposition 37**. *Equality of successor operation* + +*Let $a,b$ be sets. We have that $S\left(a\right)=S\left(b\right)$ if +and only if $a=b$.* + +*Proof:* + +*$\left(\Rightarrow\right):$ Suppose that $a,b$ are sets and +$S\left(a\right)=S\left(b\right)$. By definition of $S$ we have that* + +*$$\begin{equation*} + a\cup\left\{a\right\}=b\cup\left\{b\right\} +\end{equation*}$$* + +*Now, as $a\in S\left(a\right)$ and $S\left(a\right)=S\left(b\right)$ +then we have that $a\in b\cup\left\{b\right\}$ and so $a\in b$ or $a=b$. +Similarly, as $b\in S\left(b\right)$ we get that +$b\in a\cup\left\{a\right\}$ and so $b\in a$ or $b=a$.* + +*Now, if $a=b$ we are done, so suppose $a\neq b$, then we have that +$a\in b$ and $b\in a$. Consider the set given by* + +*$$\begin{equation*} + X=\left\{a,b\right\} +\end{equation*}$$* + +*which can be constructed by the Axiom of pairing. Now as $a\in b$ we +have that $b\cap \left\{a,b\right\}\neq\emptyset$ and likewise as +$b\in a$ we have $a\cap \left\{a,b\right\}\neq\emptyset$. This +contradicts the Axiom of Foundation, $X$ does not contain an element +that is disjoint from it. It follows that we can't have $a\neq b$ and +conclude that $a=b$.* + +*$\left(\Leftarrow\right):$ This is trivial by the definition of $S$. +$\qed$* +::: + +There are a few extra properties about the successor function that we +shall make use of + +::: corollary +**Corollary 1**. *Successor mapping is injective* + +*Let $a,b$ be sets. We have that the successor function is injective, +that is for all sets $a,b$ we have that* + +*$$\begin{equation*} + S\left(a\right)=S\left(b\right) \Rightarrow a=b +\end{equation*}$$* + +*Proof:* + +*Suppose that $a,b$ are arbitrary sets and that +$S\left(a\right)=S\left(b\right)$, by proposition +[37](#prop:EqualSuccOp){reference-type="ref" +reference="prop:EqualSuccOp"} this holds if and only if $a=b$. Hence we +have injectivity. $\qed$* +::: + +::: corollary +**Corollary 2**. *Empty-set is not the successor of any set* + +*We have that $\emptyset\neq S\left(a\right)$ for all sets $a$.* + +*Proof:* + +*Consider the definition of $S\left(a\right)$ and suppose for +contradiction that $\emptyset= S\left(a\right)$. We have by definition +of the successor mapping that* + +*$$\begin{equation*} + \emptyset=S\left(a\right)=a\cup\left\{a\right\} +\end{equation*}$$ This is a contradiction, as $a\cup\left\{a\right\}$ is +a set of two elements, namely $a$ and $\left\{a\right\}$ but the +empty-set by definition has no elements. $\qed$* +::: + +::: definition +**Definition 58**. *Recursive definition of a set* + +*A set $S$ is defined recursively if the elements of $S$ are defined in +terms of other elements $x\in S$. Moreover we have that there is some +initial element $x_0$ which is used to define the other elements of the +set.* +::: + +::: definition +**Definition 59**. *First definition of the Natural numbers* + +*We define the set $\mathbb{N}$, called the set of natural numbers, to +be the set given by* + +*$$\begin{equation} + \mathbb{N}=\left\{x: x=\emptyset\text{ or } x=S\left(y\right)\text{ for some } y\in\mathbb{N}\right\} +\end{equation}$$* +::: + +We have defined $\mathbb{N}$ recursively in terms of elements of +$\mathbb{N}$. As an example $2\in\mathbb{N}$ as $2=S\left(1\right)$ and +likewise $1=S\left(0\right)$ and we know that $0$ is really the same as +$\emptyset$, which is the initial element of $\mathbb{N}$ as defined +above. This definition allows us to get any $x\in\mathbb{N}$, however it +is not quite enough to get every element of $\mathbb{N}$ at the same +time. We know that there should be infinitely many natural numbers, +indeed for any $n\in\mathbb{N}$ we have also that $n+1\in\mathbb{N}$. In +other words we have a chain of sets of increasing size, that is we have + +$$\begin{align*} + \mathbb{N}_0&=\emptyset=0\\ + \mathbb{N}_1&=\left\{\emptyset\right\}=1\\ + \mathbb{N}_2&=\left\{\emptyset,\left\{\emptyset\right\}\right\}=2\\ + \mathbb{N}_3&=\left\{\emptyset,\left\{\emptyset\right\},\left\{\left\{\emptyset,\left\{\emptyset\right\}\right\}\right\}\right\}=3\\ +\end{align*}$$ Which satisfy +$\mathbb{N}_0\subset\mathbb{N}_1\subset\mathbb{N}_2\subset\mathbb{N}_3\subset\dots$. +So we see at each stage $\mathbb{N}_n$ is a finite set of size $n$ and +so ultimately our current definition of $\mathbb{N}$ can ultimately only +ever reach a finite $n$. although we can make this $n$ arbitrarily +large. To ensure we get every possible $n$ at once we need to invoke the +axiom of infinity. + +1. The axiom of infinity: + + The axiom of infinity asserts that there is at least one infinite + set $A$, that is at least one set with infinitely many elements. + That is we have a set $A$ such that the $\emptyset\in A$ and if + $x\in A$ then the set $x\cup\left\{x\right\}$ is also in $A$. + +There is a useful definition that we can extract from the axiom of +infinity. + +::: definition +**Definition 60**. *Inductive set* + +*Let $A$ be a set and let $f:A\rightarrow A$ be a mapping. We say that +$A$ is an inductive if it satisfies the following two properties* + +1. *$\emptyset\in A$* + +2. *If $x\in A$ then $f\left(x\right)\in A$* + +*For now, we will be focused on the case where $f=S$, the successor +mapping.* +::: + +In light of the axiom of infinity we have a set that contains the +infinitely many Natural numbers. This is nearly what we want, although +it won't be the set of Natural numbers. This set could clearly have +many, many more things than just the Natural numbers. + +We can make a new definition, which will allow us to define +$\mathbb{N}$. We will also be able to show the fact this definition +defines $\mathbb{N}$ to be the smallest such inductive set that contains +all of the Natural numbers. + +::: definition +**Definition 61**. *The set $\mathbb{N}_S$* + +*Let $S$ be an inductive set. We define $\mathbb{N}_S$ as follows* + +*$$\begin{equation} + \mathbb{N}_S=\bigcap_{\substack{A\subseteq S \\ A\text{ is inductive}}} A +\end{equation}$$* + +*This is well-defined by the axiom of specification, being an inductive +step is definable and the collection of all subsets of $S$ is a set we +can define.* +::: + +We have that all of these sets $\mathbb{N}_S$ are the same. + +::: {#thm:EveryNsSetIsSame .theorem} +**Theorem 3**. *Every $\mathbb{N}_S$ set is the same set* + +*Let $S$ and $T$ be inductive sets. Define the sets $\mathbb{N}_S$ and +$\mathbb{N}_T$ We have that* + +*$$\begin{equation} + \mathbb{N}_S=\mathbb{N}_T +\end{equation}$$* + +*Proof:* + +*By the axiom of extensionality we know that two sets are equal if and +only if they contain the same elements. To see that $\mathbb{N}_S$ and +$\mathbb{N}_T$ have the same elements consider the new set given by* + +*$$\begin{equation*} + C=\mathbb{N}_S\cap\mathbb{N}_T +\end{equation*}$$* + +*We recall from proposition +[8](#prop:PropertiesOfUnionIntersectionSetinclusion){reference-type="ref" +reference="prop:PropertiesOfUnionIntersectionSetinclusion"} that for two +sets $A$ and $B$ we have $A\cap B\subseteq A$. Hence it follows that* + +*$$\begin{equation*} + C=\mathbb{N}_S\cap\mathbb{N}_T\subseteq\mathbb{N}_S +\end{equation*}$$ That is, $C\subseteq\mathbb{N}_S$, that is to say +every element of $C$ is also an element of $\mathbb{N}_S$. Now recall +the definition of $\mathbb{N}_S$,* + +*$$\begin{equation*} + \mathbb{N}_S=\bigcap_{\substack{A\subseteq S \\ A\text{ is inductive}}} A +\end{equation*}$$ We know that $C\subseteq \mathbb{N}_S$, hence as +$\mathbb{N}_S$ is the intersection of all subsets of $S$ we must +conclude that $C\subseteq S$.* + +*Now, we know that $S$ is an inductive set. Hence $S$ satisfies the +following* + +1. *$\emptyset\in S$* + +2. *If $x\in S$ then $S\left(x\right)\in S$* + +*If we can show that $C$ is an inductive set we know that $C$ was one of +the sets we used to construct $\mathbb{N}_S$ and hence +$\mathbb{N}_S\subseteq C$, which will give the equality +$C=\mathbb{N}_S$.* + +*Now, to show that $C$ is an inductive set me must show that* + +1. *$\emptyset\in C$* + +2. *If $x\in C$ then $S\left(x\right)\in C$* + + + +1. *$\emptyset\in C$:* + + *We have that $C=\mathbb{N}_S\cap\mathbb{N}_T$ and we have that* + + *$$\begin{align*} + \mathbb{N}_S&=\bigcap_{\substack{A\subseteq S \\ A\text{ is inductive}}} A\\ + \mathbb{N}_T&=\bigcap_{\substack{A\subseteq T \\ A\text{ is inductive}}} A\\ + \end{align*}$$ In the definitions of both $\mathbb{N}_S$ and + $\mathbb{N}_T$ we have that these are the intersections of inductive + sets and so $\emptyset\in\mathbb{N}_S$ and + $\emptyset\in\mathbb{N}_T$. It hence follows that as + $C=\mathbb{N}_S\cap\mathbb{N}_T$ we must have $\emptyset\in C$.* + +2. *If $x\in C$ then $S\left(x\right)\in C$:* + + *Now suppose that $x\in C$. Like before we know that + $C=\mathbb{N}_S\cap\mathbb{N}_T$, and by the definition of the + intersection of two sets, it follows that $x\in\mathbb{N}_S$ and + $x\in\mathbb{N}_T$. Now we have that* + + *$$\begin{equation*} + \mathbb{N}_S=\bigcap_{\substack{A\subseteq S \\ A\text{ is inductive}}} A + \end{equation*}$$ hence as $x\in\mathbb{N}_S$ we have we must have + that $x\in A$ for every subset $A$ of $S$. Moreover each such $A$ is + an inductive set and so by definition of an inductive set we have + that $S\left(x\right)\in A$ for every subset $A$ of $S$. Hence + $S\left(x\right)\in\mathbb{N}_S$ and likewise a similar argument + shows that $S\left(x\right)\in\mathbb{N}_T$. It thus follows that + $S\left(x\right)\in C$.* + + *As $x\in C$ was arbitrary we must conclude that this holds for any + $x\in C$.* + +*Hence $C$ is an inductive set.* + +*Now, we know that $C\subseteq S$ and $C$ is an inductive set then it +follows that $C$ is one of the inductive sets in the definition of +$\mathbb{N}_S$. It hence follows that $\mathbb{N}_S\subseteq C$. It +follows by the axiom of extensionality that as $\mathbb{N}_S$ and $C$ +contain the same elements then $C=\mathbb{N}_S$.* + +*Likewise the a similar argument shows that $C=\mathbb{N}_T$. So it +follows that $\mathbb{N}_S = \mathbb{N}_T$. $\qed$* +::: + +In light of this theorem we can now truly define $\mathbb{N}$. + +::: definition +**Definition 62**. *The Natural numbers $\mathbb{N}$* + +*Let $S$ be an inductive set, and construct the set $\mathbb{N}_S$. The +set $\mathbb{N}_S$ is the set of Natural numbers and by theorem +[3](#thm:EveryNsSetIsSame){reference-type="ref" +reference="thm:EveryNsSetIsSame"} no matter the inductive set $S$ we +have that all such $\mathbb{N}_S$ are the same. Hence we simply refer to +the natural numbers by $\mathbb{N}$.* +::: + +We identify the elements of $\mathbb{N}$ not in terms of $\emptyset$, +and sets of sets containing $\emptyset$, but instead by the more usually +numerals that we use. We have already defined Zero and One, by +definitions [55](#def:Zero){reference-type="ref" reference="def:Zero"} +and [56](#def:One){reference-type="ref" reference="def:One"}. The other +numbers follow likewise, i.e + +$$\begin{align*} + 0&=\emptyset\\ + 1&=S\left(0\right)=\left\{\emptyset\right\}\\ + 2&=S\left(1\right)=\left\{\emptyset,\left\{\emptyset\right\}\right\}\\ + 3&=S\left(2\right)\\ + 4&=S\left(3\right)\\ + &\dots\\ + n+1&=S\left(n\right) +\end{align*}$$ + +We said that we can prove that $\mathbb{N}$ is the smallest such +inductive set that contains all the natural numbers, this is to say if +$A\subseteq\mathbb{N}$ is an inductive set we must have that +$A=\mathbb{N}$. We thankfully do not need to prove this as the previous +theorem gives this for free. This also gives us the following definition +for a minimally inductive set, we make the definition in such a way that +we argue about sets of inductive sets. + +::: definition +**Definition 63**. *Minimally inductive set of sets* + +*Let $S$ be a set whose elements are also sets satisfying some +condition, and let $f:S\rightarrow S$ be a mapping. We say that $S$ is +minimally inductive if and only if the foll lowing holds* + +1. *$S$ is an inductive set under the mapping $g$* + +2. *No proper subset of $S$ is inductive under the mapping $g$* +::: + +One of the most powerful properties of the natural numbers is the +principle of Induction. This tool is powerful in proving many statements +on the Natural numbers. It works in a similar way to how an inductive +set works[^7]. We show that the statement works for some base case, +usually $n=0$, then we assume that if it holds true for some $n$ then it +holds true for $S\left(n\right)=n+1$. + +::: theorem +**Theorem 4**. *The principle of induction* + +*Suppose we have a proposition $P\left(n\right)$ about a Natural number +$n\in\mathbb{N}$. Moreover, suppose that* + +1. *$P\left(0\right)$ is true* + +2. *$P\left(n\right)$ being true implies + $P\left(S\left(n\right)\right)=P\left(n+1\right)$ is true for any + Natural number $n$.* + +*If these two statements are true, we have that $P\left(n\right)$ is +true for any natural number $n$, and we say the proposition +$P\left(n\right)$ holds by the principle of mathematical induction.* + +*Moreover we call $P\left(0\right)$ the base case for induction and +$P\left(n\right)$ being true implies $P\left(n+1\right)$ being true is +the inductive step.* + +*Proof:* + +*Let $P\left(n\right)$ be a proposition about a Natural number +$n\in\mathbb{N}$ such that $P\left(n\right)$ satisfies* + +1. *$P\left(0\right)$ is true* + +2. *$P\left(n\right)$ being true implies + $P\left(S\left(n\right)\right)=P\left(n+1\right)$ is true for any + Natural number $n$.* + +*Consider the set given by* + +*$$\begin{equation*} + Q=\left\{n:P\left(n\right)\text{ is true}\right\} +\end{equation*}$$ That is, $Q$ is defined as the set of Natural numbers +such that that $P\left(n\right)$ is true, clearly +$Q\subseteq\mathbb{N}$. By hypothesis we know that $P\left(0\right)$ is +true, so $0\in Q$. Also by hypothesis we know that if $P\left(n\right)$ +is true for some $n\in\mathbb{N}$. then we have that +$P\left(S\left(n\right)\right)=P\left(n+1\right)$ is also true, hence we +have that every $n\in\mathbb{N}$ is also in $Q$, hence +$\mathbb{N}\subseteq Q$ and so by the axiom of extensionality we have +that $Q=\mathbb{N}$. Hence $P\left(n\right)$ is true for every Natural +number $n\in\mathbb{N}$. $\qed$.* +::: + +Now that we have induction we can make a final definition that will be +useful. This definition combines a few previously proven results into a +convenient package, this package has the strength to prove the usual +properties of the natural numbers and perhaps are an easy way to +remember the basis for deducing properties about the natural numbers. + +::: definition +**Definition 64**. *The Peano axioms* + +*We define the Peano axioms as follows. Let $A$ be a set and consider +the successor mapping on $A$, $S: A\rightarrow A$. If we have that* + +1. *$A$ is an inductive set* + + 1. *$\emptyset\in A$* + + 2. *If $x\in A$ then $S\left(x\right)\in A$* + +2. *$S$ is an injective mapping.* + +3. *$\forall x\in S$ we have that $\emptyset\neq S\left(x\right)$* + +4. *$\forall B \subseteq A$. If $0\in B$ and $S\left(n\right)\in B$ for + all $n\in B$ then $B=A$* + +*If $A$ satisfies all of the above, then we say that $A$ satisfies the +Peano axioms and induces Peano arithmetic.* +::: + +#### Properties of the natural numbers + +Although we have constructed $\mathbb{N}$ we haven't defined what we can +do with this set. We know from our intuitions that we can define +addition, a form of subtraction, multiplication and in some cases +division. We also know that there is some notion of a Natural number +being larger or smaller than another, when two Natural numbers are equal +and so. We will explore some of these properties so that we can start +doing some form of Mathematics. + +##### Equality of natural numbers + +Firstly, it is important to define when two Natural numbers are equal, +again as we have defined the natural numbers in terms of Sets, this just +comes down to the axiom of extensionality. + +::: definition +**Definition 65**. *Equality of natural numbers* + +*Let $n,m\in\mathbb{N}$ be two natural numbers. We define that two +natural numbers are equal, denoted $n=m$ if and only if $n\subseteq m$ +and $m\subseteq n$. This is simply the axiom of extensionality.* + +*If we do not have $n=m$ then we say that $n$ and $m$ are not equal and +we denote this $n\neq m$.* +::: + +This definition clearly makes sense as each natural number is a set. + +::: example +**Example 51**. *We have that $1=1$. Indeed by definition $0=\emptyset$ +and $1=\left\{\emptyset\right\}$. It is clear that +$\left\{\emptyset\right\}\subseteq \left\{\emptyset\right\}$ hence the +axiom of extensionality gives us that +$\left\{\emptyset\right\}=\left\{\emptyset\right\}$. That is $1=1$* +::: + +::: example +**Example 52**. *We have that $3=3$. Indeed by construction we have that +$3=\left\{\emptyset,\left\{\emptyset\right\},\left\{\emptyset,\left\{\emptyset\right\}\right\}\right\}$ +It is clear that +$\left\{\emptyset,\left\{\emptyset\right\},\left\{\emptyset,\left\{\emptyset\right\}\right\}\right\}$ +hence the axiom of extensionality gives us that +$\left\{\emptyset,\left\{\emptyset\right\},\left\{\emptyset,\left\{\emptyset\right\}\right\}\right\}=\left\{\emptyset,\left\{\emptyset\right\},\left\{\emptyset,\left\{\emptyset\right\}\right\}\right\}$. +i.e $3=3$* +::: + +::: example +**Example 53**. *We have $1\neq 2$. We have that +$1=\left\{\emptyset\right\}$ and +$1=\left\{\emptyset,\left\{\emptyset\right\}\right\}$. Now +$\left\{\emptyset\right\}\subseteq \left\{\emptyset,\left\{\emptyset\right\}\right\}$ +but +$\left\{\emptyset,\left\{\emptyset\right\}\right\}\not\subseteq \left\{\emptyset\right\}$.* +::: + +In particular in light of the definition of equality on the natural +numbers if $n=m$ and $m=k$ we must have that $n=k$. + +##### Inequality of natural numbers + +We can define also define what it means for natural numbers to not be +equal. We make use of the notion of set inclusion. Recall that a set $S$ +is a subset of the set $T$, written $S\subseteq T$, if for every +$s\in S$ we have that $s\in T$ and that $S$ is a proper subset of $T$, +written $S\subset T$ if $S\subseteq T$ and $S\neq T$. We will use the +proper subset notation to define the so-called less than operator. This +operation comes naturally from the definition of the natural numbers by +the successor mapping. The successor function has the following chain of +definitions for each $n\in\mathbb{N}$ + +$$\begin{align*} + 0&=\emptyset\\ + 1&=S\left(0\right)=\left\{\emptyset\right\}\\ + 2&=S\left(1\right)=\left\{\emptyset,\left\{\emptyset\right\}\right\}\\ + 3&=S\left(2\right)=\left\{\emptyset,\left\{\emptyset\right\},\left\{\emptyset,\left\{\emptyset\right\}\right\}\right\}\\ + 4&=S\left(3\right)=\left\{\emptyset,\left\{\emptyset\right\},\left\{\emptyset,\left\{\emptyset\right\}\right\},\left\{\emptyset,\left\{\emptyset\right\},\left\{\emptyset,\left\{\emptyset\right\}\right\}\right\}\right\}\\ + &\dots\\ + n+1&=S\left(n\right) +\end{align*}$$ + +From this chain of definitions and the axiom of foundation, +$0=\emptyset$ is the set element minimal element of $\mathbb{N}$, so +every natural number is contained in one that comes after. We can make +the following definition which defines when one natural number is +smaller than another. + +::: definition +**Definition 66**. *Less than Operator* + +*Let $n,m\in\mathbb{N}$. The less than operator, denoted by $nm$ and +is read as $n$ is greater than $m$, is defined as follows.* + +*We have $n>m$ if and only if $n\not\subset m$. That is, the set that +denotes the number $n$ is not an element of the set $m$. That is to say +that $>$ is a logical proposition, given by* + +*$$\begin{equation*} + >\left(n,m\right)=\begin{cases} + 1,\ \text{If } n\not\subset m\\ + 0,\ \text{Otherwise} + \end{cases} +\end{equation*}$$* +::: + +Likewise, we can define the greater than or equal to operator. + +::: definition +**Definition 69**. *Greater than or equal to operator* + +*Let $n,m\in\mathbb{N}$. The greater than or equal to operator, denoted +by $n\geq m$, and read as $n$ is greater than or equal to $m$, is +defined the same as $n>m$ except we now allow for the situation that +$n=m$. This is to say $\geq$ is a logical proposition given by* + +*$$\begin{equation*} + \geq\left(n,m\right) = \begin{cases} + 1,\ \text{If } n> m\\ + 1,\ \text{If } n=m\\ + 0,\ \text{Otherwise} + \end{cases} +\end{equation*}$$* +::: + +##### Defining addition and multiplication on the Natural numbers + +We can use the principle of induction to make definitions as well as a +proof technique. We shall use induction now to make two definitions, in +particular, we define two mappings that will allow us to start +manipulating Natural numbers as we expect them to. To do so it is enough +to specify what the mapping does when $0$ is given as an argument, and +then do define what the mapping does when given $S\left(n\right)$ as an +argument, hence defining it in terms of $n$ for each $n\in\mathbb{N}$. +This will make sense when we define these operations. + +We first recall the Cartesian product of two sets. Let $S$ and $T$ be +sets, the Cartesian product of $S$ and $T$, denoted $S\times T$ is the +set of all ordered pairs of the form $\left(S,t\right)$ where $s\in S$ +and $t\in T$. This is to say that + +$$\begin{equation*} + S\times T=\left\{\left(s,t\right):s\in S,t\in T\right\} +\end{equation*}$$ + +If $S=T$ then we denote $S\times T$ by $S^2$. + +::: definition +**Definition 70**. *Addition on the Natural numbers* + +*We define addition on the Natural numbers by the following mapping. Let +$+:\mathbb{N}^2\rightarrow\mathbb{N}$ be such that for all +$\left(m,n\right)\in\mathbb{N}^2$ we have the following* + +*$$\begin{align} + +&:\mathbb{N}^2\mathlarger{\mathlarger{\rightarrow}}\mathbb{N}\\ + \left(m,n\right)&\mapsto +\left(m,n\right)=\begin{cases} + m+0=m,\ \text{If } n=0\\ + m+S\left(n\right)=S\left(m+n\right),\ \text{If } n\neq 0 + \end{cases} +\end{align}$$* + +*We will write $+\left(m,n\right)$ as $m+n$.* +::: + +In light of this definition, we can prove that $1+1=2$ + +::: theorem +**Theorem 5**. *1+1=2* + +*We have that $1+1=2$.* + +*Proof:* + +*We know that $1=S\left(0\right)$ and $2=S\left(S\left(0\right)\right)$. +Hence, we are proving* + +*$$\begin{equation*} + S\left(0\right)+S\left(0\right)=S\left(S\left(0\right)\right) +\end{equation*}$$* + +*By the definition of the addition mapping, we know that +$\forall \left(m,n\right)\in\mathbb{N}^2$ that* + +*$$\begin{equation*} + m+S\left(n\right)=S\left(m+n\right) +\end{equation*}$$ In particular if $n=0$ we have $\forall m$ that* + +*$$\begin{equation*} + m+S\left(0\right)=S\left(m+0\right) +\end{equation*}$$* + +*and* + +*$$\begin{equation} +\label{eq:OnePlusOneProofEq1} + S\left(0\right)+S\left(0\right)=S\left(S\left(0\right)+0\right) +\end{equation}$$ Moreover, by the definition of addition, we know that +$\forall m$ that if $n=0$ then* + +*$$\begin{equation*} + m+0=m +\end{equation*}$$ Hence* + +*$$\begin{align*} + S\left(0\right)+0&=S\left(0\right)\\ + \Rightarrow S\left(S\left(0\right)+0\right)&= S\left(S\left(0\right)\right)\\ + \Rightarrow S\left(0\right)+S\left(0\right)&=S\left(S\left(0\right)\right) +\end{align*}$$* + +*This is to say. $1+1=2$. As required. $\qed$.* +::: + +::: definition +**Definition 71**. *Multiplication on the Natural numbers* + +*We define multiplication on the Natural numbers by the following +mapping. Let $*:\mathbb{N}\times\mathbb{N}\rightarrow\mathbb{N}$ be such +that for all $\left(m,n\right)\in\mathbb{N}\times\mathbb{N}$ we have the +following* + +*$$\begin{align} + *&:\mathbb{N}\times\mathbb{N}\mathlarger{\mathlarger{\rightarrow}}\mathbb{N}\\ + \left(m,n\right)&\mapsto *\left(m,n\right)=\begin{cases} + m*0=0,\ \text{If } n=0\\ + m*S\left(n\right)=m*n+m,\ \text{If } n\neq 0 + \end{cases} +\end{align}$$ We will write $*\left(m,n\right)$ as $m*n$, or more +compactly just as the juxtaposition $mn$* +::: + +As with addition we provide a proof that $2*2=4$ + +::: theorem +**Theorem 6**. *2\*2=4* + +*We have $2*2=4$.* + +*Proof:* + +*We know that $S\left(1\right)=2$ and so by definition of multiplication +we have that* + +*$$\begin{equation*} + 2*2=2*S\left(1\right)=2*1+2 +\end{equation*}$$* + +*Likewise we know that $S\left(0\right)=1$ and so by another application +of the definition of multiplication we have that* + +*$$\begin{equation*} + 2*1+2=2*S\left(0\right)+2=2*0+2+2 +\end{equation*}$$* + +*Now $2*0=0$ by definition as so we have that* + +*$$\begin{equation*} + 2*2=2*0+2+2=0+2+2=2+2 +\end{equation*}$$* + +*It is left to show that $2+2 = 4$. We use a similar proof to $1+1=2$. +As +$4=S\left(S\left(2\right)\right)=S\left(S\left(S\left(S\left(0\right)\right)\right)\right)$ +and $2=S\left(S\left(0\right)\right)$ we need to show that* + +*$$\begin{equation*} + S\left(S\left(0\right)\right)+S\left(S\left(0\right)\right)=S\left(S\left(S\left(S\left(0\right)\right)\right)\right) +\end{equation*}$$* + +*By the definition of addition we have that +$\forall\left(m,n\right)\in\mathbb{N}^2$ that* + +*$$\begin{equation*} + m+S\left(n\right)=S\left(m+n\right) +\end{equation*}$$* + +*In particular we have that if $n=0$ and $\forall n\in\mathbb{N}$ that* + +*$$\begin{equation*} + m+S\left(0\right)=S\left(m+0\right) +\end{equation*}$$* + +*So that* + +*$$\begin{align*} + S\left(S\left(0\right)\right)+S\left(S\left(0\right)\right)&=S\left(S\left(S\left(0\right)\right)+S\left(0\right)\right)\\ + &=S\left(S\left(S\left(S\left(0\right)\right)+0\right)\right)\\ + &=S\left(S\left(S\left(S\left(0\right)\right)\right)\right)\\ +\end{align*}$$* + +*That is $2+2=4$ and so the theorem is proved. $\qed$* +::: + +These two definitions are enough to prove every elementary property of +addition and multiplication that we are familiar with. However to do so +will require an upgrade to the idea of induction. This will allow us to +perform induction on both the addition and multiplication mappings. Once +we have done this we will have put the natural numbers on a firm logical +basis. This idea is called double induction, or more clearly induction +on two variables. + +For example, we know from school that $n+m=m+n$ for all natural numbers +$n$ and $m$. To show that this is true, we start by induction on $n$, so +we have to show that $m+0=0+m$ and then that $\left(m+n=n+m\right)$ +implies that $\left(m+S\left(n\right)=S\left(n\right)+m\right)$, each of +these will be proved by induction on $m$. This is the idea of double +induction. + +::: theorem +**Theorem 7**. *Double induction* + +*Let $P\left(m,n\right)$ be a proposition about a pair of natural +numbers $m,n\in\mathbb{N}$. Moreover suppose that* + +1. *$P\left(0,0\right)$ is true.* + +2. *$P\left(0,n\right)$ being true implies that + $P\left(0,S\left(n\right)\right)$ is true.* + +3. *$P\left(m,0\right)$ being true implies that + $P\left(S\left(m\right),0\right)$ is true* + +4. *For a given $m\in\mathbb{N}$, from the truth that + $P\left(m,x\right)$ is true for all $x$, and also that of + $P\left(S\left(m\right),n\right)$ for some $n$, we can infer that + $P\left(S\left(m\right),S\left(n\right)\right)$ is true.* + +*If these statements are true, we have that $P\left(m,n\right)$ is true +for any natural numbers $m,n\in\mathbb{N}$ and we say that the +proposition $P\left(m,n\right)$ hold by the principle of mathematical +double induction.* + +*Proof:* + +*Let $P\left(m,n\right)$ be a proposition about a pair of natural +numbers $m,n\in\mathbb{N}$, which satisfies* + +1. *$P\left(0,0\right)$ is true.* + +2. *$P\left(0,n\right)$ being true implies that + $P\left(0,S\left(n\right)\right)$ is true.* + +3. *$P\left(m,0\right)$ being true implies that + $P\left(S\left(m\right),0\right)$ is true* + +4. *For a given $m\in\mathbb{N}$, from the truth that + $P\left(m,x\right)$ is true for all $x$, and also that of + $P\left(S\left(m\right),n\right)$ for some $n$, we can infer that + $P\left(S\left(m\right),S\left(n\right)\right)$ is true.* + +*Statements $1$ and $2$ are the base case and the inductive step for the +proof of $P\left(0,n\right)$ for all $n\in\mathbb{N}$. Likewise +statements $1$ and $3$ are the base case and the inductive step for the +proof of $P\left(m,0\right)$ for all $m\in\mathbb{N}$.* + +*Finally, the statements $3$ and $4$ is the base case and inductive step +for a proof, by induction on $n$ for a proof of the statement that if +$P\left(m,n\right)$ holds for all $n$, then +$P\left(S\left(m\right),n\right)$ holds for all $n$, and thus by +induction we have that $P\left(m,n\right)$ is true for all $m$. $\qed$.* +::: + +We can start proving the basic properties of $\mathbb{N}$ that we are +familiar with. + +##### Closure properties of addition and multiplication + + \ +We show that addition and multiplication on the natural numbers to +produces a natural number. + +::: theorem +**Theorem 8**. *The addition and multiplication mappings on the natural +numbers are closed* + +*For all $n,m\in\mathbb{N}$. We have that* + +1. *$n+m\in\mathbb{N}$.* + +2. *$nm\in\mathbb{N}$.* + +*Proof:* + +1. *$n+m\in\mathbb{N}$:* + + *Let $n,m\in\mathbb{N}$. We need to show that* + + 1. *$0+0\in\mathbb{N}$* + + 2. *$0+n\in\mathbb{N}$ implies $0+S\left(n\right)\in\mathbb{N}$* + + 3. *$m+0\in\mathbb{N}$ implies $S\left(m\right)+0\in\mathbb{N}$* + + 4. *For some $m\in\mathbb{N}$. Suppose that $m+x\in\mathbb{N}$ for + all $x\in\mathbb{N}$, and $S\left(m\right)+n\in\mathbb{N}$ for + some $n\in\mathbb{N}$ implies that + $S\left(m\right)+S\left(n\right)\in\mathbb{N}$* + + + + 1. *$0+0\in\mathbb{N}$:* + + *We have by the definition of addition that* + + *$$\begin{equation*} + 0+0=0 + \end{equation*}$$ which is clearly in $\mathbb{N}$.* + + 2. *$0+n\in\mathbb{N}$ implies $0+S\left(n\right)\in\mathbb{N}$:* + + *Now, suppose that $0+n\in\mathbb{N}$ for some $n$, we show that + $0+S\left(n\right)\in\mathbb{N}$.* + + *By the definition of addition we have that* + + *$$\begin{equation*} + 0+S\left(n\right)=S\left(0+n\right) + \end{equation*}$$ Now $0+n\in\mathbb{N}$ by assumption, + therefore we have that $S\left(0+n\right)\in\mathbb{N}$. Hence + $0+S\left(n\right)\in\mathbb{N}$.* + + 3. *$m+0\in\mathbb{N}$ implies $S\left(m\right)+0\in\mathbb{N}$:* + + *Now, suppose that $m+0\in\mathbb{N}$ for some $m$, we show that + $S\left(m\right)+0\in\mathbb{N}$.* + + *By the definition of addition we have that* + + *$$\begin{equation*} + S\left(m\right)+0=S\left(m\right)=S\left(m+0\right) + \end{equation*}$$ Now $m+0\in\mathbb{N}$ by assumption, + therefore $S\left(m+0\right)\in\mathbb{N}$. Hence + $S\left(m\right)+0\in\mathbb{N}$* + + 4. *For some $m\in\mathbb{N}$. Suppose that $m+x\in\mathbb{N}$ for + all $x\in\mathbb{N}$, and $S\left(m\right)+n\in\mathbb{N}$ for + some $n\in\mathbb{N}$ implies that + $S\left(m\right)+S\left(n\right)\in\mathbb{N}$* + + *Now suppose that $m+x\in\mathbb{N}$ for all $x\in\mathbb{N}$ + and some fixed $m\in\mathbb{N}$, and suppose that + $S\left(m\right)+n\in\mathbb{N}$ where $n$ is some fixed value, + we show that $S\left(m\right)+S\left(n\right)\in\mathbb{N}$.* + + *So, we have that $S\left(m\right)\in\mathbb{N}$ and + $S\left(n\right)\in\mathbb{N}$ we can use the definition of + addition, doing so gives* + + *$$\begin{equation*} + S\left(m\right)+S\left(n\right)=S\left(S\left(m\right)+n\right) + \end{equation*}$$ By assumption + $S\left(m\right)+n\in\mathbb{N}$, hence as we have that + $m+x\in\mathbb{N}$ for all $x\in\mathbb{N}$, then we have that + $S\left(S\left(m\right)+n\right)\in\mathbb{N}$. Therefore we + must conclude that + $S\left(m\right)+S\left(n\right)\in\mathbb{N}$.* + + *Hence by the principle by double induction we have that + $m+n\in\mathbb{N}$ for all $m,n\in\mathbb{N}$. That is, addition is + closed.* + +2. *$nm\in\mathbb{N}$:* + + *Let $n,m\in\mathbb{N}$. We need to show that* + + 1. *$0*0\in\mathbb{N}$* + + 2. *$0*n\in\mathbb{N}$ implies $0*S\left(n\right)\in\mathbb{N}$* + + 3. *$m*0\in\mathbb{N}$ implies $S\left(m\right)*0\in\mathbb{N}$* + + 4. *For some $m\in\mathbb{N}$. Suppose that $m*x\in\mathbb{N}$ for + all $x\in\mathbb{N}$, and $S\left(m\right)*n\in\mathbb{N}$ for + some $n\in\mathbb{N}$ implies that + $S\left(m\right)*S\left(n\right)\in\mathbb{N}$* + + + + 1. *$0*0\in\mathbb{N}$:* + + *We have by the definition of multiplication that* + + *$$\begin{equation*} + 0*0=0 + \end{equation*}$$ which is clearly in $\mathbb{N}$.* + + 2. *$0*n\in\mathbb{N}$ implies $0*S\left(n\right)\in\mathbb{N}$:* + + *Now, suppose that $0*n\in\mathbb{N}$ for some $n$, we show that + $0*S\left(n\right)\in\mathbb{N}$.* + + *By the definition of multiplication we have that* + + *$$\begin{equation*} + 0*S\left(n\right)=0*n+0 + \end{equation*}$$* + + *Now $0*n\in\mathbb{N}$ by assumption, moreover we have proved + that addition is closed, so $0*n+0\in\mathbb{N}$ therefore we + have that $0*S\left(n\right)\in\mathbb{N}$* + + 3. *$m*0\in\mathbb{N}$ implies $S\left(m\right)*0\in\mathbb{N}$:* + + *Now, suppose that $m*0\in\mathbb{N}$ for some $m$, we show that + $S\left(m\right)*0\in\mathbb{N}$.* + + *By the definition of addition we have that* + + *$$\begin{equation*} + S\left(m\right)*0=0 + \end{equation*}$$ Where $S\left(m\right)*0=0$ by definition of + multiplication. Hence as $0\in\mathbb{N}$ we have that + $S\left(m\right)*0\in\mathbb{N}$.* + + 4. *For some $m\in\mathbb{N}$. Suppose that $m*x\in\mathbb{N}$ for + all $x\in\mathbb{N}$, and $S\left(m\right)*n\in\mathbb{N}$ for + some $n\in\mathbb{N}$ implies that + $S\left(m\right)*S\left(n\right)\in\mathbb{N}$:* + + *Now suppose that $m*x\in\mathbb{N}$ for all $x\in\mathbb{N}$ + and some fixed $m\in\mathbb{N}$, and suppose that + $S\left(m\right)*n\in\mathbb{N}$ where $n$ is some fixed value, + we show that $S\left(m\right)*S\left(n\right)\in\mathbb{N}$.* + + *So, we have that $S\left(m\right)\in\mathbb{N}$ and + $S\left(n\right)\in\mathbb{N}$ we can use the definition of + multiplication, doing so gives* + + *$$\begin{equation*} + S\left(m\right)*S\left(n\right)=S\left(m\right)*n+S\left(m\right) + \end{equation*}$$ By assumption + $S\left(m\right)*n\in\mathbb{N}$, moreover as $m*x\in\mathbb{N}$ + for all $x\in\mathbb{N}$ we must have + $S\left(m\right)*n+S\left(m\right)\in\mathbb{N}$ as addition is + closed.* + + *Hence $S\left(m\right)*S\left(n\right)\in\mathbb{N}$.* + + *Hence by the principle by double induction we have that + $m*n\in\mathbb{N}$ for all $m,n\in\mathbb{N}$. That is, + multiplication is closed.* + +*Hence, we have that the addition and multiplication mappings are +closed. $\qed$* +::: + +##### Commutativity of addition and multiplication + + \ +This will prove that for all $a,b\in\mathbb{N}$ that $a+b=b+a$ and +$ab=ba$. + +::: theorem +**Theorem 9**. *Addition and multiplication are commutative* + +*For all $a,b\in\mathbb{N}$ we have that* + +1. *$a+b=b+a$* + +2. *$ab=ba$* + +*Proof:* + +1. *$a+b=b+a$:* + + *We argue by double induction. We need to show that* + + 1. *$0+0=0+0$* + + 2. *$0+n=n+0$ implies $0+S\left(n\right)=S\left(n\right)+0$* + + 3. *$m+0=0+m$ implies $S\left(m\right)+0=0+S\left(m\right)$* + + 4. *If $m+x=x+m$ for all $x\in\mathbb{N}$ and + $S\left(m\right)+n=n+S\left(m\right)$ for some $n\in\mathbb{N}$, + then we have that + $S\left(m\right)+S\left(n\right)=S\left(n\right)+S\left(m\right)$* + + + + 1. *$0+0=0+0$:* + + *This is trivial by definition of addition.* + + 2. *$0+n=n+0$ implies $0+S\left(n\right)=S\left(n\right)+0$:* + + *Suppose that $0+n=n+0$, we show that + $0+S\left(n\right)=S\left(n\right)+0$. By the definition of + addition we have that* + + *$$\begin{equation*} + 0+S\left(n\right)=S\left(0+n\right) + \end{equation*}$$ We know by assumption that $0+n=n+0$. Hence* + + *$$\begin{equation*} + S\left(0+n\right)=S\left(n+0\right)=S\left(n\right)+0 + \end{equation*}$$* + + 3. *$m+0=0+m$ implies $S\left(m\right)+0=0+S\left(m\right)$:* + + *Suppose that $m+0=0+m$, we show that + $S\left(m\right)+0=0+S\left(m\right)$. By the definition of + addition we have that* + + *$$\begin{equation*} + S\left(m\right)+0=S\left(m\right)=S\left(m+0\right) + \end{equation*}$$ We know by assumption that $n+0=+m$. Hence* + + *$$\begin{equation*} + S\left(m+0\right)=S\left(0+m\right)=0+S\left(m\right) + \end{equation*}$$* + + 4. *If $m+x=x+m$ for all $x\in\mathbb{N}$ and + $S\left(m\right)+n=n+S\left(m\right)$ for some $n\in\mathbb{N}$, + then we have that + $S\left(m\right)+S\left(n\right)=S\left(n\right)+S\left(m\right)$:* + + *Suppose $m+x=x+m$ for all $x\in\mathbb{N}$ and that + $S\left(m\right)+n=n+S\left(m\right)$ for some $n\in\mathbb{N}$, + we show that + $S\left(m\right)+S\left(n\right)=S\left(n\right)+S\left(m\right)$.* + + *We have* + + *$$\begin{equation*} + S\left(m\right)+S\left(n\right)=S\left(S\left(m\right)+n\right) + \end{equation*}$$ Now we have by assumption that + $S\left(m\right)+n=n+S\left(m\right)$, for some + $n\in\mathbb{N}$, hence* + + *$$\begin{equation*} + S\left(S\left(m\right)+n\right)=S\left(n+S\left(m\right)\right)=S\left(S\left(n+m\right)\right) + \end{equation*}$$* + + *Likewise a similar chain of reasoning gives* + + *$$\begin{equation*} + S\left(n\right)+S\left(m\right)=S\left(S\left(n\right)+m\right)=S\left(m+S\left(n\right)\right)=S\left(S\left(m+n\right)\right) + \end{equation*}$$ Finally, we have that $m+n=m+n$ by assumption, + and so + $S\left(S\left(n+m\right)\right)=S\left(S\left(m+n\right)\right)$* + + *Hence by the principle of double induction we have that $a+b=b+a$ + for all $a,b\in\mathbb{N}$. That is addition is commutative.* + +2. *$ab=ba$:* + + *We need to show that* + + 1. *$0*0=0*0$* + + 2. *$0*n=n*0$ implies $0*S\left(n\right)=S\left(n\right)*0$* + + 3. *$m*0=0*m$ implies $S\left(m\right)*0=0*S\left(m\right)$* + + 4. *If $m*x=x*m$ for all $x\in\mathbb{N}$ and + $S\left(m\right)*n=n*S\left(m\right)$ for some $n\in\mathbb{N}$, + then we have that + $S\left(m\right)*S\left(n\right)=S\left(n\right)*S\left(m\right)$* + + + + 1. *$0*0=0*0$:* + + *This is trivial by the definition of multiplication.* + + 2. *$0*n=n*0$ implies $0*S\left(n\right)=S\left(n\right)*0$:* + + *Suppose that $0*n=n*0$, we show that + $0*S\left(n\right)=S\left(n\right)*0$. We have by definition of + multiplication that* + + *$$\begin{align*} + 0*S\left(n\right)&=0*n+0\\ + &=n*0+0,\ \text{By assumption}\\ + &=0+0,\ \text{By definition of multiplication}\\ + &=0,\ \text{By definition of addition}\\ + &=S\left(n\right)*0,\ \text{By definition of multiplication}\\ + \end{align*}$$* + + 3. *$m*0=0*m$ implies $S\left(m\right)*0=0*S\left(m\right)$:* + + *Suppose that $m*0=0*m$, we show that + $S\left(m\right)*0=0*S\left(m\right)$. We have by definition of + multiplication that* + + *$$\begin{align*} + 0*S\left(m\right)&=0*m+0\\ + &=m*0+0,\ \text{By assumption}\\ + &=0+0,\ \text{By definition of multiplication}\\ + &=0,\ \text{By definition of addition}\\ + &=S\left(m\right)*0,\ \text{By definition of multiplication}\\ + \end{align*}$$* + + 4. *If $m*x=x*m$ for all $x\in\mathbb{N}$ and + $S\left(m\right)*n=n*S\left(m\right)$ for some $n\in\mathbb{N}$, + then we have that + $S\left(m\right)*S\left(n\right)=S\left(n\right)*S\left(m\right)$:* + + *Suppose that $m*x=x*m$ for all $x\in\mathbb{N}$ and + $S\left(m\right)*n=n*S\left(m\right)$ for some $n\in\mathbb{N}$, + we show + $S\left(m\right)*S\left(n\right)=S\left(n\right)*S\left(m\right)$. + By definition of multiplication we have that* + + *$$\begin{equation*} + S\left(m\right)*S\left(n\right)=S\left(m\right)*n+S\left(m\right)=n*S\left(m\right)+S\left(m\right)=n*m+n+S\left(m\right)=n*m+S\left(n+m\right) + \end{equation*}$$* + + *Likewise, we have that $$\begin{equation*} + S\left(n\right)*S\left(m\right)=S\left(n\right)*m+S\left(n\right)=m*S\left(n\right)+S\left(n\right)=m*n+m+S\left(n\right)=m*n+S\left(m+n\right) + \end{equation*}$$ Now, we know that addition is commutative so + we have that $S\left(m+n\right)=S\left(n+m\right)$, moreover by + assumption we have that $n*m=m*n$. Hence* + + *$$\begin{equation*} + n*m+S\left(n+m\right)=m*n+S\left(m+n\right) + \end{equation*}$$* + + *Hence by the principle of double induction we have that $ab=ba$ for + all $a,b\in\mathbb{N}$. That is multiplication is commutative.* + +*The result now follows. $\qed$* +::: + +We can also now deduce the following property of multiplication + +##### Associativity of addition + + \ +This will prove that for all $a,b,c\in\mathbb{N}$ that +$a+\left(b+c\right)=\left(a+b\right)+c$ + +::: theorem +**Theorem 10**. *Addition is associative* + +*For all $a,b,c\in\mathbb{N}$ we have that* + +*$$\begin{equation*} + a+\left(b+c\right)=\left(a+b\right)+c +\end{equation*}$$* + +*Proof: We can show this by induction. Let $x,y\in\mathbb{N}$ be +arbitrary, and let $P\left(n\right)$ be the proposition given by* + +*$$\begin{equation*} + \left(x+y\right)+n=x+\left(y+n\right) +\end{equation*}$$* + +*For the base case we have $n=0$ and so* + +*$$\begin{align*} + \left(x+y\right)+0&=x+y ,\text{By definition of addition}\\ + &=x+\left(y+0\right) +\end{align*}$$* + +*Hence $P\left(0\right)$ is true.* + +*Now, suppose that $P\left(n\right)$ is true, that is* + +*$$\begin{equation*} + \left(x+y\right)+n=x+\left(y+n\right) +\end{equation*}$$ We show that $P\left(S\left(n\right)\right)$ is also +true, that is* + +*$$\begin{equation*} + \left(x+y\right)+S\left(n\right)=x+\left(y+S\left(n\right)\right) +\end{equation*}$$* + +*Now, we have that* + +*$$\begin{align*} + \left(x+y\right)+S\left(n\right)&=S\left(\left(x+y\right)+n\right),\text{By definition of addition}\\ + &=S\left(x+\left(y+n\right)\right),\ \text{By the induction hypothesis}\\ + &=x+\left(S\left(y+n\right)\right),\text{By definition of addition}\\ + &=x+\left(y+S\left(n\right)\right),\text{By definition of addition}\\ +\end{align*}$$* + +*Hence $P\left(S\left(n\right)\right)$ is true.* + +*It follows by mathematical induction that $\forall a,b,c\in\mathbb{N}$ +we have that $a+\left(b+c\right)=\left(a+b\right)+c$, that is addition +is associative. $\qed$* +::: + +##### Multiplication distributes over addition + + \ +This will prove that for all $a,b,c\in\mathbb{N}$ we have that +$a\left(b+c\right)=ab+ac$ and $\left(a+b\right)c=ac+bc$. + +::: theorem +**Theorem 11**. *Multiplication distributes over addition* + +*For all $a,b,c\in\mathbb{N}$ we have that* + +1. *$a\left(b+c\right)=ab+ac$* + +2. *$\left(b+c\right)a=ba+ca=ab+ac$* + +*Proof:* + +*We can be quick, and solve both problems nearly simultaneously, as we +have shown that multiplication is commutative.. To do this we show that +for all $a,b,c\in\mathbb{N}$ we have that $a\left(b+c\right)=ab+ac$.* + +*Let $a,b\in\mathbb{N}$ be arbitrary and we argue by induction on the +proposition $P\left(n\right)$ given by* + +*$$\begin{equation*} + a\left(b+n\right)=ab+an +\end{equation*}$$* + +*For the base case $n=0$ we have that* + +*$$\begin{align*} + a\left(b+0\right)&=a\left(b\right),\text{By definition of multiplication}\\ + &=ab \\ + &=ab+0,\text{By definition of addition}\\ + &=ab+a*0,\text{By definition of multiplication}\\ +\end{align*}$$* + +*Hence $P\left(0\right)$ is true.* + +*Now suppose that $P\left(n\right)$ is true, that is to say* + +*$$\begin{equation*} + a\left(b+n\right)=ab+an +\end{equation*}$$* + +*We show that $P\left(S\left(n\right)\right)$ is true, that is* + +*$$\begin{equation*} + a\left(b+S\left(n\right)\right)=ab+aS\left(n\right) +\end{equation*}$$* + +*Indeed, we have that* + +*$$\begin{align*} + a\left(b+S\left(n\right)\right)&=a\left(S\left(b+n\right)\right),\ \text{By definition of addition}\\ + &=a\left(b+n\right)+a,\ \text{By definition of multiplication}\\ + &=ab+an+a,\ \text{By assumption}\\ + &=ab+aS\left(n\right)0,\ \text{By definition of multiplication}\\ +\end{align*}$$* + +*Hence $P\left(S\left(n\right)\right)$ is true.* + +*It hence follows by the principle of mathematical induction that +$\forall a,b,c\in\mathbb{N}$ we have that $a\left(b+c\right)=ab+ac$.* + +*Now, we have shown that $a\left(b+c\right)=ab+ac$, to see that +$\left(b+c\right)a=ba+ca=ab+ac$ we simply observe that* + +*$$\begin{align*} + \left(b+c\right)a&=a\left(b+c\right),\ \text{Multiplication is commutative}\\ + &=ab+ac,\ \text{By part 1 of the theorem}\\ + &ba+ca,\ \text{Multiplication is commutative}\\ +\end{align*}$$* + +*As required. $\qed$* +::: + +##### Associativity of multiplication + + \ +This will prove that for all $a,b,c\in\mathbb{N}$ that +$a\left(bc\right)=\left(ab\right)c$ + +::: theorem +**Theorem 12**. *For all $a,b,c\in\mathbb{N}$ we have that +$a\left(bc\right)=\left(ab\right)c$* + +*Proof:* + +*We again show this by induction. Let $x,y\in\mathbb{N}$ be arbitrary, +and let $P\left(n\right)$ be the proposition given by* + +*$$\begin{equation*} + \left(xy\right)n=x\left(yn\right) +\end{equation*}$$* + +*For the base case we have $n=0$ and so* + +*$$\begin{align*} + \left(xy\right)0&=0 ,\text{By definition of multiplication}\\ + &=x\left(0\right),\text{By definition of multiplication}\\ + &=x\left(y*0\right),\text{By definition of multiplication}\\ +\end{align*}$$* + +*Hence $P\left(0\right)$ is true.* + +*Now, suppose that $P\left(n\right)$ is true, that is* + +*$$\begin{equation*} + \left(xy\right)n=x\left(yn\right) +\end{equation*}$$* + +*We show that $P\left(S\left(n\right)\right)$ is also true, that is* + +*$$\begin{equation*} + \left(xy\right)S\left(n\right)=x\left(yS\left(n\right)\right) +\end{equation*}$$* + +*Now, we have that* + +*$$\begin{align*} + \left(xy\right)S\left(n\right)&=\left(xy\right)n+xy,\ \text{Definition of multiplication}\\ + &=x\left(yn\right)+xy,\ \text{By assumption}\\ + &=xy+x\left(yn\right),\ \text{Addition is commutative}\\ + &=x\left(y+\left(yn\right)\right),\ \text{Multiplication is distributive over addition}\\ + &=x\left(\left(yn\right)+y\right),\ \text{Addition is commutative}\\ + &=x\left(yS\left(n\right)\right),\ \text{Addition is commutative}\\ +\end{align*}$$* + +*Hence $P\left(S\left(n\right)\right)$ is true.* + +*Hence, it follows by the principle of mathematical induction that for +all $a,b,c\in\mathbb{N}$ we have that +$a\left(bc\right)=\left(ab\right)c$. $\qed$* +::: + +##### The Zero and Identity laws + + \ +These two laws allow us to note that adding zero to any natural number +$n$ gives back $n$ and multiplying $n$ by $1$ gives $n$. + +::: theorem +**Theorem 13**. *The zero and Identity laws* + +*Let $n\in\mathbb{N}$. We have that* + +1. *$n+0=n=0+n$* + +2. *$1*n=n=n*1$* + +*Proof:* + +*By commutativity, it is enough to only prove* + +1. *$n+0=n$* + +2. *$n*1=n$* + + + +1. *$n+0=n$:* + + *This is true by the definition of addition.* + +2. *$n*1=n$:* + + *We have by the definition of multiplication that* + + *$$\begin{equation*} + n*1=n*S\left(0\right)=n*0+n=0+n=n + \end{equation*}$$ Where the last equality comes from the zero law + and the fact addition is commutative.* + +*The result follows. $\qed$* +::: + +##### The cancellation laws + + \ +These laws allow us to deduce that if $a+b=a+c$ then we must have $b=c$, +and if $a\neq 0$ that $ab=ac$ gives $b=c$ + +::: theorem +**Theorem 14**. *The cancellation laws* + +*Let $a,b,c\in\mathbb{N}$. We have that* + +1. *If $a+b=a+c$ then we have $b=c$.* + +2. *For $a\neq 0$, if $ab=ac$ then we have that $b=c$* + +*Proof:* + +1. *If $a+b=a+c$ then we have $b=c$:* + + *We argue by induction, let $b,c\in\mathbb{N}$ be arbitrary and let + $P\left(n\right)$ be the proposition given by* + + *$$\begin{equation*} + n+b=n+c \Rightarrow b=c + \end{equation*}$$* + + *For the base case $P\left(0\right)$ this holds trivially. Now + suppose the proposition $P\left(n\right)$ holds that is* + + *$$\begin{equation*} + n+b=n+c \Rightarrow b=c + \end{equation*}$$* + + *We show that $P\left(S\left(n\right)\right)$ holds, that is* + + *$$\begin{equation*} + S\left(n\right)+b=S\left(n\right)+c \Rightarrow b=c + \end{equation*}$$* + + *Now, we have that* + + *$$\begin{align*} + S\left(n\right)+b&=S\left(n\right)+c\\ + S\left(n+0\right)+b&=S\left(n+0\right)+c\\ + n+S\left(0\right)+b&=n+S\left(0\right)+c\\ + n+\left(S\left(0\right)+b\right)&=n+\left(S\left(0\right)+c\right),\ \text{By associativity}\\ + \left(S\left(0\right)+b\right)&=\left(S\left(0\right)+c\right),\ \text{By hypothesis, as $P\left(n\right)$ has $b,c$ being arbitrary}\\ + b+S\left(0\right)&=c+S\left(0\right),\ \text{By commutativity}\\ + S\left(b+0\right)&=S\left(c+0\right)\\ + S\left(b\right)&=S\left(c\right)\\ + \end{align*}$$* + + *Hence we have $b=c$ by proposition + [37](#prop:EqualSuccOp){reference-type="ref" + reference="prop:EqualSuccOp"}. So $P\left(S\left(n\right)\right)$ is + true.* + + *Hence by mathematical induction we have that if $a+b=a+c$ we must + have that $b=c$.* + +2. *For $a\neq 0$, if $ab=ac$ then we have that $b=c$:* + + *We again argue by induction, let $b,c\in\mathbb{N}$ be arbitrary + and let $P\left(n\right)$ be the proposition given by* + + *$$\begin{equation*} + nb=nc\Rightarrow b=c + \end{equation*}$$* + + *Moreover, we do induction starting at $n=1$ as the case $n=0$ is + vacuously true. So for $P\left(1\right)$ we have that this holds + trivially. Now suppose that $P\left(n\right)$ holds. that is* + + *$$\begin{equation*} + nb=nc\Rightarrow b=c + \end{equation*}$$* + + *We show that $P\left(S\left(n\right)\right)$ is true* + + *$$\begin{equation*} + S\left(n\right)b=S\left(n\right)c\Rightarrow b=c + \end{equation*}$$* + + *Indeed we have that* + + *$$\begin{align*} + S\left(n\right)b&=S\left(n\right)c\\ + bS\left(n\right)&=cS\left(n\right),\ \text{By commutativity}\\ + bn+b&=cn+c,\ \text{By commutativity}\\ + a+b&=a+c,\ nb=nc \text{ by assumption, so let } nb=nc=a \text{ for some } a\\ + b&=c,\ \text{By the cancellation law for addition}\\ + \end{align*}$$* + + *Hence $P\left(S\left(n\right)\right)$ is true.* + + *Hence by mathematical induction we have that for $a\neq 0$ if + $ab=ac$ we must have that $b=c$.* + +*As required. $\qed$.* +::: + +##### Summation and product notation + +Now that we have a well-defined notion of addition and multiplication we +can define a shorthand to can be useful in avoiding writing out longer +chains of additions (or multiplications) in certain situations. We will +require the following mapping. Let $s\in\mathbb{N}^{n+1}$ be an ordered +$n+1$-tuple of Natural numbers where +$s=\left(s_0,s_1,s_1,s_2,\dots,s_n\right)$ and define +$\mathbb{N}_n=\left\{0,1,2,3,\dots,n\right\}$. Let +$f:\mathbb{N}_n\rightarrow\mathbb{N}$ be a mapping defined by + +$$\begin{align*} + f:\mathbb{N}_n&\rightarrow\mathbb{N}\\ + i&\mapsto f\left(i\right) =s_i +\end{align*}$$ + +This is to say that $f$ simply gets the value of $s_i$ which is an +element of the ordered tuple $s$. + +::: definition +**Definition 72**. *Summation notation* + +*Let $s\in\mathbb{N}^{n+1}$ be an ordered $n+1$-tuple of Natural numbers +where $s=\left(s_0,s_1,s_1,s_2,\dots,s_n\right)$ and define +$\mathbb{N}_n=\left\{0,1,2,3,\dots,n\right\}$. Let +$f:\mathbb{N}_n\rightarrow\mathbb{N}$ be a mapping defined by* + +*$$\begin{align*} + f:\mathbb{N}_n&\rightarrow\mathbb{N}\\ + i&\mapsto f\left(i\right) =s_i +\end{align*}$$* + +*We define the summation notation by* + +*$$\begin{equation*} + \sum_{i=0}^n f\left(i\right)=f\left(0\right)+f\left(1\right)+f\left(2\right)+\dots+f\left(n\right) +\end{equation*}$$ This can also be written as* + +*$$\begin{equation*} + \sum_{i=0}^n s_i=s_0+s_1+s_2+\dots+s_n +\end{equation*}$$* + +*We call $i$ the index of the summation and that $i=0$ as the starting +index of the summation for some $a\in\mathbb{N}$ and that $n$ is the +ending index of the summation. In the case that $s\in\emptyset$ then we +define the summation to be $0$ and call such a summation an empty sum.* + +*We can also define the summation over a subset of $\mathbb{N}_n$ which +allows for starting the summation at a starting point other than $i=0$. +Let $T\subseteq\mathbb{N}$. We can define the summation over the set $T$ +by* + +*$$\begin{equation*} + \sum_{i\in T} s_i +\end{equation*}$$* + +*If we have a mapping $g:\mathbb{N}\rightarrow\mathbb{N}$ for some +mapping $g$ then we can define a summation over $g$ by +$$\begin{equation*} + \sum_{i\in T} g\left(s_i\right) +\end{equation*}$$* + +*Finally, we can define a summation over a predicate $P\left(i\right)$ +for $i\in T$ giving* + +*$$\begin{equation*} + \sum_{P\left(i\right)} g\left(s_i\right) +\end{equation*}$$ which means to take the sum of the $g\left(s_i\right)$ +where $i$ satisfies the predicate $P$. If the predicate is not satisfied +by any $i$ then the summation is also said to be an empty summation and +given a value of $0$.* + +*In light of definition a summation of a predicate we have that if $a>n$ +where $a$ is the index lower of summation and $n$ the upper point of +summation then the sum would be by definition equal to $0$. That is to +say* + +*$$\begin{equation*} + \sum_{i=a}^n s_i = 0 ,\ \text{If } a>n +\end{equation*}$$* +::: + +::: example +**Example 56**. *Let $s=\left(2,3,4,8\right)\in\mathbb{N}^4$ then we +have that* + +*$$\begin{equation*} + \sum_{i=0}^3 s_i = 2+3+4+8 = 17 +\end{equation*}$$* +::: + +::: example +**Example 57**. *Let $g\left(n\right)=n$ and let $k=4$ then we have +that* + +*$$\begin{equation*} + \sum_{i=0}^4-1 g\left(i\right) = \sum_{i=0}^3 i = 1+2+3+4 = 10 +\end{equation*}$$* +::: + +::: example +**Example 58**. *Let $s_1\in\mathbb{N}$ then we have* + +*$$\begin{equation*} + \sum_{i=1}^1 s_1 = s_1 +\end{equation*}$$* +::: + +::: example +**Example 59**. *Let $g\left(n\right) = n*n$ and let +$T=\left\{2,6,11\right\}\subseteq\mathbb{N}^{11}$ then* + +*$$\begin{equation*} + \sum_{i\in T} g\left(i\right) = g\left(2\right)+g\left(6\right)+g\left(11\right)=2*2+6*6+11*11=4+36+121=161 +\end{equation*}$$* +::: + +::: example +**Example 60**. *Let $g\left(n\right) = n$, let $P\left(n\right)$ be the +predicate such that* + +*$$\begin{equation*} + P\left(n\right)=\begin{cases} + 1,\ \text{If } n=2,4,6\\ + 0,\ \text{Otherwise } + \end{cases} +\end{equation*}$$ Let $T=\left\{2,6,11\right\}\subseteq\mathbb{N}^{11}$ +then we have for the $i\in T$ that satisfies $P\left(i\right)$ is given +by* + +*$$\begin{equation*} + \sum_{P\left(i\right)} i = 2+4=6 +\end{equation*}$$* +::: + +::: example +**Example 61**. *Let $f\left(n\right)= n+5$. Consider the sum* + +*$$\begin{equation*} + \sum_{i=3}^6 n+5 = \left(3+5\right)+\left(4+5\right)+\left(5+5\right)+\left(6+5\right)=8+9+10+11=38 +\end{equation*}$$* + +*We can re-express this sum as* + +*$$\begin{equation*} + \sum_{i=0}^3 n+5 = \left(\left(0+3\right)3+5\right)+\left(\left(1+3\right)+5\right)+\left(\left(2+3\right)+5\right)+\left(\left(3+3\right)+5\right)=38 +\end{equation*}$$* + +*We have re-indexed the sum into an equivalent form.* +::: + +We can make some observations about summation notation. + +::: {#prop:summation_properties_naturals .proposition} +**Proposition 38**. *Properties of summation notation* + +*Let $n,m\in\mathbb{N}$ such that $m 0 +\end{equation*}$$* + +*A contradiction to the hypothesis.* + +*A similar result holds for $\displaystyle ab = \sum_{i=1}^a b$. Finally +if both $a$ and $b$ are zero the result is trivial.* + +*The result has been shown. $\qed$.* +::: + +A similar definition can be made for multiplication, called product +notation + +::: definition +**Definition 73**. *Product notation* + +*Let $s\in\mathbb{N}^{n+1}$ be an ordered $n+1$-tuple of Natural numbers +where $s=\left(s_0,s_1,s_1,s_2,\dots,s_n\right)$ and define +$\mathbb{N}_n=\left\{0,1,2,3,\dots,n\right\}$. Let +$f:\mathbb{N}_n\rightarrow\mathbb{N}$ be a mapping defined by* + +*$$\begin{align*} + f:\mathbb{N}_n&\rightarrow\mathbb{N}\\ + i&\mapsto f\left(i\right) =s_i +\end{align*}$$* + +*We define the product notation by* + +*$$\begin{equation*} + \prod_{i=0}^n f\left(i\right)=f\left(0\right)*f\left(1\right)*f\left(2\right)*\dots*f\left(n\right) +\end{equation*}$$ This can also be written as* + +*$$\begin{equation*} + \prod_{i=0}^n s_i=s_*s_1*s_2*\dots*s_n +\end{equation*}$$* + +*We call $i$ the index of the product and that $i=0$ as the lower +starting point of the product for some $a\in\mathbb{N}$ and that $n$ is +the upper point of the product. In the case that $s\in\emptyset$ then we +define the product to be $1$ and call such a product an empty product.* + +*We can also define the product over a subset of $\mathbb{N}_n$ which +allows for starting the product at a starting point other than $i=0$. +Let $T\subseteq\mathbb{N}$. We can define the product over the set $T$ +by* + +*$$\begin{equation*} + \prod_{i\in T} s_i +\end{equation*}$$* + +*If we have a mapping $g:\mathbb{N}\rightarrow\mathbb{N}$ for some +mapping $g$ then we can define a product over $g$ by $$\begin{equation*} + \prod_{i\in T} g\left(s_i\right) +\end{equation*}$$* + +*Finally, we can define a product over a predicate $P\left(i\right)$ for +$i\in T$ giving* + +*$$\begin{equation*} + \sum_{P\left(i\right)} g\left(s_i\right) +\end{equation*}$$ which means to take the product of the +$g\left(s_i\right)$ where $i$ satisfies the predicate $P$. If the +predicate is not satisfied by any $i$ then the product is also said to +be an empty product and given a value of $1$. In light of definition a +product of a predicate we have that if $a>n$ where $a$ is the lower +index of the product and $n$ the upper point of product then the product +would be by definition equal to $1$. That is to say* + +*$$\begin{equation*} + \sum_{i=a}^n s_i = 1 ,\ \text{If } a>n +\end{equation*}$$* +::: + +::: example +**Example 62**. *Let $s=\left(2,3,4,8\right)\in\mathbb{N}^4$ then we +have that* + +*$$\begin{equation*} + \prod_{i=0}^3 s_i = 2*3*4*8 = 192 +\end{equation*}$$* +::: + +::: example +**Example 63**. *Let $g\left(n\right)=n$ and let $k=4$ then we have +that* + +*$$\begin{equation*} + \prod_{i=0}^{4-1} g\left(i\right) = \prod_{i=0}^3 i = 1*2*3*4 = 24 +\end{equation*}$$* +::: + +::: example +**Example 64**. *Let $s_1\in\mathbb{N}$ then we have* + +*$$\begin{equation*} + \prod_{i=1}^1 s_1 = s_1 +\end{equation*}$$* +::: + +::: example +**Example 65**. *Let $g\left(n\right) = n*n$ and let +$T=\left\{2,6,11\right\}\subseteq\mathbb{N}^{11}$ then* + +*$$\begin{equation*} + \prod_{i\in T} g\left(i\right) = g\left(2\right)*g\left(6\right)*g\left(11\right)=\left(2*2\right)+\left(6*6\right)+\left(11*11\right)=4*36*121=17424 +\end{equation*}$$* +::: + +::: example +**Example 66**. *Let $g\left(n\right) = n$, let $P\left(n\right)$ be the +predicate such that* + +*$$\begin{equation*} + P\left(n\right)=\begin{cases} + 1,\ \text{If } n=2,4,6\\ + 0,\ \text{Otherwise } + \end{cases} +\end{equation*}$$ Let $T=\left\{2,6,11\right\}\subseteq\mathbb{N}^{11}$ +then we have for the $i\in T$ that satisfies $P\left(i\right)$ is given +by* + +*$$\begin{equation*} + \sum_{P\left(i\right)} i = 2*4=12 +\end{equation*}$$* +::: + +There is an some immediate properties of product notation that are clear + +::: proposition +**Proposition 40**. *Properties of product notation* + +*Let $n,m\in\mathbb{N}$ such that $m0$ and $m>0$. We have by +definition of exponentiation that* + +*$$\begin{align*} + a^n*a^m=\prod_{i=1}^n a * \prod_{i=1}^m a=\underbrace{a*a*\dots*a}_{n\text{ times}}*\underbrace{a*a*\dots*a}_{m\text{ times}} + &=\underbrace{a*a\dots*a}_{n+m\text{ times}} =a^{n+m} +\end{align*}$$ as required. $\qed$* +::: + +We also have the following result that combines multiplying two numbers +and raising that result to a power. As an example consider +$\left(2*3\right)^2= 6^2=36$. Now consider $2^2=4$ and $3^2=9$ and we +clearly have $4*9=36$. The powers can come through to each of the +numbers of the multiplication. + +::: {#prop:ExponentiationPowerOfProductIsProductOfPowers .proposition} +**Proposition 45**. *Power of product is product of powers* + +*Let $a,b,n\in\mathbb{N}$. We have that* + +*$$\begin{equation*} + \left(a*b\right)^n=a^n*b^n +\end{equation*}$$* + +*Proof:* + +*If $n=0$ then $\left(a*b\right)^0=1$ by definition and $a^0*b^0=1$. So +suppose that $n>0$ then we have that* + +*$$\begin{align*} + \left(a*b\right)^n=\prod_{i=1}^n ab &= \underbrace{ab*ab*ab\dots*ab}_{n\text {times}}\\ + &= \left(\underbrace{a*a*a\dots*a}_{n\text {times}}\right)*\left(\underbrace{b*b*b\dots*b}_{n\text {times}}\right),\ \text{By commutativity of multiplication}\\ + &= a^n*b^n\\ +\end{align*}$$* + +*The proposition has been shown. $\qed$* +::: + +##### Subtraction + +We can define an operation that will allow us to at least partially undo +addition. To define this operation we need to make use of the less than +operator. + +::: definition +**Definition 75**. *Subtraction of natural number* + +*Let $n,m\in\mathbb{N}$ such that $m\leq n$. Let $d\in\mathbb{N}$ such +that $n=m+d$. We define subtraction by* + +*$$\begin{equation*} + d=n-m +\end{equation*}$$* + +*We call $d$ the difference between $n$ and $m$.* +::: + +There is an immediate result from the definition of subtraction + +::: {#prop:NaturalAddDifference .proposition} +**Proposition 46**. *$a+\left(b-c\right)=\left(a+b\right)-c$* + +*Let $a,b,c\in\mathbb{N}$ with $b\geq c$. We have that* + +*$$\begin{equation*} + a+\left(b-c\right)=\left(a+b\right)-c +\end{equation*}$$* + +*Proof:* + +*We argue by induction. Let $P\left(n\right)$ denote the proposition* + +*$$\begin{equation*} + a+\left(n-c\right)=\left(a+n\right)-c +\end{equation*}$$* + +*For the base case $n=0$ we have by definition $c=0$ and so* + +*$$\begin{equation*} + a+\left(0-0\right)=a=\left(a+0\right)-0 +\end{equation*}$$* + +*Now suppose that $P\left(n\right)$ holds, we show that +$P\left(n+1\right)$ is true that is* + +*$$\begin{equation*} + a+\left(\left(n+1\right)-c\right)=\left(a+\left(n+1\right)\right)-c +\end{equation*}$$* + +*We have that $n+1=\left(n+0\right)+1=n+\left(0+1\right)$ and so* + +*$$\begin{align*} + a+\left(\left(n+1\right)-c\right)&=a+\left(n+\left(0+1\right)-c\right)\\ + &=a+\left(n+\left(1-c\right)\right)\\ + &=\left(a+n\right)+1-c\\ + &=a+\left(n+1\right)-c +\end{align*}$$* + +*As required. $\qed$* +::: + +We immediately see that subtraction is not commutative that is +$a-b\neq b-a$ in fact it is not even defined for $b-a$ unless $b\geq a$ +but then it is not defined for $a-b$ and visa-versa. Likewise it is not +associative as for example $\left(8-4\right)-2=2$ but +$8-\left(4-2\right)=6$. We do however retain the fact that +multiplication is commutative over subtraction + +::: proposition +**Proposition 47**. *Multiplication distributes over subtraction* + +*Let $a,b,c\in\mathbb{N}$ with $b\geq c$ and let $a\in\mathbb{N}$. We +have that* + +1. *$a\left(b-c\right)=ab-ac$* + +2. *$\left(b-c\right)a=ba-ca=ab-ac$* + +*Proof:* + +1. *$a\left(b-c\right)=ab-ac$:* + + *Let $a\in\mathbb{N}$ be arbitrary. We argue by induction of the + proposition $P\left(n\right)$ given by* + + *$$\begin{equation*} + a\left(n-m\right)=an-am + \end{equation*}$$ where by definition $m\leq n$. For the base case + we have $P\left(0\right)$ we have that $n=m=0$ and so* + + *$$\begin{equation*} + a\left(0-0\right)=a*0=0=a*0-a*0 + \end{equation*}$$* + + *Showing the base case. Now suppose that $P\left(n\right)$ holds we + show that $P\left(n+1\right)$ is true, that is we show* + + *$$\begin{equation*} + a\left(\left(n+1\right)-m\right)=a\left(n+1\right)-am + \end{equation*}$$ where $m\leq \left(n+1\right)$. There are two + cases to consider $if m=n+1$ then we have* + + *$$\begin{equation*} + a\left(\left(n+1\right)-m\right)=a*0=0=a\left(n-1\right)-am + \end{equation*}$$* + + *Now suppose that $m<\left(n+1\right)$ then* + + *$$\begin{equation*} + a\left(\left(n+1\right)-m\right)=a\left(n+1\right)-am + \end{equation*}$$ by the induction hypothesis. The result follows by + induction.* + +2. *$\left(b-c\right)a=ba-ca=ab-ac$:* + + *As multiplication is commutative we have that* + + *$$\begin{align*} + \left(b-c\right)a&=a\left(b-c\right)\\ + &=ab-ac\\ + &=ba-ca + \end{align*}$$* + +*The result follows. $\qed$* +::: + +##### The principle of strong induction + + \ +The final property of the natural we shall look at is that of the +principle of strong induction, although as we will see, this is actually +equivalent to usual induction. There is one more version of induction +that is sometimes useful, this is the so-called principle of strong +induction, this is instead of assuming $P\left(n\right)$ is true and +showing that $P\left(n+1\right)$. We instead assume that for all +$n\leq k$ for some $k\in\mathbb{N}$ we have that $P\left(n\right)$ is +true for all $n\leq k$ and we show that this implies that +$P\left(k+1\right)$ is true. + +::: theorem +**Theorem 16**. *The principle of strong induction* + +*Let $P\left(n\right)$ be a proposition about a natural number +$n\in\mathbb{N}$. Moreover, suppose that* + +1. *$P\left(0\right)$ is true.* + +2. *$\forall k\in\mathbb{N}:P\left(0\right)\wedge P\left(1\right)\wedge P\left(2\right)\wedge\dots\wedge P\left(k\right)$ + all being true implies that $P\left(k+1\right)$ is true.* + +*If these two statements are true, we have that $P\left(n\right)$ is +true for any natural number $n$, and we say the proposition +$P\left(n\right)$ holds by the principle of strong mathematical +induction.* + +*Proof:* + +*Define $\Tilde{P}\left(n\right)$ to be the following proposition* + +*$$\begin{equation*} + \Tilde{P}\left(n\right)=P\left(0\right)\wedge P\left(1\right)\wedge P\left(2\right)\wedge\dots\wedge P\left(n\right) +\end{equation*}$$* + +*We show that $\Tilde{P}\left(n\right)$ for all $n\geq 0$. By assumption +$\Tilde{P}\left(n\right)$ is true as +$\Tilde{P}\left(n\right)=P\left(0\right)$. Now suppose that +$\Tilde{P}\left(n\right)$ is true for some $n\in\mathbb{N}$, that is* + +*$$\begin{equation*} + \Tilde{P}\left(n\right)=P\left(0\right)\wedge P\left(1\right)\wedge P\left(2\right)\wedge\dots\wedge P\left(n\right) +\end{equation*}$$ is true, we show that $\Tilde{P}\left(n+1\right)$ is +true, that is* + +*$$\begin{equation*} + \Tilde{P}\left(n+1\right)=P\left(0\right)\wedge P\left(1\right)\wedge P\left(2\right)\wedge\dots\wedge P\left(n\right)\wedge P\left(n+1\right) +\end{equation*}$$* + +*By assumption 2. as we have that +$\forall n\in\mathbb{N}:P\left(0\right)\wedge P\left(1\right)\wedge P\left(2\right)\wedge\dots\wedge P\left(n\right)$ +implies that $P\left(n+1\right)$ is true. Hence we have that* + +*$$\begin{equation*} + \Tilde{P}\left(n+1\right)=\Tilde{P}\left(n\right)\wedge P\left(n+1\right)=\Tilde{P}\left(n+1\right) +\end{equation*}$$ is true.* + +*Hence by the principle of mathematical induction we have that +$\Tilde{P}\left(n\right)$ is true for all $n\geq 0$. $\qed$* +::: + +As mentioned earlier, we said that strong induction and the usual +induction are equivalent, we shall prove this. We used induction to +prove strong induction so it is left to show that given the assumptions +for strong induction, we can deduce the truth $\forall n\in\mathbb{N}$ +of the proposition $P\left(n\right)$ only using induction. + +::: theorem +**Theorem 17**. *Strong induction is equivalent to the usual induction* + +*Suppose that the assumptions of strong induction hold. That is suppose +$P\left(n\right)$ be a proposition about a natural number +$n\in\mathbb{N}$ and moreover suppose that* + +1. *$P\left(0\right)$ is true.* + +2. *$\forall k\in\mathbb{N}:P\left(0\right)\wedge P\left(1\right)\wedge P\left(2\right)\wedge\dots\wedge P\left(k\right)$ + all being true implies that $P\left(k+1\right)$ is true.* + +*We have that the truth of $P\left(n\right)$ for all $n\in\mathbb{N}$ +can be deduced using only regular induction.* + +*Proof:* + +*Let $\Tilde{P}\left(n\right)$ be the proposition be given by* + +*$$\begin{equation*} + \forall k\leq n\text{ we have } P\left(k\right) \text{ is true} +\end{equation*}$$* + +*We show by the principle of induction that* + +1. *$\Tilde{P}\left(0\right)$ is true* + +2. *$\Tilde{P}\left(n\right)$ being true implies + $\Tilde{P}\left(n+1\right)$ is true for any natural number $n$.* + + + +1. *$\Tilde{P}\left(0\right)$ is true:* + + *To see this, we have that $\Tilde{P}\left(0\right)$ is given by* + + *$$\begin{equation*} + \forall k\leq 0\text{ we have } P\left(0\right) \text{ is true} + \end{equation*}$$* + + *This clearly holds as the only natural number that is less than or + equal to zero is zero. Hence $P\left(0\right)$ is true and so + $\Tilde{P}\left(0\right)$.* + +2. *$\Tilde{P}\left(n\right)$ being true implies + $\Tilde{P}\left(n+1\right)$ is true for any Natural number $n$:* + + *Suppose that $\Tilde{P}\left(n\right)$ is true, that is* + + *$$\begin{equation*} + \forall k\leq n\text{ we have } P\left(k\right) \text{ is true} + \end{equation*}$$* + + *we show that $\Tilde{P}\left(n+1\right)$ is true, that is* + + *$$\begin{equation*} + \forall k\leq n+1\text{ we have } P\left(k\right) \text{ is true} + \end{equation*}$$* + + *Let $k\leq n+1$ be a natural number, have two cases to consider.* + + 1. *If $ka$* + +3. *If $a\leq b$ and $b\leq c$ then $a\leq c$* + +4. *If $ab$ and $b\geq c$ then $a>c$* + +9. *If $a\geq b$ and $b>c$ then $a>c$* + +10. *If $a>b$ and $b>c$ then $a>c$* + +11. *If $a\leq b$ then $a+c\leq b+c$* + +12. *If $ab$ then $a+c>b+c$* + +15. *If $a\leq b$ then $ac\leq bc$* + +16. *If $ab$ then $ac>bc$* + +*Proof:* + +1. *$a\leq b$ is the same as $b\geq a$:* + + *Suppose that $a\leq b$ then by definition of $a\leq b$ we have that + $a\subseteq b$. We then clearly have that $b\not\subset a$ and so + either $b>a$ by definition or $b=a$. In other words $b\geq a$.* + +2. *$aa$:* + + *Similar to the first part. If $aa$ by definition of greater than.* + +3. *If $a\leq b$ and $b\leq c$ then $a\leq c$:* + + *Suppose that $a\leq b$ and $b\leq c$. By definition, we have that + $a\subseteq b$ and $b\subseteq a$ and so by proposition + [2](#prop:SetInclusionTransitivityProp){reference-type="ref" + reference="prop:SetInclusionTransitivityProp"} we have + $a\subseteq c$ which is to say $a\leq c$.* + +4. *If $ab$ and $b\geq c$ then $a>c$:* + + *Applying part 2. of this proposition to $a>b$ and $a>c$ and part 1. + to $b\geq c$ gives the equivalent statement $bc$ then $a>c$:* + + *Applying part 1. of this proposition to $a\geq b$ and part 1. to + $b>c$ and $a>c$ gives the equivalent statement $b\leq a$ and $c< b$ + then $cb$ and $b>c$ then $a>c$:* + + *Applying part 2. to $a>b$, $b>c$ and $c>a$ gives the equivalent + statement $bb$ then $a+c>b+c$:* + + *Applying part 2. of the proposition give the equivalent statement + of $b< a$ then $b+c< a+c$ and so we can apply part 11.* + +14. *If $a\geq b$ then $a+c\geq b+c$:* + + *Applying part 1. of the proposition give the equivalent statement + of $b\leq a$ then $b+c\leq a+c$ and so we can apply part 12.* + +15. *If $ab$ then $ac>bc$:* + + *Applying part 2. of the proposition gives the equivalent statement + of $b< a$ then $bc + +1. *$P\left(0\right)$ is true:* + + *We have that $\left|T\right|=0$ and so $T=\emptyset$. As + $T=\emptyset$ then there are no subsets $S\subset \emptyset$ for if + there were then $T\neq\emptyset$. Hence the base case is vacuously + true.* + +2. *If $P\left(n\right)$ is true then $P\left(n+1\right)$ is true:* + + *Suppose that $P\left(n\right)$ holds for some $n\in\mathbb{N}$ + which is the statement* + + *$$\begin{equation*} + \text{If }T\text{ is a finite set with } S\subset T \text{ and }\left|T\right|=n \text{ then } S\text{ is a finite set and} \left|S\right|<\left|T\right|=n + \end{equation*}$$ We need to show that $P\left(n+1\right)$ also + holds that is we show that* + + *$$\begin{equation*} + \text{If }T\text{ is a finite set with } S\subset T \text{ and }\left|T\right|=n+1 \text{ then } S\text{ is a finite set and} \left|S\right|<\left|T\right|=n+1 + \end{equation*}$$* + + *So suppose that $\left|T\right|=n+1$ for some $n\in\mathbb{N}$ such + that $S\subset T$. As $S$ is a strict subset of $T$ we know that + $\exists t\in T$ with $t\not\in S$. Hence we have that + $S\subseteq T\setminus\left\{t\right\}$. We need to now show that + $\left|T\setminus\left\{t\right\}\right|=n$* + + ::: lemma + ***Lemma 4**. *Set of cardinality $n+1$ minus an element has + cardinality $n$** + + Let $S$ be a finite set with cardinality $n+1$. Consider the set + $S\setminus\left\{s\right\}$ where $s\in S$ is an arbitrary element + of $S$. We have that $\left|S\setminus\left\{s\right\}\right|=n$ + + Proof: + + We need to show that for the set $S\setminus\left\{s\right\}$ that + there exists a bijective mapping to a set of $n$ elements. We know + that $S$ has cardinality $n+1$, hence there exists a bijection + $f:S\rightarrow n+1$. We know by construction that + $n+1=n\cup\left\{n\right\}$, hence we have that + $n=n+1\setminus\left\{n\right\}$. + + Consider the mapping given by $g$ defined as follows + + $$\begin{align*} + g:S\setminus\left\{s\right\}&\rightarrow n=n+1\setminus\left\{n\right\}\\ + x&\mapsto g\left(x\right)=\begin{cases} + f\left(x\right): \text{If }f\left(x\right)\neq \left\{n\right\}\\ + f\left(s\right): \text{If }f\left(x\right)=\left\{n\right\} + \end{cases} + \end{align*}$$ + + This is to say $g$ is a mapping that takes each $x\in S$ and maps it + to $f\left(x\right)$ if + $f\left(x\right)\neq \left\{n\right\}\in n+1$, that is if $f$ + doesn't map $x$ to the removed element of the set $n+1$, otherwise + if $f$ does map an element $x\in S$ to $\left\{n\right\}$ then $g$ + maps $x$ to whatever $f$ takes the removed element $s$ to. + + For example suppose that $S=\left\{0,1,2\right\}$, i.e we are + considering the case $n=2$, let $f:S\rightarrow 3$ be the identity + mapping, this is a bijection. Suppose we now consider + $S\setminus\left\{2\right\}=\left\{0,1\right\}$ and consider the + mapping $g:S\setminus\left\{2\right\}\rightarrow 2$ given by + + $$\begin{align*} + g:S\setminus\left\{2\right\}&\rightarrow 2=3\setminus\left\{2\right\}=\left\{\emptyset,\left\{\emptyset\right\}\right\}\\ + x&\mapsto g\left(x\right)=\begin{cases} + f\left(x\right): \text{If }f\left(x\right)\neq \left\{2\right\}\\ + f\left(2\right): \text{If }f\left(x\right)=\left\{2\right\} + \end{cases} + \end{align*}$$ We have that $g\left(0\right)=0=\emptyset$ and + $g\left(1\right)=1=\left\{\emptyset\right\}$. We could have instead + considered $S\setminus\left\{1\right\}=\left\{0,2\right\}$ again + with $f$ being the identity mapping. We have that in this case $g$ + is the mapping given by + + $$\begin{align*} + g:S\setminus\left\{1\right\}&\rightarrow 2=3\setminus\left\{2\right\}=\left\{\emptyset,\left\{\emptyset\right\}\right\}\\ + x&\mapsto g\left(x\right)=\begin{cases} + f\left(x\right): \text{If }f\left(x\right)\neq \left\{2\right\}\\ + f\left(1\right): \text{If }f\left(x\right)=\left\{2\right\} + \end{cases} + \end{align*}$$ In this case we have that + $g\left(0\right)=0=\emptyset$ but + $g\left(2\right)=f\left(1\right)=1=\left\{\emptyset\right\}$. + + Now, we need to show the general case where $g$ is given by + + $$\begin{align*} + g:S\setminus\left\{s\right\}&\rightarrow n=n+1\setminus\left\{n\right\}\\ + x&\mapsto g\left(x\right)=\begin{cases} + f\left(x\right): \text{If }f\left(x\right)\neq \left\{n\right\}\\ + f\left(s\right): \text{If }f\left(x\right)=\left\{n\right\} + \end{cases} + \end{align*}$$ is a bijection. + + 1. $g$ is an injection: + + To see that $g$ is an injection, suppose that + $x,y\in S\setminus\left\{s\right\}$ and that $x\neq y$. There + are three cases to consider. + + 1. $f\left(x\right)\neq \left\{n\right\}$ and + $f\left(y\right)\neq\left\{n\right\}$: + + We have by definition of the mapping $g$ that + $f\left(x\right)=g\left(x\right)$ and + $f\left(y\right)=g\left(y\right)$. Moreover we know that $f$ + is a bijection and in particular an injection, hence as + $f\left(x\right)\neq f\left(y\right)$ we must have that + $g\left(x\right)\neq g\left(y\right)$ + + 2. $f\left(x\right)=\left\{n\right\}$: + + By the definition of the mapping $g$ we have that + $g\left(x\right)=f\left(s\right)$. Now, recall that + $y\in S\setminus\left\{s\right\}$, thus it follows that + $y\neq s$. Now, by the injectivity of $f$ we have that + $f\left(y\right)\neq f\left(s\right)=g\left(x\right)$. + Moreover by the injectivity of $f$ we have that + $f\left(y\right)\neq \left\{n\right\}$. It now follows by + definition of $g$ that + + $$\begin{equation*} + g\left(y\right)=f\left(y\right)\neq f\left(x\right)=g\left(x\right) + \end{equation*}$$ That is + $g\left(y\right)\neq g\left(x\right)$. + + 3. $f\left(y\right)=\left\{n\right\}$: + + This is the same as $f\left(x\right)=\left\{n\right\}$ + except the roles of $x$ and $y$ are swapped, for + completeness we give the details. + + By the definition of the mapping $g$ we have that + $g\left(y\right)=f\left(s\right)$. Now, as + $x\in S\setminus\left\{s\right\}$ it follows that $x\neq s$. + By the injectivity of $f$ we have that + $f\left(x\right)\neq f\left(s\right)=g\left(y\right)$. + Moreover by the injectivity of $f$ we have that + $f\left(x\right)\neq \left\{n\right\}$. It now follows by + definition of $g$ that + + $$\begin{equation*} + g\left(x\right)=f\left(x\right)\neq f\left(y\right)=g\left(y\right) + \end{equation*}$$ That is + $g\left(y\right)\neq g\left(x\right)$. + + This shows that $g$ is an injection. + + 2. $g$ is a surjection: + + We need to show that $\forall y\in n,\exists x\in S$ such that + $g\left(x\right)=y$. Let $y\in n$. We know that $f$ is a + bijection and in particular it is a surjection and so by + definition we know we must have + + $$\begin{equation*} + \forall y\in n+1,\exists x\in S : f\left(x\right)=y + \end{equation*}$$ + + Consider the definition of $g$. We know that + $g:S\setminus\left\{s\right\}\rightarrow n$, hence to show that + $g$ is surjective we need to show that any $y\in n$ has an + element $x'\in S$ with $f\left(x'\right)=y$. Moreover as $S$ + doesn't have the element $s$ we can't use $x=s$ in the + surjectivity of $f$ to show surjectivity of $g$. + + $$\begin{equation*} + \forall y\in n=n+1\setminus\left\{n\right\},\exists x'\in S\setminus\left\{s\right\}: x\neq a\text{ and } f\left(x'\right)=y + \end{equation*}$$ + + Finally, we need to consider $f\left(s\right)$ and in particular + the two cases of $f\left(s\right)\neq\left\{n\right\}$ and + $f\left(s\right)=\left\{n\right\}$, from the definition of $g$. + + 1. $f\left(s\right)\neq\left\{n\right\}$: + + Suppose that $f\left(s\right)\neq\left\{n\right\}$. As $f$ + is a bijection we have that $f$ is invertible, in particular + we must have that + $f^{-1}\left(\left\{n\right\}\right)\neq s$. There are two + additional cases to consider now, + $f\left(s\right)=y=f\left(x\right)$ and + $f\left(s\right)\neq y=f\left(x\right)$. + + 1. $f\left(s\right)=y=f\left(x\right)$: + + Suppose that $f\left(s\right)=y$, by definition of $g$ + we have that + + $$\begin{equation*} + g\left(f^{-1}\left(\left\{n\right\}\right)\right)=y + \end{equation*}$$ as + $f^{-1}\left(\left\{n\right\}\right)\neq s$. So let + $x'=f^{-1}\left(\left\{n\right\}\right)$. + + 2. $f\left(s\right)\neq y=f\left(x\right)$: + + Suppose that $f\left(s\right)\neq y$, by assumption of + surjectivity of $f$ we have that $f\left(x\right)=y$. + Hence $f\left(s\right)\neq f\left(x\right)$ and so by + injectivity of $f$ we have that $x\neq s$., hence we can + simply take $x'=x$, + + 2. $f\left(s\right)=\left\{n\right\}$: + + Now suppose that $f\left(s\right)=\left\{n\right\}$ We know + that $\left\{n\right\}\not\in n$ and so by assumption we + have that $f\left(x\right)=y\neq \left\{n\right\}$. Thus we + conclude that $x\neq s$ so we let $x'=x$ + + In each case we have found a valid choice for $x'$ and so + surjectivity has been shown. + + It follows that $g:S\setminus\left\{s\right\}\rightarrow n$ is a + bijection and by definition of set cardinality we conclude that + $S\setminus\left\{s\right\}$ has cardinality $n$. As required. + $\qed$ + ::: + + *Now, by the lemma we have that $T\setminus\left\{t\right\}$ is set + of cardinality $n$. Now if $S=T\setminus\left\{t\right\}$ then + $\left|S\right|=n + +1. *$\forall x,y\in K$ we have that $x\cap y=\emptyset$ whenever + $x\neq y$* + + *We can make use of the fact that $g$ is a bijection. If + $g\left(x\right)=g\left(y\right)$ then $x=y$ and so + $x\cap y=x=y\neq\emptyset$. Now if + $g\left(x\right)\neq g\left(y\right)$ then $x\neq y$ say + $x=\left\{s_1\right\}\times T$ and $y=\left\{s_2\right\}\times T$ + with $s_1\neq s_2$. It follows that $x\cap y = \emptyset$.* + +2. *$\forall x\in K$ we have that* + + *$$\begin{equation*} + S\times T=\bigcup_{x\in K} x + \end{equation*}$$* + + *By definition we have that any $x\in K$ has the form + $\left\{s\right\}\times T$ where $s\in S$. Let + $y\in \left\{s\right\}\times T$ then $y=\left(s,t\right)$ for some + $t\in T$ and so $y\in S\times T$ therefore* + + *$$\begin{equation*} + \bigcup_{x\in K} x\subseteq S\times T + \end{equation*}$$* + + *Likewise suppose that $x\in S\times T$ then $x = \left(s,t\right)$ + for some $s\in S$ and $t\in T$. This implies that + $x\in \left\{s\right\}\times T$ and as + $\left\{s\right\}\times T\in K$ then $x\in K$ so that* + + *$$\begin{equation*} + S\times T\subseteq\bigcup_{x\in K} x + \end{equation*}$$* + + *It follows that* + + *$$\begin{equation*} + S\times T=\bigcup_{x\in K} x + \end{equation*}$$ for all $x\in K$.* + +3. *$\forall x\in K$ we have that $x\neq \emptyset$* + + *Let $x\in K$ then $x\neq\emptyset$ as $S\neq\emptyset$ and + $T\neq\emptyset$. Hence $\forall x\in K$ $x\neq \emptyset$.* + +*It follows that $K$ partitions $S\times T$. Now as $K$ is a set +containing $n$ elements and $K$ partitions $S\times T$ and each element +of $K$ is a set containing $m$ elements. We have that the cardinality of +$S\times T$ is the sum of the cardinalities of each set $x\in K$ which +is $m*n$. That is to say* + +*$$\begin{equation*} + \left|S\times T\right|=nm +\end{equation*}$$ and the result is shown. $\qed$* +::: + +#### Countability + +::: definition +**Definition 78**. *Countable Set* + +*Let $S$ be a set. Let $T\subseteq\mathbb{N}$ allowing for the +possibility that $T=\mathbb{N}$. We say that $S$ is a countable set if +and only if the mapping $f:S\rightarrow T$ is a bijection.* + +*If $T$ is a finite subset of $\mathbb{N}$ we say that $S$ is a finitely +countable set and thus countable. If $T=\mathbb{N}$ we say that $S$ is a +countably infinite set. If $S$ is not a finitely countable set or a +countably infinite set we say that $S$ is a uncountably infinite set.* +::: + +Informally, a set $S$ is finitely countable or countably infinite if we +have some process for which we can enumerate each element of $S$, that +is to say list out each element in some way. We have an immediate +result. We can make the notion of an enumeration rigorous + +::: definition +**Definition 79**. *Enumeration* + +*Let $S$ be a finitely countable set with cardinality $\left|S\right|=n$ +and define $\mathbb{N}_n=\left\{1,2,3,\dots,n\right\}$ for some +$n\in\mathbb{N}$. We define an enumeration of $S$ to be a bijective +mapping $f:\mathbb{N}_n\rightarrow S$ or a bijective mapping +$g:S\rightarrow\mathbb{N}_n$.* + +*If $S$ is a countably infinite we define an enumeration of $S$ to be +the bijection $f:\mathbb{N}\rightarrow S$ or a bijective mapping +$g:S\rightarrow\mathbb{N}$.* +::: + +It is clear that in either case the if $f$ is a enumeration of a +countable set $S$ then so is $f^{-1}$ is also an enumeration of $S$. + +::: proposition +**Proposition 53**. *Inverse of an enumeration mapping is an enumeration +mapping* + +1. *Let $S$ be a finitely countable set with cardinality + $\left|S\right|=n$ have enumeration $f:\mathbb{N}_n\rightarrow S$ + then $f^{-1}:S\rightarrow\mathbb{N}_n$ is an enumeration of $S$ + where $f$ and $f^{-1}$ define the same enumeration of the elements + of $S$* + +2. *Let $S$ be a countable set have enumeration + $f:\mathbb{N}\rightarrow S$ then $f^{-1}:S\rightarrow\mathbb{N}$ is + an enumeration of $S$ where $f$ and $f^{-1}$ define the same + enumeration of the elements of $S$* + +*Proof:* + +1. *Let Let $S$ be a finitely countable set with cardinality + $\left|S\right|=n$ have enumeration $f:\mathbb{N}_n\rightarrow S$ + then $f^{-1}:S\rightarrow\mathbb{N}_n$ is an enumeration of $S$ + where $f$ and $f^{-1}$ define the same enumeration of the elements + of $S$:* + + *As $f$ is a bijection then it has an inverse + $f^{-1}:S\rightarrow\mathbb{N}_n$ which is also a bijection. Hence + $f^{-1}$ is an enumeration. To show that $f$ and $f^{-1}$ define the + same enumeration of the elements of $S$ we note that + $f\circ f^{-1}=\mathop{\mathrm{id}}_{\mathbb{N}_n}$ and + $f^{-1}\circ f = \mathop{\mathrm{id}}_S$.* + +2. *Let Let $S$ be a countable set have enumeration + $f:\mathbb{N}\rightarrow S$ then $f^{-1}:S\rightarrow\mathbb{N}$ is + an enumeration of $S$ where $f$ and $f^{-1}$ define the same + enumeration of the elements of $S$:* + + *As $f$ is a bijection then it has an inverse + $f^{-1}:S\rightarrow\mathbb{N}$ which is also a bijection. Hence + $f^{-1}$ is an enumeration. To show that $f$ and $f^{-1}$ define the + same enumeration of the elements of $S$ we note that + $f\circ f^{-1}=\mathop{\mathrm{id}}_{\mathbb{N}}$ and + $f^{-1}\circ f = \mathop{\mathrm{id}}_S$.* + +*The result is shown. $\qed$* +::: + +::: proposition +**Proposition 54**. *The natural numbers are countably infinite* + +*We have that $\mathbb{N}$ is a countably infinite set.* + +*Proof:* + +*To show that $\mathbb{N}$ is countable we need to find a bijective +mapping $f:\mathbb{N}\rightarrow\mathbb{N}$. We can clearly take +$\mathop{\mathrm{id}}_\mathbb{N}$, that is the identity mapping on +$\mathbb{N}$. That is to say* + +*$$\begin{align*} + \mathop{\mathrm{id}}_\mathbb{N}:\mathbb{N}&\rightarrow\mathbb{N}\\ + x&\mapsto\mathop{\mathrm{id}}_\mathbb{N}\left(x\right)=x +\end{align*}$$ As required. $\qed$* +::: + +We also have the following immediate result. + +::: proposition +**Proposition 55**. *Any subset of $\mathbb{N}$ is countable* + +*Let $S\subseteq\mathbb{N}$ then $S$ is countable.* + +*Proof:* + +*Let $S\subseteq\mathbb{N}$ and suppose that $S$ is not finite, for if +it is by definition it is countable. As $\mathbb{N}$ is well-ordered we +have by theorem [18](#thm:WOP){reference-type="ref" reference="thm:WOP"} +that $S$ is well-ordered and so have a set inclusion minimal element say +$s_0$. As $S$ is infinite then $S\setminus\left\{s_0\right\}$. We will +use this as the basis for induction.* + +*Suppose we have +$s_n\in S\setminus\left\{s_0,s_1,s_2,\dots,s_{n-1}\right\}$ then another +application of the well-order principle means there is some set +inclusion minimal element $s_{n+1}$ with +$s_{n+1}\in S\setminus\left\{s_0,s_1,s_2,\dots,s_n\right\}$. This holds +for all $n\in\mathbb{N}$ and so we conclude that +$S=\left\{s_0,s_1,s_2,\dots\right\}$ is countable by defining the +bijective mapping mapping* + +*$$\begin{align*} + f:\mathbb{N}&\rightarrow S\\ + x&\mapsto f\left(x\right)=s_x +\end{align*}$$* + +*The result follows. $\qed$* +::: + +::: proposition +**Proposition 56**. *The empty-set is countable* + +*We have that $\emptyset$ is a countable set.* + +*Proof:* + +*The empty-set has cardinality $0$ which is finite. $\qed$* +::: + +There are some results that can be deduced which give equivalent +conditions for a set to be countable. Two of these results follow by +definition of a countable set. + +::: {#prop:EquivalelntDefinitionsOfCountable .proposition} +**Proposition 57**. *Equivalence definitions of a countable set* + +*Let $S$ be a set. The following hold.* + +1. *$S$ is countable if and only if there is an injection + $f:S\rightarrow T$ for some subset $T\subseteq\mathbb{N}$* + +2. *$S$ is countable if and only if $S=\emptyset$ or there is a + surjection $f:T\rightarrow S$ for some subset + $T\subseteq\mathbb{N}$* + +*Proof:* + +1. *$S$ is countable if and only if there is an injection + $f:S\rightarrow T$ for some subset $T\subseteq\mathbb{N}$:* + + *$\left(\Rightarrow\right)$: Suppose that $S$ is countable then by + definition there is a bijection $f:S\rightarrow T$ for some + $T\subseteq\mathbb{N}$. As $f$ is a bijection then $f$ is an + injection and we are done.* + + *$\left(\Leftarrow\right)$: Suppose that there is an injection + $f:S\rightarrow T$ for some $T\subseteq\mathbb{N}$. Consider the + mapping $g:S\rightarrow\mathop{\mathrm{Image}}\left(f\right)$. By + proposition + [15](#prob:RestOfCodomainToImageIsSurjective){reference-type="ref" + reference="prob:RestOfCodomainToImageIsSurjective"} we have that $g$ + is a surjection. By definition of a surjection we have that + $\forall y\in\mathop{\mathrm{Image}}\left(f\right)$ there is some + $x\in S$ such that $f\left(x\right)=y$. It follows that $g$ is a + bijection as $g$ is also an injection by definition of the image of + a mapping. Therefore + $\left|S\right|=\left|\mathop{\mathrm{Image}}\left(f\right)\right|$ + and as + $\mathop{\mathrm{Image}}\left(f\right)\subseteq T\subseteq\mathbb{N}$ + we have that $S$ is countable.* + +2. *$S$ is countable if and only if $S=\emptyset$ or there is a + surjection $f:T\rightarrow S$ for some subset + $T\subseteq\mathbb{N}$:* + + *$\left(\Rightarrow\right)$: Suppose that $S$ is countable then + there is a bijection $f:T\rightarrow S$ and by definition is + therefore a surjection.* + + *$\left(\Leftarrow\right)$: Suppose that $f:T\rightarrow S$ is a + surjection. If $S=\emptyset$ then $f:T\rightarrow S$ is vacuously + injective and surjective and therefore + $\left|S\right|=\left|\emptyset\right|=\left|T\right|$ and therefore + countable. So suppose that $S\neq\emptyset$. By proposition + [14](#prop:PropertyImagePreImage){reference-type="ref" + reference="prop:PropertyImagePreImage"} we have for any mapping + $g:X\rightarrow Y$ that the pre-image of $g^{-1}\left(Y\right)=X$, + therefore $f^{-1}\left(S\right)=T$. By assumption + $T\subseteq \mathbb{N}$ and is therefore either finite or some + countably infinite subset of $\mathbb{N}$ possibly being + $\mathbb{N}$ itself. If $T$ is finite then we have that + $\left|S\right|\leq\left|T\right|$ by definition of $f$ being + surjective and there for $\left|S\right|$ is finite and therefore + countable. So suppose that $\left|T\right|=\aleph_0$ then $T$ is + either a countable subset of $\mathbb{N}$ or $\mathbb{N}$ itself.* + + *Let $g:T\rightarrow\mathbb{N}$ be a bijection then + $g^{-1}:\mathbb{N}\rightarrow T$ is an bijection by proposition + [35](#prop:InverseBijectionIsBijection){reference-type="ref" + reference="prop:InverseBijectionIsBijection"} and we have that + $f\circ g^{-1}:\mathbb{N}\rightarrow S$ is a surjection by + proposition + [20](#prop: PropInjecSurjecBijecMapping){reference-type="ref" + reference="prop: PropInjecSurjecBijecMapping"}. It is left to show + that $f\circ g^{-1}$ being surjective implies $S$ is countable. + Proposition + [28](#prop:RightInverseIffSurjective){reference-type="ref" + reference="prop:RightInverseIffSurjective"} gives that + $f\circ g^{-1}$ being surjective means there exists a right inverse + $h$ such that $h:S\rightarrow \mathbb{N}$. By proposition + [30](#RightInverseOfSurjecctionisInection){reference-type="ref" + reference="RightInverseOfSurjecctionisInection"} we have that $h$ is + injective. It follows by part 1 that $S$ is countable.* + +*The result is shown. $\qed$* +::: + +::: proposition +**Proposition 58**. *Set is countable if cardinality of set equals +cardinality of a countable set* + +*Let $S,T$ be sets such that $\left|S\right|=\left|T\right|$ then if $S$ +is countable so is $T$.* + +*Proof:* + +*Suppose that $S$ is countable. We have that as +$\left|S\right|=\left|T\right|$ then there exists a bijection +$f:S\rightarrow T$, in particular there exists a bijection +$g:T\rightarrow S$. Now as $S$ is countable there exists and injection +$h:S\rightarrow\mathbb{N}$. Now as $g$ is a bijection we have that $g$ +is an injection. The mapping $h\circ g:T\rightarrow \mathbb{N}$ is an +injection as $h$ and $g$ are. Hence as $h\circ g$ is an injection it +follows that $T$ is countable by proposition +[57](#prop:EquivalelntDefinitionsOfCountable){reference-type="ref" +reference="prop:EquivalelntDefinitionsOfCountable"}. $\qed$* +::: + +#### Relations + +##### Definition of a relation + +So far we have seen a few notations that relate elements of a set to +another. An example that relates elements of a set is equality of +natural numbers, two natural numbers are equal if and only if there are +the same element. Another example that we have seen on the natural +numbers is the less than operator $<$. A natural number $x$ is less than +$y$ if and only if $x\subseteq y$. A more fundamental example of a +relation is that of a mapping $f:S\rightarrow T$. We can consider a +function as relating any $s\in S$ and $t\in T$ to the pair +$\left(s,t\right)$ where $f\left(s\right)=t$. + +In a sense, we have that the idea of relations is somehow as fundamental +as sets and mappings, in fact we just described a mapping as some form +of relation so the idea of relations is more fundamental than that of a +mapping. Using the examples of the comparison operators on $\mathbb{N}$ +we can motivate a definition for a relation. + +::: definition +**Definition 80**. *Relation* + +*Let $S$ be a set and consider the Cartesian product $S\times S$. A +relation is a subset $R\subseteq S\times S$. We write an element +$\left(a,b\right)\in R$ as $aRb$ or we also write $a\sim b$ and we say +that $a$ relates to $b$. If $\left(a,b\right)\not\in R$ we write +$a\slashed{R} b$ or we write $a\not\sim b$* +::: + +We can recast the ideas at the start of this section into the language +of relations. + +::: example +**Example 72**. *Consider equality on $\mathbb{N}$. We can define +equality as a relation $\mathbb{N}\times \mathbb{N}$ where $a\sim b$ if +and only if $a\subseteq b$ and $b\subseteq a$. Explicitly we have that +$R$ is a subset of $\mathbb{N}\times\mathbb{N}$ given by* + +*$$\begin{equation*} + R=\left\{\left(0,0\right),\left(1,1\right),\left(2,2\right),\dots\right\} +\end{equation*}$$* +::: + +::: example +**Example 73**. *Consider the less than operator on $\mathbb{N}$. We +have that the less than operator is a relation where $a\sim b$ is given +by $a\subset b$. To see this consider $T=\left\{0,1,2\right\}$. Then the +less than relation on $T$ is given by the relation* + +*$$\begin{equation*} + R=\left\{\left(0,1\right),\left(0,2\right),\left(1,2\right)\right\} +\end{equation*}$$* +::: + +::: example +**Example 74**. *Let $S=\left\{0,1\right\}\subseteq\mathbb{N}$ and +define $T=P\left(S\right\}$ be the power set of $S$ given by* + +*$$\begin{equation*} + T=\left\{\emptyset,\left\{0\right\}, \left\{1\right\}, \left\{0,1\right\}, S\right\} +\end{equation*}$$* + +*We can define a relation $R\subseteq T\times T$ by* + +*$$\begin{align*} + R = \{ + &\left(\emptyset,\emptyset\right),\left(\emptyset,\left\{0\right\}\right),\left(\emptyset,\left\{1\right\}\right),\left(\emptyset,\left\{0,1\right\}\right),\left(\emptyset,S\right),\left(\left\{0\right\},\left\{0\right\}\right),\left(\left\{0\right\},\left\{0,1\right\}\right),\left(\left\{0\right\},S\right),\\ + &\left(\left\{1\right\},\left\{1\right\}\right),\left(\left\{1\right\},\left\{0,1\right\}\right),\left(\left\{1\right\},S\right),\left(\left\{0,1\right\},\left\{0,1\right\}\right),\left(\left\{0,1\right\},S\right),\left(S,S\right)\} +\end{align*}$$ This relation expresses inclusive subset inclusion, +$\subseteq$, on $S$.* +::: + +::: example +**Example 75**. *Let $S=\left\{0,1,2\right\}$ and $T=S$. Define +$T\times T$ by* + +*$$\begin{equation*} + T\times T = \left\{\left(0,0\right),\left(0,1\right),\left(0,2\right),\left(1,0\right),\left(1,1\right),\left(1,2\right),\left(2,0\right),\left(2,1\right),\left(2,2\right)\right\} +\end{equation*}$$ We can use the less than or equal to operator, $\leq$, +to define a relation. We have that* + +*$$\begin{equation*} + R=\left\{\left(0,0\right),\left(0,1\right),\left(0,2\right),\left(1,1\right),\left(1,2\right),\left(2,2\right)\right\} +\end{equation*}$$* +::: + +##### Reflexive Relation + +All of the examples from the previous section, except the strictly less +than example, share a common property. Each element is related to +itself, that is in each example there is some element $s\in S$ such that +$\left(s,s\right)\in R\subseteq S\times S$. We formalise this in the +following definition. + +::: definition +**Definition 81**. *Reflexive relation* + +*Let $S$ be a set with a relation $R\subseteq S\times S$. We say that +the relation $R$ is reflexive if and only if $\forall s\in S$ we have +that $\left(s,s\right)\in R$. If there is an $s\in S$ such that +$\left(s,s\right)\not\in R$ then we say that the relation is +anti-reflexive.* +::: + +We have given examples of reflexive relations and one example of an +anti-reflexive relation. We give an additional example of an +anti-reflexive relation. + +::: example +**Example 76**. *We have for $a,b\in\mathbb{N}$ that $a=b$ if and only +if $a\subseteq b$ and $b\subseteq a$. If this doesn't hold then +$a\neq b$ and either one of $a\subseteq b$ or $b\subseteq a$ is true but +not both. It follows that the relation $a\sim b$ meaning $a\neq b$ is +anti-reflexive. This also implies that if $a\neq b$ then either +$a\leq b$ or $b\leq a$.* +::: + +The examples given so far have allowed us to see some examples of +relations and one particular type of relation, a reflexive relation. +Unfortunately only considering relations on elements a single set $S$ +currently gives us few practical examples to work with. A simple +extension to the idea of a relation can fix this. + +::: definition +**Definition 82**. *Binary Relation* + +*Let $S$ and $T$ be sets. We define a binary relation to be a subset +$R\subseteq S\times T$. We write an element $\left(s,t\right)\in R$ as +$sRt$ or write $s\sim t$ and we say that $s$ relates to $t$. If +$\left(s,t\right)\not\in R$ we write $s\slashed{R} t$ or we write +$s\not\sim t$.* +::: + +We can extend this the notion of a relation and binary relation to that +of any finite Cartesian product + +::: definition +**Definition 83**. *$n$-ary Relation* + +*Let $S_1,S_2,S_3,\dots,S_n$ be sets. We define an $n$-ary relation to +be a subset +$R\subseteq S_1\times S_2\times S_3\times\dots\times S_n=\mathbb{S}$. An +element of $R$ has the form $r=\left(r_1,r_2,r_3,\dots,r_n\right)$ and +we say that the elements of $r$ relate. We write this as +$R\left(r\right)=R\left(r_1,r_2,r_3\dots,r_n\right)$* +::: + +In light of these previous definitions we would like to extend the +definition of a reflexive relation to binary and $n$-ary relations. To +see how we could extend a reflexive relation to a binary relation +suppose we have two sets $S$ and $T$. The definition of a reflexive +relation of a set $Z$ is that +$\left(z,z\right)\in R_z\subseteq Z\times Z$ where $z\in Z$ and $R_z$ is +the relation defined on $Z$. A natural way to extend this two $S$ and +$T$ is to have either $\left(s,s\right)\in R\subseteq S\times T$ or +$\left(t,t\right)\in R\subseteq S\times T$ where $R$ is a binary +relation for $S$ and $T$. Hence for a reflexive binary relation to makes +sense we must have that $s,t\in S\cap T$ and therefore the relation +would have to be defined on $S\cap T$. + +In the first case $\left(s,s\right)\in R\subseteq S\times T$ we have by +definition of an ordered tuple that $\left(s,s\right)\in R$ if and only +if $s\in S$ and $s\in T$. Likewise for +$\left(t,t\right)\in R\subseteq S\times T$ we must have $s\in S$ and +$t\in T$ which is to say $s,t\in S\cap T$. If $S\neq T$ then there will +exist at least one element $\left(s,t\right)\in R\subseteq S\times T$ +where either $s\in S$ and $s\not\in T$ or $t\in T$ and $t\not\in S$, in +this case it is not possible for a reflexive relation to exist. + +::: definition +**Definition 84**. *Reflexive binary relation* + +*Let $S$ and $T$ with relation $R\subseteq S\times T$. We say that the +relation $R$ is reflexive if and only if $S=T$.* +::: + +A similar argument shows there can be no reflexive $n$-ary relation +unless all of the sets that make the relation are the same. For example +consider the sets $X,Y$ and $Z$. The natural way to represent a relation +$R\subseteq X\times Y\times Z$ would be to have either +$\left(x,x,x\right)\in R$, $\left(y,y,y\right)\in R$ or +$\left(z,z,z\right)\in R$ where $x\in X$, $y\in Y$ and $z\in Z$. If +$\left(x,x,x\right)\in R$ then by definition we must have $x\in Y$ and +$x\in Z$, likewise if $\left(y,y,y\right)\in R$ then $y\in X$ and +$y\in Z$ and finally if $\left(z,z,z\right)\in R$ then $z\in X$ and +$z\in Y$. Any of the cases implies that $x,y,z\in X\cap Y\cap Z$ + +::: definition +**Definition 85**. *Reflexive $n$-ary relation* + +*Let $S_1,S_2,S_3,\dots,S_n$ be sets with relation +$R\subseteq S_1\times S_2\times S_3\times\dots\times S_n$. We say that +the relation $R$ is reflexive if and only if $S_i=S_j$ for all +$i,j\in\left\{1,2,3,\dots,n\right\}$* +::: + +This means when talking about a reflexive relation we only need to +consider a single set. + +An example of a binary relation is a mapping. + +::: example +**Example 77**. *Let $S=T=\mathbb{N}$ and define the mapping +$f:S\rightarrow T$ given by $f\left(s\right)=s$. We have that $f$ +defines a relation as we have that* + +*$$\begin{equation*} + R=\left\{\left(0,0\right),\left(1,1\right),\left(2,2\right),\left(3,3\right),\dots\right\}\subseteq\mathbb{N}\times\mathbb{N} +\end{equation*}$$* +::: + +::: example +**Example 78**. *Let $S=\left\{1,2\right\}$ and $T=\left\{3,4\right\}$. +Define the mapping $f:S\rightarrow T$ by $f\left(1\right)=4$ and +$f\left(2\right)=3$. We have $f$ defines a relation as* + +*$$\begin{equation*} + R=\left\{\left(1,4\right),\left(2,3\right)\right\}\subseteq S\times T +\end{equation*}$$* +::: + +We can consider operators as relations by using the $n$-aray notion of a +relation + +::: example +**Example 79**. *Let $X=Y=Z=\mathbb{N}$. We can consider the operator +$+$ as a mapping given by* + +*$$\begin{align*} + f:X\times Y &\rightarrow Z\\ + \left(x,y\right)&\mapsto f\left(x,y\right) = x+y +\end{align*}$$* + +*A relation can be defined by $f$. A sample of this relation $R$ looks +as follows* + +*$$\begin{equation*} + R=\left\{\left(0,0,0\right), \left(0,1,1\right),\left(4,3,7\right),\left(3,4,7\right),\left(2,2,4\right),\dots,\right\}\subseteq\mathbb{N}\times\mathbb{N}\times\mathbb{N} +\end{equation*}$$* + +*In general, $R$ has the following definition* + +*$$\begin{equation*} + R=\left\{\left(x,y,x+y\right):x,y\in\mathbb{N}\right\} +\end{equation*}$$* + +*We note that as $X=Y$ then for any $x\in X$ we have $x\in Y$ and +likewise for any $y\in Y$ we have that $y\in X$. We therefore have that +$R\left(x,y,x+y\right)=R\left(y,x,y+x\right)$. This is confirming the +fact that addition is commutative.* +::: + +::: example +**Example 80**. *Let $X=Y=Z=\mathbb{N}$. We can consider the operator +$*$ as a mapping given by* + +*$$\begin{align*} + f:X\times Y &\rightarrow Z\\ + \left(x,y\right)&\mapsto f\left(x,y\right) = x*y +\end{align*}$$* + +*The relation defined by $f$ looks as follows* + +*$$\begin{equation*} + R=\left\{\left(0,0,0\right), \left(0,1,0\right),\left(4,3,12\right),\left(3,4,12\right),\left(2,2,4\right),\dots,\right\}\subseteq\mathbb{N}\times\mathbb{N}\times\mathbb{N} +\end{equation*}$$* + +*In general, $R$ has the following definition* + +*$$\begin{equation*} + R=\left\{\left(x,y,x*y\right):x,y\in\mathbb{N}\right\} +\end{equation*}$$* + +*As before, we have that as $X=Y$ then for any $x\in X$ we have $x\in Y$ +and likewise, for any $y\in Y$ we have that $y\in X$. We, therefore, +have that $R\left(x,y,x*y\right)=R\left(y,x,y*x\right)$, again +confirming the fact that multiplication is commutative.* +::: + +::: example +**Example 81**. *Let $X=Y=Z=\mathbb{N}$. We can consider the operator +$\wedge$ as a mapping given by* + +*$$\begin{align*} + f:X\times Y &\rightarrow Z\\ + \left(x,y\right)&\mapsto f\left(x,y\right) = \wedge\left(x,y\right)=x^y +\end{align*}$$* + +*The relation defined by $f$ looks as follows* + +*$$\begin{equation*} + R=\left\{\left(0,0,1\right), \left(0,1,0\right),\left(2,3,8\right),\left(8,2,64\right),\left(3,2,9\right),\dots,\right\}\subseteq\mathbb{N}\times\mathbb{N}\times\mathbb{N} +\end{equation*}$$* + +*In general, $R$ has the following definition* + +*$$\begin{equation*} + R=\left\{\left(x,y,x^y\right):x,y\in\mathbb{N}\right\} +\end{equation*}$$* + +*As before, we have that as $X=Y$ then for any $x\in X$ we have $x\in Y$ +and likewise, for any $y\in Y$ we have that $y\in X$. We, therefore, +have that $R\left(x,y,x^y\right)\neq R\left(y,x,y^x\right)$, which +confirms that in general exponentiation is not commutative.* +::: + +The last three examples expose another property that relations can have. +If two or more elements relate then it doesn't matter which way the +relation is written, that is if $x\sim y$ then we can have the case that +$y\sim x$. Such a relation is called symmetric. + +::: definition +**Definition 86**. *Symmetric relation* + +*Let $S$ be a set with relation $R\subseteq S\times S$. We say that $R$ +is a symmetric relation if and only if $\forall x,y\in S$ we have that +$xRy$ implies $yRx$, equivalently we can write $R$ is symmetric if and +only if $x\sim y$ implies $y\sim x$. If $R$ is not symmetric we say that +$R$ is an anti-symmetric relation.* +::: + +As with reflexive relations we can show that trying to extend a the idea +of a symmetric relation on a single set to multiple sets we have to +conclude the sets have to be the same. + +Indeed suppose that $S$ and $T$ are sets with a relation +$R\subseteq S\times T$. The natural extension for a symmetric relation +would be $\forall s\in S$ that $sRt\Rightarrow tRs$ for $t\in T$. This +implies that $t\in S$ and $s\in T$ and therefore $s,t\in S\cap T$. + +::: definition +**Definition 87**. *Symmetric binary relation* + +*Let $S$ and $T$ be sets with relation $R\subseteq S\times T$. We say +that $R$ is symmetric if and only if $S=T$* +::: + +Likewise a similar argument holds for $n$-ary symmetric relations + +::: definition +**Definition 88**. *Symmetric $n$-ary relation* + +*Let $S_1,S_2,S_3,\dots,S_n$ be sets with relation +$R\subseteq S_1\times S_2\times S_3\times\dots\times S_n$. We say that +the relation $R$ is symmetric if and only if $S_i=S_j$ for all +$i,j\in\left\{1,2,3,\dots,n\right\}$* +::: + +The comparison, less than, less than or equal to, greater than, and +greater than or equal to operators on the naturals also give insight +into another interesting property. The following examples will make it +more clear + +::: example +**Example 82**. *Let $S=T=\mathbb{N}$ and define $x\sim y$ by $x\leq y$. +Consider $a,b,c\in\mathbb{N}$ with $a=2$, $b=4$ and $c=6$. We have that +$a\sim b$ as $2\leq 4$ and we have that $b\sim c$ as $4\leq 6$, we +clearly also have $a\sim c$ as $2\leq 6$. In general if we have +$a,b,c\in\mathbb{N}$ with $a\leq b\leq c$ we have that $a\sim b$ and +$b\sim c$ implies $a\sim c$.* +::: + +::: example +**Example 83**. *Let $S=T=\mathbb{N}$ and define $x\sim y$ by $x\geq y$. +Consider $a,b,c\in\mathbb{N}$ with $a=8$, $b=3$ and $c=1$. We have that +$a\sim b$ as $8\geq 3$ and we have that $b\sim c$ as $3\leq 1$, we also +have $a\sim c$ as $8\geq 1$. More generally if we have +$a,b,c\in\mathbb{N}$ with $a\geq b\geq c$ we have that $a\sim b$ and +$b\sim c$ implies $a\sim c$.* +::: + +::: example +**Example 84**. *Let $S=T=\mathbb{N}$ and define $x\sim y$ by $x= y$. +Consider $a,b,c\in\mathbb{N}$ with $a=2$, $b=2$ and $c=2$. We have that +$a\sim b$ as $2=2$ and we have that $b\sim c$ as $2=2$, we also have +$a\sim c$ as $2=2$. More generally if we have $a,b,c\in\mathbb{N}$ with +$a= b= c$ we have that $a\sim b$ and $b\sim c$ implies $a\sim c$.* +::: + +We see that with certain relations that if $a\sim b$ is true and +$b\sim c$ is true then we can conclude that $a\sim b$ is true. Such a +relation is called a Transitive relation. + +::: definition +**Definition 89**. *Transitive relation* + +*Let $S$ be a set with relation $R\subseteq S\times S$. We say that $R$ +is a transitive relation if and only if $\forall a,b,c\in S$ we have +that if $aRb$ and $bRc$ then we have that $aRc$.* +::: + +We again consider if a transitive relation can be extended to multiple +sets. Suppose that we have a binary relation $R\subseteq S\times T$ for +some sets $S$ and $T$. The natural extension to make $R$ a transitive +relation is to have $s\sim t$ and $t\sim u$ implies $s\sim u$ for +$s,t\in S$ and $t,u\in T$. Hence we must have $s,t\in S$ but need not +have $u\in S$. As we aren't assuming anything else about the relation +$R$ there is nothing more we can deduce about a binary transitive +relation. + +::: definition +**Definition 90**. *Transitive binary relation* + +*Let $S$ and $T$ be sets with relation $R\subseteq S\times T$. We say +that $R$ is transitive if and only if the set $\tilde{R}$ given by* + +*$$\begin{equation*} + \tilde{R} = \left\{\left(x,z\right) \in S \times T:\forall x\in S\wedge\forall z\in T: \exists y \in S \cap T: \left(x, y\right) \in R \wedge \left(y, z\right) \in R\right\} +\end{equation*}$$ is non-empty.* +::: + +A definition can be made for a transitive $n$-ary relation. I AM NOT +SURE HOW TO DEFINE THIS YET, PAIR-WISE RELATION OF EACH +SET????????????????? We can make use of a binary relation in order to +define + +::: definition +**Definition 91**. *Transitive $n$-ary relation* + +*Let $S_1,S_2,S_3,\dots,S_n$ be sets with relation +$R\subseteq S_1\times S_2\times S_3\times\dots\times S_n$. We say that +the relation $R$ is transitive if and only if the set $\tilde{R}$ given +by* + +*$$\begin{equation*} + \tilde{R}=\left\{\left(x,z\right)\in \right\} +\end{equation*}$$ is non-empty* +::: + +##### Equivalence Relations + +Of all the examples of relations we have seen so far there is one in +particular that is special, the equality operator $=$. This relation is +reflexive, symmetric and transitive. + +::: {#prop:EqualityOnNaturalsIsEquivRelation .proposition} +**Proposition 59**. *The equality relation on the natural numbers is +reflexive, symmetric and transitive* + +*Let $S=T=\mathbb{N}$ and for $x,y\in\mathbb{N}$ define the relation +$x\sim y$ by $x=y$. We have that* + +1. *$\sim$ is reflexive, that is $\forall x\in\mathbb{N}$ we have + $x\sim x$* + +2. *$\sim$ is symmetric, that is $\forall x,y\in\mathbb{N}$ we have + $x\sim y\Rightarrow y\sim x$* + +3. *$\sim$ is transitive, that is $\forall x,y,z\in\mathbb{N}$ we have + that if $x\sim y$ and $y\sim z$ then $x\sim z$* + +*Proof:* + +1. *$\sim$ is reflexive, that is $\forall x\in\mathbb{N}$ we have + $x\sim x$:* + + *Let $x\in\mathbb{N}$ then by definition of equality we have that + for $y,z\in\mathbb{N}$ that $y=z$ if and only if $y\subseteq z$ and + $z\subseteq y$. It is clear that $x=x$ and therefore $x\sim x$ + proving reflexivity.* + +2. *$\sim$ is symmetric, that is $\forall x,y\in\mathbb{N}$ we have + $x\sim y\Rightarrow y\sim x$:* + + *Let $x,y\in\mathbb{N}$ with $x\sim y$. We have that as $x\sim y$ + then $x=y$. By definition of equality we also have $y=x$ and so + $y\sim x$ showing that $\sim$ is symmetric.* + +3. *$\sim$ is transitive, that is $\forall x,y,z\in\mathbb{N}$ we have + that if $x\sim y$ and $y\sim z$ then $x\sim z$:* + + *Let $x,y,z\in\mathbb{N}$ such that $x\sim y$ and $y\sim z$, then + $x=y$ and $y=z$. By definition of equality it follows that $x=z$ and + so $x\sim z$ showing transitivity.* + +*The result follows. $\qed$* +::: + +What does it mean for a relation to be reflexive, symmetric and +transitive? In the case of equality on the natural numbers we see that +reflexivity tells us that an element is equal to itself. Equality being +symmetric tells us that if $x=y$ then $y=x$ that is it does not matter +which we we say the two numbers are equal. Finally transitivity tells us +that if $x=y$ and $y=z$ we are able to deduce that $x=z$. In this +context, equality being reflexive, symmetric and transitive allows us to +quantify which elements are equivalent. In the case of equality it is +clear which elements are equivalent, the ones that are equal! + +::: example +**Example 85**. *Consider $X=Y=\mathbb{N}$ and for $x,y\in\mathbb{N}$ +define the relation $R=\mathbb{N}\times\mathbb{N}$. We have that $R$ is +reflexive as for any $x\in\mathbb{N}$ we have that +$\left(x,x\right)\in R$. Likewise $R$ symmetric as +$\forall x,y\in\mathbb{N}$ we have that +$\left(x,y\right)\in R\Rightarrow\left(y,x\right)\in R$. as $X=Y$. +Finally $R$ is transitive as $\forall x,y,z\in\mathbb{N}$ we have that +$\left(x,y\right)\in R$ and $\left(y,z\right)\in R$ and +$\left(x,z\right)\in R$.* + +*What does $R$ being reflexive, symmetric and transitive mean? In this +case $R$ being reflexive, symmetric and transitive means that every +$x\in X$ and $y\in Y$ are related and we can see $R$ as a relation +meaning \"is an element of $\mathbb{N}$\". This means that we have shown +that $X$ and $Y$ are equivalent, which we already know by the fact we +set $X=Y=\mathbb{N}$.* +::: + +Based on the two examples we motivate the following definition. + +::: definition +**Definition 92**. *Equivalence relation* + +*Let $S$ be a set and $R\subseteq S\times S$ a relation. We say that $R$ +is an equivalence relation if and only if* + +1. *$R$ is reflexive* + +2. *$R$ is symmetric* + +3. *$R$ is transitive* +::: + +Proposition +[59](#prop:EqualityOnNaturalsIsEquivRelation){reference-type="ref" +reference="prop:EqualityOnNaturalsIsEquivRelation"} is equivalent to +saying that equality is an equivalence relation on $\mathbb{N}$. The two +examples also show a disparity between the two equivalence relations +shown. In the case of the equality the relation $R$ was a strict subset +of $\mathbb{N}\times\mathbb{N}$ where as in the second example $R$ was +equal to $\mathbb{N}\times \mathbb{N}$. This raises the question what is +different? We can answer this by looking at the set of elements that +relate to a given element. Such a set is called an equivalence class. + +::: definition +**Definition 93**. *Equivalence class* + +*Let $S$ be a set, let $x\in S$ and let $R$ be an equivalence relation +on $S$. We define an equivalence class, denoted $\left[x\right]$ to be +the set* + +*$$\begin{equation*} + \left[x\right]=\left\{y\in S:xRy\right\} +\end{equation*}$$* + +*If the context doesn't make clear the relation we are referring we +explicitly write $\left[x\right]_R$ to be the equivalence class of $x$ +under the relation $R$.* + +*We say that an element $y\in\left[x\right]$ is a representative of the +equivalence class of $x$* +::: + +To get a feel for equivalence classes we consider the, non-mathematical, +following example. + +::: example +**Example 86**. *Consider the set $X$ to be the set of all people +currently alive. Define a relation, $\sim$, on $X$ by* + +*$$\begin{equation*} + \forall\left(x,y\right)\in X\times X: x\sim y\iff x\text{ and }y\text{ where born in the same year} +\end{equation*}$$* + +*We have that $\sim$ is an equivalence relation. Clearly if $x\sim x$ as +$x$ was born in some year $D$. We have that if $x\sim y$ then $x$ and +$y$ are born in the same year and clearly $y\sim x$. Now if $x\sim y$ +and $y\sim z$ then $x$ and $y$ are born in the same year and $y$ and $z$ +are born in the same year. This therefore means $x$ and $z$ are born in +the same year so $x\sim z$ showing transitivity.* + +*Now let $x\in X$ and consider the equivalence class +$\left[x\right]_\sim$. By definition of an equivalence class we have +that* + +*$$\begin{equation*} + \left[x\right]_\sim=\left\{y\in x:x\sim y\right\} +\end{equation*}$$* + +*This means that the equivalence class $\left[x\right]_\sim$ is the set +of all people currently alive that were born in the same year. As $X$ +was the set of all currently alive people we have found a way to extract +a subset of $X$ such that they are all born in the same year. If we now +pick another element of $X$, say a, such that $x\not\sim a$ then by +definition $a$ was not born in the same year as $x$ and +$\left[a\right]_\sim$ is another subset of $X$ of currently alive people +born in the same year. Moreover we have that +$\left[x\right]_\sim\neq\left[a\right]_\sim$. We can do this for every +element of $X$ and get a collection of sets that correspond to all of +the possible different years that anyone currently alive could possibly +be in.* +::: + +The previous example has shown that we are able to construct a partition +of a set $S$ which has an equivalence relation $\sim$. We can prove this +more generally, firstly we recall the definition of a set partition. + +Let $S$ be a set and define $\mathbb{S}$ to be the set of subsets of +$S$. We say that $\mathbb{S}$ is a partition of $S$ if the following +hold. + +1. $\forall S_1,S_2\in\mathbb{S}$ we have $S_1\cap S_2=\emptyset$ + whenever $S_1\neq S_2$ + +2. Taking the union of every $T\in\mathbb{S}$ gives us $S$ that is + + $$\begin{equation*} + S=\bigcup_{T\in\mathbb{S}} T + \end{equation*}$$ + +3. $\forall T\in\mathbb{S}$ we have that $T\neq\emptyset$. + +Before we can show that the equivalence classes partition the set we +must first show that there can be no empty equivalence class. + +::: {#prop:EquivClassNonEmpty .proposition} +**Proposition 60**. *Equivalence class is non-empty* + +*Let $S$ be a set with an equivalence relation $\sim$. Let $x\in S$ then +we have that $\left[x\right]_\sim\neg\emptyset$* + +*Proof:* + +*Let $S$ be a set with an equivalence relation $\sim$. By definition of +an equivalence relation we have that $\forall x,y,z\in S$ that* + +1. *$\sim$ is reflexive, that is $x\sim x$* + +2. *$\sim$ is symmetric, that is $x\sim y\Rightarrow y\sim x$* + +3. *$\sim$ is transitive, that is $x\sim y$ and $y\sim x$ implies that + $x\sim z$* + +*Consider the equivalence class $\left[x\right]_\sim$. By definition of +an equivalence class we know that* + +*$$\begin{equation*} + \left[x\right]_\sim=\left\{y\in S:x\sim y\right\} +\end{equation*}$$* + +*As $\sim$ is reflexive we have that $x\mathop{\mathrm{Im}}x$ and so +$x\in\left[x\right]_\sim$ and therefore +$\left[x\right]_\sim\neq\emptyset$. $\qed$* +::: + +We can prove that an equivalence relation partitions the set it is +defined on. + +::: {#thm:EquivClassesOfRelationPartitionSet .theorem} +**Theorem 19**. *Equivalence classes of a relation partitions the set* + +*Let $S$ be a set with an equivalence relation $\sim$. Let $\mathbb{S}$ +denote the equivalence classes of $\sim$ for each $s\in S$. We have that +$\mathbb{S}$ is a partition of $S$.* + +*Proof:* + +*Let $S$ be a set with an equivalence relation $\sim$ and let +$\mathbb{S}$ be the set of equivalence classes of $\sim$ for each +$s\in S$. Let $x\in S$ then $x$ belongs to at least one equivalence +class by proposition [60](#prop:EquivClassNonEmpty){reference-type="ref" +reference="prop:EquivClassNonEmpty"}. We therefore have that* + +*$$\begin{equation*} + S=\bigcup_{x\in S}\left[x\right]_\sim +\end{equation*}$$* + +*It is left to show that if $\left[x\right]_\sim\neq\left[y\right]_\sim$ +for $x,y\in S$ then we have that +$\left[x\right]_\sim\cap\left[y\right]_\sim=\emptyset$. This is +equivalent to saying that if +$\left[x\right]_\sim\cap\left[y\right]_\sim\neq\emptyset$ then +$\left[x\right]_\sim=\left[y\right]_\sim$. So suppose that +$\left[x\right]_\sim\cap\left[y\right]_\sim\neq\emptyset$ then +$\left[x\right]_\sim\cap\left[y\right]_\sim$ has at least one element +$z$. Suppose that $z\in\left[x\right]_\sim$ then by definition we have +that $x\sim z$. Let $a\in\left[x\right]_\sim$ be an arbitrary element of +the equivalence class of $x$. We have that $a\sim x$ then by +transitivity of $\sim$ we conclude that $a\sim z$. However as +$z\in\left[x\right]_\sim\cap\left[y\right]_\sim$ then we have that +$z\in\left[y\right]_\sim$ and so $y\sim z$. As $\sim$ is symmetric we +have $z\sim y$ and again by transitivity we conclude that $a\sim y$. +Hence $a\in\left[y\right]_\sim$ and so +$\left[x\right]_\sim\subseteq\left[y\right]_\sim$.* + +*A similar argument shows +$\left[y\right]_\sim\subseteq\left[x\right]_\sim$ and therefore we have +that $\left[x\right]_\sim=\left[y\right]_\sim$. Finally we conclude that +unequal equivalence classes are disjoint and therefore the set of +equivalence classes $\mathbb{S}$ is a partition for $S$.* + +*The result is shown. $\qed$* +::: + +### Construction of the Integers + +::: epigraph +The trouble with integers is that we have examined only the very small +ones. Maybe all the exciting stuff happens at really big numbers, ones +we can't even begin to think about in any very definite way. + +*Ronald Graham* +::: + +We now have enough theory to consider extending the natural numbers +$\mathbb{N}$. One reason to do this is to provide a completion to the +idea of subtraction. Recall that $n-m$ is only defined in $\mathbb{N}$ +if and only if $m\leq n$. This is a limiting idea. For example, the idea +of debt can't be explained using only $\mathbb{N}$. We know that if the +balance on your bank account is negative then you owe money to someone, +if your balance is positive you have money to spend[^8]. The natural +numbers don't have a concept of \"negative\" or debt, we can only deal +with \"positive\" values. To keep the financial institutions happy we +should resolve this issue. + +To do this we need to consider exactly what it is we want to achieve. +Firstly we want to be able to define $n-m$ for all $n,m\in\mathbb{N}$. +Clearly, if $n\geq m$ then such a number already exists in $\mathbb{N}$. +Secondly, such a number $n-m$ could have many different representations, +for example, $6-2=4$ and $5-1=4$. We need a way to say that any of these +different representations actually represents the same thing. Formally +if we have $a,b,c,d\in\mathbb{N}$ such that $a-b=c-d$ then $a-b$ and +$c-d$ represent the same number, this is equivalent to $a+d=b+c$. +Thinking of $-$ as a relation we can use the language of equivalence +relations to solve this issue. That is a relation where +$\left(a,b\right)\sim\left(c,d\right)$ + +#### Defining the Integers + +We start by recasting the defining of subtraction to be defined as an +ordered tuple. + +::: definition +**Definition 94**. *Subtraction as an ordered tuple* + +*Let $a,b\in\mathbb{N}$. We define the subtraction as an ordered tuple +$\left(a,b\right)\in\mathbb{N}^2$ to mean $\left(a-b\right)$. We will +call an element $x\in\mathbb{N}^2$ a subtraction tuple. We note that if +$a\geq b$ we have $\left(a-b\right)\in\mathbb{N}$* +::: + +From this we can define a relation + +::: definition +**Definition 95**. *Relation on subtraction* + +*Let $\left(a,b\right),\left(c,d\right)\in\mathbb{N}^2$ be subtraction +tuples. We define the relation $\sim$ such that +$\left(a,b\right)\sim\left(c,d\right)$ if and only if $a+d=b+c$* +::: + +We have that this relation is an equivalence relation. + +::: proposition +**Proposition 61**. *Relation on subtraction ordered tuples is an +equivalence relation* + +*Let $x,y\in\mathbb{N}^2$ be subtraction tuples and define the relation +$x\sim y$ as above. We have that $\sim$ is an equivalence relation.* + +*Proof:* + +*Let $x,y,z\in\mathbb{N}^2$ be subtraction tuples such that +$x=\left(a,b\right)$, $y=\left(c,d\right)$ and $z=\left(e,f\right)$. We +need to show that $\sim$ is an equivalence relation, that is* + +1. *$\sim$ is reflexive* + +2. *$\sim$ is symmetric* + +3. *$\sim$ is transitive* + + + +1. *$\sim$ is reflexive:* + + *We have that $x=\left(a,b\right)$ and by definition of $\sim$ we + know that $x\sim x$ if and only if $a+b=a+b$ which is clear by + definition of equality on the natural numbers. Hence $x\sim x$ and + $\sim$ is reflexive.* + +2. *$\sim$ is symmetric:* + + *We have that $x=\left(a,b\right)$ and $y=\left(c,d\right)$. Suppose + that $x\sim y$ then we have that $a+d=b+c$. By commutativity of + equality of natural numbers that $a+d=b+c\Rightarrow b+c=a+d$. By + commutativity of addition on the natural numbers we have that + $b+c=a+d$ is the same as $c+b=d+a$. Hence we have that + $\left(c,d\right)\sim\left(a,b\right)$ by definition of $\sim$ and + so $y\sim x$ showing that $\sim$ is symmetric.* + +3. *$\sim$ is transitive:* + + *We know that $x=\left(a,b\right)$, $y=\left(c,d\right)$ and + $z=\left(e,f\right)$. Now suppose that $x\sim y$ and $y\sim z$ then + by definition we have that $\left(a,b\right)\sim\left(c,d\right)$ + and $\left(c,d\right)\sim\left(e,f\right)$ and hence by definition + of $\sim$ we have $a+d=c+b$ and $c+f=e+d$.* + + *Consider $a+c+f$ we have* + + *$$\begin{equation*} + a+c+f=a+e+d=a+d+e=c+b+e + \end{equation*}$$* + + *That is to say $a+c+f=c+b+e$. We have by the cancellation laws on + the natural numbers that $a+f=b+e$ which implies that + $\left(a,b\right)\sim\left(e,f\right)$. Which is to say $x\sim z$. + Hence transitivity has been shown.* + +*It follows that $\sim$ is an equivalence relation. $\qed$* +::: + +Now that we have shown that $\sim$ is an equivalence relation we can +solve the multiple representation problem by considering the equivalence +classes of $\mathbb{N}^2$ with the relation $\sim$. Let +$x\in\mathbb{N}^2$ with $x=\left(a,b\right)$ then the equivalence class +of $x$ is given by + +$$\begin{equation*} + \left[x\right]_\sim=\left[\left(a,b\right)\right]_\sim=\left\{\left(c,d\right)\in\mathbb{N}^2 : \left(a,b\right)\sim\left(c,d\right)\right\} +\end{equation*}$$ + +We know by theorem +[19](#thm:EquivClassesOfRelationPartitionSet){reference-type="ref" +reference="thm:EquivClassesOfRelationPartitionSet"} that for each +$x\in\mathbb{N}^2$ there is set of equivalence classes partition +$\mathbb{N}^2$ and that each equivalence class is disjoint. This is to +say if $x,y\in\mathbb{N}^2$ then we have that if +$\left[x\right]_\sim\cap\left[y\right]_\sim\neq\emptyset$ then +$\left[x\right]_\sim =\left[y\right]_\sim$. This solves the multiple +representation issue. + +Let us have a look at some equivalence classes + +::: example +**Example 87**. *Let $x\in\mathbb{N}^2$ with $x=\left(1,3\right)$ by +definition we have that $x$ represents $1-3$. Consider the equivalence +class of $x$, $\left[x\right]=\left\{y\in\mathbb{N}:x\sim y\right\}$ and +let $y\in\left[x\right]$. We have that $y=\left(c,d\right)$ and that +$1+d=3+c$, one possible $y$ where this is true is given by +$y=\left(0,2\right)$ and $y$ represents $0-2$, As we have +$y\in\left[x\right]$ then we have that $\left[x\right]=\left[y\right]$ +so we shall take $y$ to be the canonical representative of this +equivalence class.* +::: + +Now that we have that the subtraction tuples are in equivalence classes +we can consider the following. Suppose that $a,b,c\in\mathbb{N}$ then +what is $a-\left(b-c\right)$? For example if $a=10, b=6$ and $c=3$ then +we have that $10-\left(6-3\right)=10-3=7$. This is also the same as +$10+3-6=13*6=7$. This holds in general where we have that +$\left(a,b-c\right)\sim\left(a+c,b\right)$ + +::: {#lem:NaturalMinusDifferenceOfNatural .lemma} +**Lemma 6**. *$\left(a,b-c\right)\sim\left(a+c,b\right)$* + +*Let $a,b,c\in\mathbb{N}$ with $a> b\geq c$. We have that* + +*$$\begin{equation*} + \left(a,b-c\right)\sim\left(a+c,b\right) +\end{equation*}$$* + +*Proof:* + +*Let $a,b,c\in\mathbb{N}$ be as given. By definition of $\sim$ we have +$\left(x,y\right)\sim\left(u,v\right)$ if and only if $x+v=u+y$. We +argue by contradiction, suppose that +$\left(a,b-c\right)\not\sim\left(a+c,b\right)$ then by definition we +have that* + +*$$\begin{align*} + a+b&\neq a+c+\left(b-c\right)\\ + b&\neq c+\left(b-c\right),\ \text{By the cancellation law}\\ + b&\neq \left(c+b\right)-c,\ \text{By proposition}\ref{prop:NaturalAddDifference}\\ + b&\neq\left(b+c\right)-c,\ \text{By commutativity}\\ + b&\neq b+\left(c-c\right),\ \text{By proposition}\ref{prop:NaturalAddDifference}\\ + 0&\neq \left(c-c\right),\ \text{By the cancellation law}\\ + 0&\neq 0 +\end{align*}$$* + +*A contradiction. $\qed$* +::: + +By this lemma it follows that $a-\left(b-c\right)=\left(a+c\right)-b$. + +We now look at the definition of what the set of equivalence relations +looks like. We make the following definition + +::: {#def:QuotientSet .definition} +**Definition 96**. *Quotient set* + +*Let $S$ be a set with an equivalence relation $\sim$. Let $x\in S$ and +consider the equivalence class $\left[x\right]_\sim$. We define the +quotient set of $S$, denoted by $S/\sim$ by* + +*$$\begin{equation*} + S/\sim=\left\{\left[x\right]_\sim :x\in S\right\} +\end{equation*}$$* +::: + +Why have we called the set of the equivalence classes a quotient set? We +can see why with a few examples. + +::: example +**Example 88**. *We reconsider the example where $X$ is the set of all +people currently alive with the relation $\sim$ given by* + +*$$\begin{equation*} + \forall\left(x,y\right)\in X\times X: x\sim y\iff x\text{ and }y\text{ where born in the same year} +\end{equation*}$$* + +*We know that $\sim$ is an equivalence relation and we know that the +equivalence classes define a set of all people currently alive born in a +certain year. We can identify the quotient set $X/\sim$ as the set of +all of the possible years that all people currently alive could live in. +As an example suppose that person $x\in X$ was born in 1983. Then by the +definition of $\sim$ we have that $x\sim y$ if and only if $y$ is also +born in 1983 and that $\left[x\right]_\sim$ is the equivalence class of +all people born in 1983. As $\left[x\right]_\sim\in X/\sim$ then +$\left[x\right]_\sim$ is the set in $X/\sim$ that represents the year +1983. That is the quotient set has taken the set $X$ of all currently +alive people who were born in a certain year and turned it into the set +of all possible years.* +::: + +::: example +**Example 89**. *Let $X$ be the set of all possible cars and define the +equivalence relation $\sim$ such that $x\sim y$ if and only if $x$ and +$y$ are the same colour. We have that $sim$ is an equivalence relation. +Reflexivity is clear as if $x$ is a certain colour then clearly +$x\sim x$ will be true. Now if $x\sim y$ then both $x$ and $y$ are the +same colour and so $y\sim x$. Finally if $x\sim y$ and $y\sim z$ then +$x$ and $y$ are the same colour and so are $y$ and $z$ so it follows +that $x\sim z$.* + +*Suppose now that $x\in X$, then the equivalence class +$\left[x\right]_\sim$ is the set where all cars are the same colours. +Hence the quotient set $X/\sim$ will be the set of all possible car +colours. The quotient set has taken the set of all possible cars and +turned it into the set of all possible car colours.* + +*If we had a different relation $R$ where $xRy$ if and only if $x$ and +$y$ have exactly two doors then $R$ is also an equivalence relation and +$X/R$ would take all of the possible cars $X$ and turn it into the set +of all of the cars that have exactly two doors.* +::: + +These examples show that the quotient set takes a set of objects $S$ and +extracts a given property defined by the equivalence relation $\sim$ +defined on $S$. How can we use the quotient set on the equivalence +classes of the subtraction tuples? + +We have that the the quotient set of $\mathbb{N}^2/\sim$ is given by + +$$\begin{equation*} + \mathbb{N}^2/\sim=\left\{\left[x\right]_\sim:x\in\mathbb{N}^2\right\} +\end{equation*}$$ + +What do these elements actually look like? Let +$\left(a,b\right)=x\in\mathbb{N}^2$ and consider the equivalence class +$\left[x\right]_\sim$. Firstly, in the naturals, we know that $0=0-0$ +and more generally that $0=a-a$ for any $a\in\mathbb{Z}$. Hence +$0\in\left[\left(0,0\right)\right]$. + +Now, consider $\left[\left(a,0\right)\right]$ then we would have that +any $\left(c,d\right)=y\in\left[\left(a,0\right)\right]$ is such that +$\left(a,0\right)\sim\left(c,d\right)$ if and only if $a-0=c-d$. Hence +each $a$ is equivalent to some subtraction tuple. Moreover each +$\left(a,0\right)=a\in\mathbb{N}$, therefore we have a canonical +representation for each element $a\in\mathbb{N}$. What happens if we +have a tuple $\left(a,b\right)$ where $a\geq b$? We can see that if +$\left(a,b\right)\sim\left(c,d\right)$ then $a+d=c+b$. For example we +have that $\left(0,3\right)\sim\left(1,4\right)$ which gives + +$\left(8,11\right)\sim\left(0,3\right) = 8-11 = 0-3 8+3 = 11$ +$$\begin{equation*} + 0-3=1-4 \Rightarrow 0+4=1+3 \Rightarrow 4=4 +\end{equation*}$$ + +Hence we can define a canonical representation for each +$\left(0,a\right)$ where $a\in\mathbb{N}$. We will write the element +$\left(0 ,a\right)$ by $-a$ for each $a\in\mathbb{N}$. We have define +the set of Integers. + +::: definition +**Definition 97**. *Integers* + +*Let $\mathbb{N}^2$ have the equivalence relation $\sim$ defined by +$\left(a,b\right)\sim\left(c,d\right)$ if and only if $a+d=b+c$. We +define the set of Integers, denoted $\mathbb{Z}$, as the quotient set +$\mathbb{N}^2/\sim$. The set $\mathbb{Z}$ has the form* + +*$$\begin{equation} + \mathbb{Z}=\left\{\dots,-4,-3,-2,-1,0,1,2,3,4,\dots\right\} +\end{equation}$$* +::: + +We make two additional definitions based on the definition of the +canonical form the equivalence classes + +::: definition +**Definition 98**. *Positive Integer* + +*Let $a\in\mathbb{Z}$. We say that $a$ is a positive integer if and only +if $a\in\left[\left(b,0\right)\right]$ for some $b\in\mathbb{N}$ with +$b\neq 0$.* +::: + +::: definition +**Definition 99**. *Negative Integer* + +*Let $a\in\mathbb{Z}$. We say that $a$ is a negative integer if and only +if $a\in\left[\left(0,b\right)\right]$ for some $b\in\mathbb{N}$ with +$b\neq 0$.* +::: + +We can use these two definitions to define an occasionally useful idea. + +::: definition +**Definition 100**. *Sign of an integer* + +*Let $x\in\mathbb{Z}$. We define the sign of $x$, denoted by +$\mathop{\mathrm{sgn}}\left(x\right)$ to be the following function* + +*$$\begin{align*} + \mathop{\mathrm{sgn}}:\mathbb{Z}&\rightarrow\left\{-1,0,1\right\}\\ + x&\mapsto\mathop{\mathrm{sgn}}\left(x\right)=\begin{cases} + 1,\ \text{If } x\text{ is a positive integer}\\ + -1,\ \text{If } x\text{ is a negative integer}\\ + 0,\ \text{Otherwise} + \end{cases} +\end{align*}$$* +::: + +We also have the following, clear result + +::: proposition +**Proposition 62**. *The natural numbers are a subset of the integers* + +*We have that $\mathbb{N}\subseteq\mathbb{Z}$* + +*Proof:* + +*We have that the elements of the equivalence class +$\left[\left(x,0\right)\right]$ have the form $x-0=x\in\mathbb{N}$. Let +$a\in\mathbb{N}$ then we have that $a\in\left[\left(a,0\right)\right]$. +This holds for every $a\in\mathbb{N}$ and so +$\mathbb{N}\subseteq\mathbb{Z}$. $\qed$* +::: + +We will let $\left[\left(a,b\right)\right]$ be denoted by +$\left[a,b\right]$ and extend the operations of addition and +multiplication to the integers by defining how they work on the +equivalence classes. + +#### Extending equality to the integers + +Equality for the integers is easy to define. + +::: definition +**Definition 101**. *Equality of integers* + +*Let $x,y\in\mathbb{Z}$ be two integer numbers. We define that two +integers are equal, denoted $x=y$ if and only if $x\sim y$. This is the +same as saying both $x$ and $y$ belong to the same equivalence class. In +the case where $x\not\sim y$, we say that $x$ is not equal to $y$ and +write $x\neq y$.* +::: + +#### Extending inequality operators to the integers + +Inequality operators extend in a natural way. + +::: definition +**Definition 102**. *Less than operator* + +*Let $x,y\in\mathbb{Z}$ where $x\in\left[a,b\right]$ and +$y\in\left[c,d\right]$ for some $a,b,c,d\in\mathbb{N}$. The less than +operator, denoted by $xy$ is defined by the logical proposition* + +*$$\begin{equation*} + >\left(x,y\right)=\begin{cases} + 1,\ \text{If } a+d>b+c\\ + 0,\ \text{Otherwise} + \end{cases} +\end{equation*}$$* + +*This can equivalently be express as* + +*$$\begin{equation*} + x>y \iff a+d>b+c +\end{equation*}$$* +::: + +::: definition +**Definition 105**. *Greater than or equal to operator* + +*Let $x,y\in\mathbb{Z}$ where $x\in\left[a,b\right]$ and +$y\in\left[c,d\right]$ for some $a,b,c,d\in\mathbb{N}$. The greater than +or equal to operator, denoted by $x\geq y$ is defined by the logical +proposition* + +*$$\begin{equation*} + \geq\left(x,y\right)=\begin{cases} + 1,\ \text{If } a+d\geq b+c\\ + 0,\ \text{Otherwise} + \end{cases} +\end{equation*}$$* + +*This can equivalently be express as* + +*$$\begin{equation*} + x\geq y \iff a+d\geq b+c +\end{equation*}$$* +::: + +#### Extending addition to the integers + +We have an understanding of addition on the natural numbers, mainly the +recursive definition given by + +$$\begin{align*} + +&:\mathbb{N}^2\mathlarger{\mathlarger{\rightarrow}}\mathbb{N}\\ + \left(m,n\right)&\mapsto +\left(m,n\right)=\begin{cases} + m+0=m,\ \text{If } n=0\\ + m+S\left(n\right)=S\left(m+n\right),\ \text{If } n\neq 0 + \end{cases} +\end{align*}$$ + +Now if we take $a,b\in\mathbb{Z}$ with $a,b$ being positive integers +then we have that $a\in\left[\left(a,0\right)\right]$ and +$b\in\left[\left(b,0\right)\right]$. We then have that $a+b$ will be in +$\left[\left(a+b,0\right)\right]$. Now suppose that $a,b\in\mathbb{N}$ +with $a,b$ being negative integers then we have that +$a\in\left[\left(0,a\right)\right]$ and +$b\in\left[\left(0,b\right)\right]$. Intuitively we know that $-2+-3=-5$ +so we want these to add like in the positive integer case. This is to +say we have $a+b$ will be in the class +$\left[\left(0,a+b\right)\right]$. + +We can combine these two observations to define addition on the +integers. + +::: definition +**Definition 106**. *Addition on the Integers* + +*Let $x,y\in\mathbb{Z}$ with $x=\left(a,b\right)$ and +$y=\left(c,d\right)$. We define addition on the integers by* + +*$$\begin{equation} + \left[a,b\right]+\left[c,d\right]=\left[a+c,b+d\right] +\end{equation}$$* +::: + +To check this definition makes sense consider $x=4,y=3$. Both $x$ and +$y$ belong to some equivalence class, for example +$x\in\left[\left(5,1\right)\right]$ and +$y\in\left[\left(8,5\right)\right]$. Then we have that $x+y=7$ and + +$$\begin{equation*} + \left(5,1\right)+\left(8,5\right)=\left(5+8,1+5\right)=\left(13,6\right) \Rightarrow 13-6=7 +\end{equation*}$$ + +#### Extending multiplication to the integers + +We also extend multiplication to the integers. We have the definition of +multiplication on the naturals given by + +$$\begin{align*} + *&:\mathbb{N}\times\mathbb{N}\mathlarger{\mathlarger{\rightarrow}}\mathbb{N}\\ + \left(m,n\right)&\mapsto *\left(m,n\right)=\begin{cases} + m*0=0,\ \text{If } n=0\\ + m*S\left(n\right)=m*n+m,\ \text{If } n\neq 0 + \end{cases} +\end{align*}$$ + +As before, if we take $x,y\in\mathbb{Z}$ with $x,y$ being positive +integers then we have that $x\in\left[\left(x,0\right)\right]$ and +$b\in\left[\left(x,0\right)\right]$ we have that +$x*y\in\left[\left(x*y,0\right)\right]$. + +Suppose that $x,y\in\mathbb{Z}$ with $x=\left(a,b\right)$ and +$y=\left(c,d\right)$. We have that + +$$\begin{align*} + \left(a-b\right)*\left(c-d\right)&=\left(a-b\right)c-\left(a-b\right)d\\ + &=ac-bc-\left(ad-bd\right)\\ + &=ac-bc+bd-ad\\ + &= ac+bd-bc-ad\\ + &=ac+bd-\left(ad+bc\right) +\end{align*}$$ This is +$\left(a,b\right)*\left(c,d\right)=\left(ac+bd,ad+bc\right)$ + +This well be the definition of multiplication of the integers. + +::: definition +**Definition 107**. *Multiplication on the Integers* + +*Let $x,y\in\mathbb{Z}$ with $x=\left(a,b\right)$ and +$y=\left(c,d\right)$. We define multiplication on the integers by* + +*$$\begin{equation} + \left[a,b\right]*\left[c,d\right]=\left[ac+bd,ad+bc\right] +\end{equation}$$* +::: + +#### Closure properties of addition and multiplication + +As with the natural numbers we need to show that the operations of +addition and multiplication are closed. Additionally we want to prove +our claim at the start of this section that the integers allow us to +completely perform subtraction. + +::: theorem +**Theorem 20**. *Addition and multiplication on the integers are +well-defined operators and closed* + +*We have that $\forall x,y\in\mathbb{Z}$ that* + +1. *$x+y\in\mathbb{Z}$* + +2. *$x*y\in\mathbb{Z}$* + +*Proof:* + +1. *$x+y\in\mathbb{Z}$:* + + *We need to show that if $\left(a,b\right)\sim\left(a',b'\right)$ + and $\left(c,d\right)\sim\left(c',d'\right)$ then + $\left(a+c,b+d\right)\sim\left(a'+c',b'+d'\right)$ as this will show + equivalent elements produce the same result when added and therefore + integer addition is well-defined.* + + *We have by definition that $\left(a,b\right)\sim\left(a',b'\right)$ + that $a+b'=a'+b$, likewise we have + $\left(c,d\right)\sim\left(c',d'\right)$ gives $c+d'=c'+d$.* + + *Now, we have that* + + *$$\begin{align*} + a+b'+c+d'&=a'+b+c'+d\\ + a+c+b'+d'&=a'+c'+b+d\\ + \Rightarrow \left(a+c,b+d\right)&\sim\left(a'+c',b'+d'\right) + \end{align*}$$ Hence + $\left[\left(a+c,b+d\right)\right]=\left[\left(a'+c',b'+d'\right)\right]$ + and so addition is well-defined.* + + *It is left to prove closure. Let $x,y\in\mathbb{Z}$ with + $x=\left(a,b\right)$ and $y=\left(c,d\right)$. By definition of + integer addition we have that $x+y=\left(a+c,b+d\right)$ and + moreover we have $a+c\in\mathbb{N}$ and $b+d\in\mathbb{N}$. Hence + $\left(a+c,b+d\right)\in\left[a+c,b+d\right]$ and therefore + $x+y\in\mathbb{Z}$ showing closure.* + +2. *$x*y\in\mathbb{Z}$:* + + *As with addition we need to show that if + $\left(a,b\right)\sim\left(a',b'\right)$ and + $\left(c,d\right)\sim\left(c',d'\right)$ then + $\left(a,b\right)*\left(c,d\right) \sim \left(a',b'\right)*\left(c',d'\right)$. + As before we have that* + + *We have that* + + *$$\begin{equation*} + \left(a,b\right)*\left(c,d\right)=\left(ac+bd,ad+bc\right)\iff ac+bd-\left(ad+bc\right) + \end{equation*}$$* + + *Now as $\left(a,b\right)\sim\left(a',b'\right)$ then $a+b'=b+a'$ + and $\left(c,d\right)\sim\left(c',d'\right)$ then $c+d'=d+c'$. + Hence* + + *$$\begin{align*} + ac+bd-\left(ad+bc\right)&=\left(ac-ad\right)+\left(bd-bc\right)\\ + &=a\left(c-d\right)+b\left(d-c\right)\\ + &=a\left(c'-d'\right)+b\left(d'-c'\right), \text{ By assumption as} c+d'=d+c'\Rightarrow c-d=c'-d'\\ + &=ac'-ad'+bd'-bc'\\ + &=\left(ac'-bc'\right)+\left(bd'-ad'\right), \text{ By commutativity of the Naturals}\\ + &=c'\left(a-b\right)+d'\left(b-a\right)\\ + &=c'\left(a'-b'\right)+d'\left(b'-a'\right), \text{ By assumption as } a+b'=b+a'\Rightarrow a-b=a'-b'\\ + &=\left(c'a'-c'b'\right)+\left(d'b'-d'a'\right)\\ + &=c'a'-c'b'+d'b'-d'a'\\ + &=a'c'-b'c'+b'd'-a'd', \text{ By commutativity of the Naturals}\\ + &=\left(a'c+b'd'\right)-b'c'-a'd'\\ + &=\left(a'c+b'd'\right)-\left(a'd'+b'c'\right), \text{ By lemma \ref{lem:NaturalMinusDifferenceOfNatural}}\\ + \end{align*}$$* + + *This shows that multiplication is well-defined. It is left to show + closure. Let $x,y\in\mathbb{Z}$ with $x=\left(a,b\right)$ and + $y=\left(c,d\right)$. By the definition of multiplication on the + integers we have that $x*y=\left(ac+bd,ad+bc\right)$ with + $ac+bd\in\mathbb{N}$ and $ad+bc\in\mathbb{N}$. Hence we conclude + that $\left(ac+bd,ad+bc\right)\in\left[ac+bd,ad+bc\right]$, and so + by definition $x*y\in\mathbb{Z}$.* + +*The result is shown. $\qed$* +::: + +Now that we have shown closure we can deduce an immediate property. + +::: {#prop:multiplication_by_negative_one_for_integers .proposition} +**Proposition 63**. *Multiplication of an integer by $-1$* + +*Let $x\in\mathbb{Z}$ where $x\in\left[a,b\right]$ for some +$a,b\in\mathbb{N}$. We have that* + +1. *$-1*x = -1*\left(a,b\right)=\left(b,a\right)$* + +2. *$x*-1 = \left(a,b\right)*-1=\left(b,a\right)$* + +*Proof:* + +1. *$-1*x = -1*\left(a,b\right)=\left(b,a\right)$:* + + *We have that $-1\in\left[0,1\right]$ and so* + + *$$\begin{align*} + -1*x&=\left(0,1\right)*\left(a,b\right)\\ + &=\left(0*a+1*b,0*b+1*a\right)\\ + &=\left(b,a\right) + \end{align*}$$* + +2. *$x*-1 = \left(a,b\right)*-1=\left(b,a\right)$:* + + *Likewise we have* + + *$$\begin{align*} + x*-1&=\left(a,b\right)*\left(0,1\right)\\ + &=\left(a*0+b*1,a*1+b*0\right)\\ + &=\left(b,a\right) + \end{align*}$$* + +*As required. $\qed$* +::: + +::: {#cor:multiplication_by_negative_one_changes_integer_sign .corollary} +**Corollary 3**. *Multiplication of a positive integer by $-1$ makes it +a negative integer and multiplication of a negative integer by $-1$ +makes it a positive integer* + +1. *If $x$ is a positive integer then $-1*x$ is a negative integer.* + +2. *If $x$ is a negative integer then $-1*x$ is a positive integer.* + +*Proof:* + +*By definition if $x\in\mathbb{Z}$ is positive then +$x\in\left[a,0\right]$ for some $a\in\mathbb{N}$. By proposition +[63](#prop:multiplication_by_negative_one_for_integers){reference-type="ref" +reference="prop:multiplication_by_negative_one_for_integers"} we have +that $-1*x=\left(0,a\right)=x*-1$, which is by definition a negative +integer.* + +*Likewise if $x\in\mathbb{Z}$ is negative then $x\in\left[0,a\right]$ +for some $a\in\mathbb{N}$. By proposition +[63](#prop:multiplication_by_negative_one_for_integers){reference-type="ref" +reference="prop:multiplication_by_negative_one_for_integers"} we have +that $-1*x=\left(a,0\right)=x*-1$, which is by definition a positive +integer.* + +*$\qed$* +::: + +#### Associativity of integer addition and multiplication + +The associativity of addition and multiplication of the naturals also +extends to the integers. + +::: theorem +**Theorem 21**. *Let $x,y,z\in\mathbb{Z}$. We have that* + +1. *$x+\left(y+z\right)=\left(x+y\right)+z$* + +2. *$x\left(yz\right)=\left(xy\right)z$* + +*Proof:* + +1. *$x+\left(y+z\right)=\left(x+y\right)+z$:* + + *Let $x,y,z\in\mathbb{Z}$ be such that + $x=\left(a,b\right), y=\left(c,d\right)$ and $z=\left(e,f\right)$ + where $a,b,c,d,e,f\in\mathbb{N}$ and we have that + $\left(a,b\right)\in\left[a,b\right], \left(c,d\right)\in\left[c,d\right]$ + and $\left(e,f\right)\in\left[e,f\right]$. We have that* + + *$$\begin{align*} + x+\left(y+z\right)&=\left(a,b\right)+\left(\left(c,d\right)+\left(e,f\right)\right)\\ + &=\left(a,b\right)+\left(c+e,d+f\right)\\ + &=\left(a+\left(c+e\right),b+\left(d+f\right)\right)\\ + &=\left(\left(a+c\right)+e,\left(b+d\right)+f\right),\text{ By associativity of addition for natural numbers}\\ + &=\left(a+c,b+d\right)+\left(e,f\right)\\ + &=\left(\left(a,b\right)+\left(c,d\right)\right)+\left(e,f\right)\\ + &=\left(x+y\right)+z + \end{align*}$$* + + *Which shows associativity of addition.* + +2. *$x\left(yz\right)=\left(xy\right)z$:* + + *As with addition, let $x,y,z\in\mathbb{Z}$ be such that + $x=\left(a,b\right), y=\left(c,d\right)$ and $z=\left(e,f\right)$ + where $a,b,c,d,e,f\in\mathbb{N}$ and we have that + $\left(a,b\right)\in\left[a,b\right], \left(c,d\right)\in\left[c,d\right]$ + and $\left(e,f\right)\in\left[e,f\right]$. We then have that* + + *$$\begin{align*} + x\left(yz\right)&=\left(a,b\right)*\left(\left(c,d\right)\left(e,f\right)\right)\\ + &=\left(a,b\right)\left(ce+df,cf+de\right)\\ + &=\left(a\left(ce+df\right)+b\left(cf+de\right),a\left(cf+de\right)+b\left(ce+df\right)\right)\\ + &=\left(ace+adf+bcf+bde,acf+ade+bce+bdf\right)\\ + &=\left(ace+bde+adf+bcf,acf+bdf+ade+bce\right),\ \text{By associativity of addition for natural numbers}\\ + &=\left(\left(ac+bd\right)e+\left(ad+bc\right)f,\left(ac+bd\right)f+\left(ad+bc\right)e\right)\\ + &=\left(ac+bd,ad+bc\right)\left(e,f\right)\\ + &=\left(\left(a,b\right)\left(c,d\right)\right)\left(e,f\right)\\ + &=\left(xy\right)z + \end{align*}$$* + + *Showing associativity of multiplication.* + +*The result follows. $\qed$* +::: + +#### Commutativity of integer addition and multiplication + +As with the naturals, addition and multiplication in the integers both +satisfy commutativity. + +::: theorem +**Theorem 22**. *Addition and multiplication are commutative* + +*For all $x,y\in\mathbb{Z}$ we have that* + +1. *$x+y=y+x$* + +2. *$xy=yx$* + +*Proof:* + +1. *$x+y=y+x$:* + + *Let $x,y\in\mathbb{Z}$. By definition we have that + $x\in\left[a,b\right]$ and $y\in\left[c,d\right]$ for some + $a,b,c,d\in\mathbb{N}$. Let $x=\left(a,b\right)$ and + $y=\left(c,d\right)$. We then have by definition of addition that* + + *$$\begin{align*} + x+y&=\left(a,b\right)+\left(c,d\right)\\ + &=\left(a+c,b+d\right)\\ + &=\left(c+a,d+b\right),\ \text{By commutativity of addition for natural numbers}\\ + &= \left(c,d\right)+\left(a,b\right) + &=y+x + \end{align*}$$* + + *Showing commutativity holds for addition in the integers.* + +2. *$xy=yx$:* + + *Let $x,y\in\mathbb{Z}$ by definition we have that + $x\in\left[a,b\right]$ and $y\in\left[c,d\right]$ for some + $a,b,c,d\in\mathbb{N}$. So let $x=\left(a,b\right)$ and + $y=\left(c,d\right)$. By definition of multiplication we have* + + *$$\begin{align*} + xy&=\left(a,b\right)*\left(c,d\right)\\ + &=\left(ac+bd,ad+bc\right)\\ + &=\left(ca+db,da+bc\right), \text{By commutativity of multiplication of the naturals}\\ + &=\left(ca+db,da+bc\right), \text{By commutativity of addition of the naturals}\\ + &=\left(c,d\right)*\left(a,b\right)\\ + &=yx + \end{align*}$$* + + *Showing commutativity for integer multiplication.* + +*The result has been shown. $\qed$* +::: + +#### Multiplication distributes over addition + +Another result that extends from the naturals is that multiplication +distributes over addition. + +::: theorem +**Theorem 23**. *Multiplication distributes over addition* + +*For all $x,y,z\in\mathbb{Z}$ we have that* + +1. *$x\left(y+z\right)=xy+xz$* + +2. *$\left(y+z\right)x=yx+zx=xy+xz$* + +*Proof:* + +*Let $x,y,z\in\mathbb{Z}$ then +$x\in\left[a,b\right],y\in\left[c,d\right]$ and $z\in\left[e,f\right]$ +for some $a,b,c,d,e,f\in\mathbb{N}$.* + +*So let $x=\left(a,b\right), y=\left(c,d\right)$ and +$z=\left(e,f\right)$.* + +1. *$x\left(y+z\right)=xy+xz$:* + + *We have that* + + *$$\begin{align*} + x\left(y+z\right)&=\left(a,b\right)\left(\left(c,d\right)+\left(e,f\right)\right)\\ + &=\left(a,b\right)\left(c+e,d+f\right)\\ + &=\left(a\left(c+e\right)+b\left(d+f\right),a\left(d+f\right)+b\left(c+e\right)\right)\\ + &=\left(ac+ae+bd+bf,ad+af+bc+be\right)\\ + &=\left(ac+bd+ae+bf,ad+bc+af+be\right)\\ + &=\left(ac+bd,ad+bc\right)+\left(ae+bf,af+be\right)\\ + &=\left(a,b\right)\left(c,d\right)+\left(a,b\right)\left(e,f\right)\\ + &=xy+xz + \end{align*}$$* + +2. *$\left(y+z\right)x=yx+zx=xy+xz$:* + + *Now that we have the previous part the proof of this part is quick. + We have* + + *$$\begin{align*} + \left(y+z\right)x&=x\left(y+z\right), \text{By commutativity of multiplication}\\ + &=xy+xz, \text{By part }1.\\ + &=yx+zx, \text{By commutativity of multiplication} + \end{align*}$$* + +*As required. $\qed$* +::: + +#### The Zero and Identity laws + +The zero and identity laws from the naturals extend to the integers. + +::: theorem +**Theorem 24**. *The zero and Identity laws* + +*Let $x\in\mathbb{Z}$. We have that* + +1. *$x+0=x=0+x$* + +2. *$1*x=x=x*1$* + +*Proof:* + +*Let $x\in\mathbb{Z}$ then we have that $x=\left(a,b\right)$ for some +$a,b\in\mathbb{N}$* + +1. *$x+0=x=0+x$:* + + *We have that $0\in\left[0,0\right]$. Hence we have that* + + *$$\begin{equation*} + x+0=\left(a,b\right)+\left(0,0\right)=\left(a+0,b+0\right)=\left(a+b\right)=\left(0+a,0+b\right)=\left(0,0\right)+\left(a,b\right)=0+x + \end{equation*}$$* + +2. *$x*1=x=1*x$:* + + *As $1\in\left[1,0\right]$ then* + + *$$\begin{align*} + x*1&=\left(a,b\right)*\left(1,0\right)\\ + &=\left(a*1+b*0,b*1+a*0\right)\\ + &=\left(a+0,b+0\right)\\ + &=\left(a,b\right)=x\\ + &=\left(1*a+0*b,0*a+1*b\right)\\ + &=\left(1,0\right)\left(a,b\right)\\ + &=1*x + \end{align*}$$* + +*The result follows. $\qed$* +::: + +#### Extending subtraction to the integers + +As we have a notion of subtraction on the naturals, we can ask about +extending this to the integers. We defined subtraction on the naturals +as follows. Let $n,m\in\mathbb{N}$ such that $n\leq m$. Let +$d\in\mathbb{N}$ such that $n=m+d$. We define subtraction by + +$$\begin{equation*} + d=n-m +\end{equation*}$$ + +Where we called $d$ the difference between $n$ and $m$. We also have the +notion of a positive and negative integer. Recall that $x\in\mathbb{Z}$ +is a positive integer if and only if x Let $x\in\mathbb{Z}$. We say that +$x$ is a positive integer if and only if +$x\in\left[\left(b,0\right)\right]$ for some $b\in\mathbb{N}$. Likewise +$x$ is a negative integer if and only if +$x\in\left[\left(0,b\right)\right]$ for some $b\in\mathbb{N}$. In order +to extend subtraction to the integers we need to consider a few things. + +::: definition +**Definition 108**. *Negation of an natural number* + +*Let $x\in\mathbb{Z}$ so that $x$ is a positive integer, i.e a natural +number. We define the negation of $x$, denoted $-x$ by* + +*$$\begin{equation*} + -x=-1*x=\left(0,1\right)*x +\end{equation*}$$* + +*where $\left(0,1\right)\in\left[\left(0,-1\right)\right]$. That is +$\left(0,1\right)$ is an element of the equivalence class +$\left[\left(0,1\right)\right]$ which represents all possible elements +that are $-1$.* +::: + +We can extend this result to include a general integer. + +::: proposition +**Proposition 64**. *Negation of an integer* + +*Let $x\in\mathbb{Z}$ so that $x\in\left[\left(a,b\right)\right]$ for +some $a,b\in\mathbb{N}$. We have that* + +*$$\begin{equation*} + -1*x=-1*\left(a,b\right)=\left(b,a\right) +\end{equation*}$$* + +*Proof:* + +*Let $x\in\mathbb{Z}$ be as given by the hypothesis. We have that* + +*$$\begin{align*} + -1*x&=-1*\left(a,b\right)\\ + &=\left(0,1\right)*\left(a,b\right)\\ + &=\left(0*a+b*1,0*b+1*a\right)\\ + &=\left(b,a\right) +\end{align*}$$* + +*As required. $\qed$* +::: + +In light of this, we can define subtraction for integers. + +::: definition +**Definition 109**. *Integer subtraction* + +*Let $x,y\in\mathbb{Z}$. We define the subtraction of $y$ from $x$, +denoted $x-y$ by* + +*$$\begin{equation*} + x-y=x+\left(-y\right)=x+\left(-1*y\right) +\end{equation*}$$* +::: + +We immediately get that subtraction is closed, from the fact that both +addition and multiplication are closed. We do not have associativity of +subtraction in general. + +::: proposition +**Proposition 65**. *Integer subtraction is not associative* + +*Let $x,y,z\in\mathbb{Z}$. We have that* + +*$$\begin{equation*} + x-\left(y-z\right)\neq \left(x-y\right)-z +\end{equation*}$$* + +*Proof:* + +*Let $x=2, y=4$ and $z=6$, we have +$x\in\left[2,0\right], y\in\left[4,0\right]$ and $z\in\left[0,6\right]$ +so $x\in\left(2,0\right), y\in\left(4,0\right)$ and +$z\in\left(0,6\right)$ . We have that* + +*$$\begin{align*} + x-\left(y-z\right)&=\left(2,0\right)-\left(\left(4,0\right)-\left(6,0\right)\right)\\ + &=\left(2,0\right)-\left(\left(4,0\right)+\left(-1*\left(6,0\right)\right)\right)\\ + &=\left(2,0\right)-\left(\left(4,0\right)+\left(0,6\right)\right)\\ + &=\left(2,0\right)-\left(4,6\right)\\ + &=\left(2,0\right)+\left(-1*\left(4,6\right)\right)\\ + &=\left(2,0\right)+\left(6,4\right)\\ + &=\left(8,4\right)\\ +\end{align*}$$* + +*On the other side we have* + +*$$\begin{align*} + \left(x-y\right)-z&=\left(\left(2,0\right)-\left(4,0\right)\right)-\left(6,0\right)\\ + &=\left(\left(2,0\right)+\left(-1*\left(4,0\right)\right)\right)-\left(6,0\right)\\ + &=\left(\left(2,0\right)+\left(0,4\right)\right)-\left(6,0\right)\\ + &=\left(2,4\right)-\left(6,0\right)\\ + &=\left(2,4\right)+\left(-1*\left(6,0\right)\right)\\ + &=\left(2,4\right)+\left(0,6\right)\\ + &=\left(2,10\right) +\end{align*}$$* + +*Clearly $\left(8,4\right)\neq \left(2,10\right)$. Indeed they are not +even equivalent. Suppose that $\left(8,4\right)\sim\left(2,10\right)$ +then we have that $8+10=4+2$. However $18\neq 6$. $\qed$* +::: + +We can also immediately see the following result, which allows us to +formally show that subtraction is an inverse to addition. + +::: {#prop:IntegerAdditiveInverse .proposition} +**Proposition 66**. *Subtracting an integer from itself gives zero* + +*Let $x\in\mathbb{Z}$. We have that* + +*$$\begin{equation*} + x-x=0 +\end{equation*}$$* + +*Proof:* + +*Let $x\in\mathbb{Z}$ where $x\in\left[a,b\right]$ for some +$a,b\in\mathbb{N}$. We have* + +*$$\begin{align*} + x-x&=\left(a,b\right)-\left(a,b\right)\\ + &=\left(a,b\right)+\left(b,a\right)\\ + &=\left(a+b,b+a\right) +\end{align*}$$* + +*It is left to show that $\left(a+b,b+a\right)\sim\left(0,0\right)$. +Indeed* + +*$$\begin{equation*} + \left(a+b\right)+0=\left(b+a\right)+0 \Rightarrow a+b=b+a +\end{equation*}$$* + +*The result is shown. $\qed$* +::: + +#### The cancellation laws + +We can now deduce that the cancellation laws also extend to the +integers. + +::: theorem +**Theorem 25**. *The cancellation laws* + +*Let $x,y,z\in\mathbb{Z}$.* + +1. *If $x+y=x+z$ then we have $y=z$.* + +2. *For $x\neq 0$, if $xy=xz$ then we have that $y=z$* + +*Proof:* + +1. *If $x+y=x+z$ then we have $y=z$:* + + *Let $x,y,z\in\mathbb{Z}$. We have that* + + *$$\begin{align*} + x+y&=x+z\\ + \Rightarrow -x+x+y&=-x+x+z,\ \text{Adding the negative of } x \text{ to both sides}\\ + \Rightarrow \left(-x+x\right)+y*&=\left(-x+x\right)+z,\ \text{Associativity of integers}\\ + \Rightarrow 0+y&=0+z,\ \text{By proposition \ref{prop:IntegerAdditiveInverse}}\\ + \Rightarrow y&=z + \end{align*}$$* + +2. *For $x\neq 0$, if $xy=xz$ then we have that $y=z$:* + + *Let $x,y,z\in\mathbb{Z}$ where $x\neq 0$. Suppose that + $x\in\left[a,b\right], y\in\left[c,d\right]$ and + $z\in\left[e,f\right]$. We have* + + *$$\begin{align*} + xy&=\left(a,b\right)\left(c,d\right)=\left(ac+bd,ad+bc\right)\\ + xz&=\left(a,b\right)\left(e,f\right)=\left(ae+bf,af+be\right) + \end{align*}$$* + + *Now assume $xy=xz$ then we have that + $\left(ac+bd,ad+bc\right)\sim\left(ae+bd,ad+be\right)$ which is to + say* + + *$$\begin{equation*} + ac+bd+af+be=ae+bf+ad+bc + \end{equation*}$$* + + *Observe that* + + *$$\begin{align*} + ac+bd+af+be&=a\left(c+f\right)+b\left(d+e\right)\\ + ae+bf+ad+bc&=a\left(e+d\right)+b\left(f+c\right) + \end{align*}$$* + + *Which gives* + + *$$\begin{equation*} + a\left(c+f\right)+b\left(d+e\right)=a\left(e+d\right)+b\left(f+c\right) + \end{equation*}$$* + + *There are now two cases to consider, $ab$. Firstly + suppose that $a0$, + this is well-defined as $a,b\in\mathbb{N}$. We then have* + + *$$\begin{align*} + a\left(c+f\right)+b\left(d+e\right)&=a\left(e+d\right)+b\left(f+c\right)\\ + a\left(c+f\right)+\left(a+h\right)\left(d+e\right)&=a\left(e+d\right)+\left(a+h\right)\left(f+c\right)\\ + a\left(c+f\right)+a\left(d+e\right)+h\left(d+e\right)&=a\left(e+d\right)+a\left(f+c\right)+h\left(f+c\right)\\ + a\left(d+e\right)+h\left(d+e\right)&=a\left(e+d\right)+h\left(f+c\right),\text{ Cancelling }a\left(c+f\right)\\ + h\left(d+e\right)&=h\left(f+c\right),\text{ Cancelling }a\left(d+e\right)\\ + \left(d+e\right)&=\left(f+c\right),\text{ Cancelling }h\\ + \end{align*}$$* + + *Now as $d+e=f+c$ we have that + $c-d=e-f\Rightarrow \left(c,d\right)\sim\left(e,f\right)$ which is + the same as saying $y=z$.* + + *Now if $a>b$ then we write $b=a-h$ for some $h>0$, again being + well-defined as $a,b\in\mathbb{N}$. Thus* + + *$$\begin{align*} + a\left(c+f\right)+b\left(d+e\right)&=a\left(e+d\right)+b\left(f+c\right)\\ + a\left(c+f\right)+\left(a-h\right)\left(d+e\right)&=a\left(e+d\right)+\left(a-h\right)\left(f+c\right)\\ + a\left(c+f\right)+a\left(d+e\right)-h\left(d+e\right)&=a\left(e+d\right)+a\left(f+c\right)-h\left(f+c\right)\\ + a\left(d+e\right)-h\left(d+e\right)&=a\left(e+d\right)-h\left(f+c\right),\text{ Cancelling }a\left(c+f\right)\\ + -h\left(d+e\right)&=-h\left(f+c\right),\text{ Cancelling }a\left(d+e\right)\\ + \left(f+c\right)&=\left(d+e\right),\text{By adding each side to the other and cancelling }h\\ + \end{align*}$$* + + *As $f+c=d+e$ then we have by similar logic to before the $y=z$* + +*The result is shown. $\qed$* +::: + +#### Extending the summation and product notations to integers + +Summation and product notation has been defined on the naturals. As with +the theme of this section the notations extend in a natural way to +integers. As before we need to define a few things. + +Let $z\in\mathbb{Z}^{n+m+1}$ be an ordered $n+m+1$ tuple of integers +where $z=\left(z_{-m},z_{-m+1},\dots,z_{-1},z_0,z_1,z\dots, z_n\right)$ +and define +$\mathbb{Z}_m^n=\left\{-m,-m+1,-m+2,\dots,-1,0,1,\dots,n-1,n\right\}$. +Define $f:\mathbb{Z}_m^n\rightarrow\mathbb{Z}$ by + +$$\begin{align*} + f:\mathbb{Z}_m^n&\rightarrow \mathbb{Z}\\ + i&\mapsto f\left(i\right)=z_i +\end{align*}$$ + +As before, $f$ simply maps gets the value of $z_i$ from the ordered +tuple $z$. + +::: definition +**Definition 110**. *Summation notation for the integers* + +*Let $z\in\mathbb{Z}^{n+m+1}$ be ordered $n+m+1$ tuple of integers where +$z=\left(z_{-m},z_{-m+1},\dots,z_{-1},z_0,z_1,z\dots, z_n\right)$. +Define $\mathbb{Z}_m^n$ by +$\mathbb{Z}_m^n=\left\{-m,-m+1,-m+2,\dots,-1,0,1,\dots,n-1,n\right\}$. +Let $f:\mathbb{Z}^{n+m+1}:\mathbb{Z}$ defined by* + +*$$\begin{align*} + f:\mathbb{Z}^{m+n+1}&\rightarrow\mathbb{Z}\\ + i&\mapsto f\left(i\right)=z_i +\end{align*}$$* + +*We define the summation notation for integers by* + +*$$\begin{equation*} + \sum_{i=-m}^n f\left(i\right)=f\left(-m\right)+f\left(-m+1\right)+\dots+f\left(-1\right)+f\left(0\right)+f\left(1\right)+\dots+f\left(n\right) +\end{equation*}$$* + +*Alternatively this is written* + +*$$\begin{equation*} + \sum_{i=-m}^n z_i = z_{-m}+z_{-m+1}+\dots+z_{-1}+z_0+z_1+\dots+z_n +\end{equation*}$$* + +*We have that $i$ is called the index of summation and that $i=-m$ is +the starting index of the summation, and $n$ the ending index of the +summation. If $z=\emptyset$ then we define the summation to be $0$ and +call a summation an empty sum.* + +*We can also define the summation of some subset of $\mathbb{Z}_m^n$ +which allows for starting a summation at some starting point other than +$i=-m$. Let $T\subseteq\mathbb{Z}_m^n$. We define the summation over the +set $T$ by* + +*$$\begin{equation*} + \sum_{i\in T} z_i +\end{equation*}$$* + +*If we have a mapping $g:\mathbb{Z}\rightarrow\mathbb{Z}$ we can define +a summation over $g$ by* + +*$$\begin{equation*} + \sum_{i\in T} g\left(z_i\right) +\end{equation*}$$* + +*Finally we can define a summation over a predicate $P\left(i\right)$ +for $i\in T$ by* + +*$$\begin{equation*} + \sum_{P\left(i\right)}g\left(z_i\right) +\end{equation*}$$* + +*where we take the sum of the $g\left(z_i\right)$ for the $i$ that +satisfy the predicate $P$. We note that if we have $k>n$ for some +$k\in\mathbb{N}$ then the sum* + +*$$\begin{equation*} + \sum_{i=k}^n z_i=0 +\end{equation*}$$* +::: + +The proprieties shown for summations with natural numbers also extend to +the integer version. + +::: proposition +**Proposition 67**. *Properties of summation notation* + +*Let $n,m\in\mathbb{Z}$ such that $m0$ and $k>=0$ then the result is the same as for natural + numbers. So suppose that $k<0$. Consider the following set of the + indices given by* + + *$$\begin{equation*} + S=\left\{k,k+1,k+2,\dots,-1,0,1,\dots,n-1,n\right\} + \end{equation*}$$* + + *We have that the cardinality of $S$ is $n+1-k$. Indeed consider the + following mapping* + + *$$\begin{align*} + f:S&\rightarrow \mathbb{N}\\ + s&\mapsto f\left(s\right)=s-k + \end{align*}$$* + + *Define the mapping + $g:S\rightarrow\mathop{\mathrm{Image}}\left(f\right)$ then we have + that $g$ is a bijection. Suppose that + $g\left(x\right)=g\left(y\right)$ for some $x,y\in S$ then* + + *$$\begin{align*} + g\left(x\right)&=g\left(y\right)\\ + x-k&=y-k\\ + x&=y + \end{align*}$$* + + *showing injectivity. Now as $g$ is a mapping from $S$ to the image + of $f$ we have by proposition + [15](#prob:RestOfCodomainToImageIsSurjective){reference-type="ref" + reference="prob:RestOfCodomainToImageIsSurjective"} that $g$ is + surjective. Hence we conclude that $g$ is a bijection.* + + *Now we have that* + + *$$\begin{align*} + \mathop{\mathrm{Image}}\left(f\right)&=\left\{f\left(x\right):x\in S\right\}\\ + &= \left\{k-k,\left(k+1\right)-k,\left(k+2\right)-k,\dots,-1-k,0-k,1-k,\dots,\left(n-1\right)-k,n-k\right\}\\ + &=\left\{0,1,2,\dots,k-1,k,k-1,\dots,n-1-k,n-k\right\} + \end{align*}$$* + + *Hence + $\left|S\right|=\left|\mathop{\mathrm{Image}}\left(f\right)\right|=n-k+1$. + Hence the sum is adding $c$ to itself $n+1-k$ times. This is to say* + + *$$\begin{equation*} + \sum_{i=k}^n c= c\left(n+1-k\right) + \end{equation*}$$* + +5. *$\displaystyle\sum_{i=k}^n s_i+t_i = \sum_{i=k}^n s_i + \sum_{i=k}^n t_i$:* + + *This follows by the definition. We have* + + *$$\begin{align*} + \sum_{i=k}^n s_i+t_i&= \left(s_k+t_k\right)+\left(s_{k+1}+t_{k+1}\right)+\dots\\ + &+\left(s_{-1}+t_{-1}\right)+\left(s_{0}+t_{0}\right)+\left(s_{1}+t_{1}\right)+\dots+\left(s_{n-1}+t_{n-1}\right)+\left(s_{n}+t_{n}\right)\\ + &=\left(s_k+s_{k+1}+\dots+s_{-1}+s_0+s_1+\dots+s_{n-1}+s_n\right)+\\ + &+\left(t_k+t_{k+1}+\dots+t_{-1}+t_0+t_1+\dots+t_{n-1}+t_n\right)\\ + &= \sum_{i=k}^n s_i + \sum_{i=k}^n t_i + \end{align*}$$* + +*$\qed$* +::: + +We make a similar definition for product notation. + +::: definition +**Definition 111**. *Product notation for the integers* + +*Let $z\in\mathbb{Z}^{n+m+1}$ be ordered $n+m+1$ tuple of integers where +$z=\left(z_{-m},z_{-m+1},\dots,z_{-1},z_0,z_1,z\dots, z_n\right)$. +Define $\mathbb{Z}_m^n$ by +$\mathbb{Z}_m^n=\left\{-m,-m+1,-m+2,\dots,-1,0,1,\dots,n-1,n\right\}$. +Let $f:\mathbb{Z}^{n+m+1}:\mathbb{Z}$ defined by* + +*$$\begin{align*} + f:\mathbb{Z}^{m+n+1}&\rightarrow\mathbb{Z}\\ + i&\mapsto f\left(i\right)=z_i +\end{align*}$$* + +*We define the summation notation for integers by* + +*$$\begin{equation*} + \prod_{i=-m}^n f\left(i\right)=f\left(-m\right)*f\left(-m+1\right)*\dots*f\left(-1\right)*f\left(0\right)*f\left(1\right)*\dots+f\left(n\right) +\end{equation*}$$* + +*Alternatively this is written* + +*$$\begin{equation*} + \prod_{i=-m}^n z_i = z_{-m}*z_{-m+1}*\dots*z_{-1}*z_0*z_1*\dots*z_n +\end{equation*}$$* + +*We have that $i$ is called the index of the product and that $i=-m$ is +the starting index of the product, and $n$ the ending index of the +product. If $z\in\emptyset$ then we define the product to be $1$ and +call a product an empty sum.* + +*We can also define the product of some subset of $\mathbb{Z}_m^n$ which +allows for starting a product at some starting point other than $i=-m$. +Let $T\subseteq\mathbb{Z}_m^n$. We define the product over the set $T$ +by* + +*$$\begin{equation*} + \prod_{i\in T} z_i +\end{equation*}$$* + +*If we have a mapping $g:\mathbb{Z}\rightarrow\mathbb{Z}$ we can define +a product over $g$ by* + +*$$\begin{equation*} + \prod_{i\in T} g\left(z_i\right) +\end{equation*}$$* + +*Finally we can define a product over a predicate $P\left(i\right)$ for +$i\in T$ by* + +*$$\begin{equation*} + \prod_{P\left(i\right)}g\left(z_i\right) +\end{equation*}$$* + +*where we take the sum of the $g\left(z_i\right)$ for the $i$ that +satisfy the predicate $P$. We note that if we have $k>n$ for some +$k\in\mathbb{N}$ then the product* + +*$$\begin{equation*} + \prod_{i=k}^n z_i=1 +\end{equation*}$$* +::: + +::: proposition +**Proposition 68**. *Properties of product notation* + +*Let $n,m\in\mathbb{Z}$ such that $md$ then we have +that $\exists p\in\mathbb{N}$ such that $d+p=c$. We hence have* + +*$$\begin{align*} + ac+bd&=ad+bc\\ + a\left(d+p\right)+bd&=ad+b\left(d+p\right)\\ + ad+ap+bd&=ad+bd+bp\\ + ap&=bp\\ + a&=b ,\text{By the cancellation laws for the natural numbers}\\ + a+0&=b+0 \Rightarrow \left(a,b\right)=\left(0,0\right) +\end{align*}$$* + +*A similar argument applies for $c -1*y$. This can be shown in general. + +::: {#prop:MultiplicationByNegativeOneFlipsInequalitySign .proposition} +**Proposition 70**. *Multiplication by $-1$ changes the inequality sign* + +*Let $x,y\in\mathbb{Z}$. We have the following* + +1. *If $x-y$* + +2. *If $x\leq y$ then $-x\geq -y$* + +3. *If $x>y$ then $-x<-y$* + +4. *If $x\geq y$ then $-x\leq-y$* + +*Proof:* + +1. *If $x-y$:* + + *Let $x,y\in\mathbb{Z}$ so that $x + + 1. *$x\geq 0$ and $y\geq 0$:* + + *Suppose that $x\geq 0$ and $y\geq 0$ then + $x\in\left[\left(a,0\right)\right]$ for some $a\in\mathbb{N}$ + and $y\in\left[\left(b,0\right)\right]$ for some + $b\in\mathbb{N}$. As $xa$. Now we have $-x>-y$ by definition of + greater than for integers as we have* + + *$$\begin{equation*} + -x>-y \iff 0+b>a+0 + \end{equation*}$$* + + 2. *$x<0$ and $y\geq 0$:* + + *Now suppose that $x<0$ and $y\geq 0$ then we have that + $x\in\left[\left(0,a\right)\right]$ and + $y\in\left[\left(b,0\right)\right]$ where $a,b\in\mathbb{N}$.* + + *$$\begin{align*} + -x=-1*x=-1*\left(0,a\right)&=\left(a,0\right)\\ + -y=-1*y=-1*\left(b,0\right)&=\left(0,b\right) + \end{align*}$$* + + *Now, we have that if $-x>-y$ then we have* + + *$$\begin{equation*} + a+b>0+0 + \end{equation*}$$* + + *However as $a,b\in\mathbb{N}$ and $x<0 \implies a> 0$. We + conclude that $a+b\geq a > 0$ and so $-x>-y$.* + + 3. *$x<0$ and $y<0$:* + + *Now suppose that $x<0$ and $y< 0$ then + $x\in\left[\left(0,a\right)\right]$ for some $a\in\mathbb{N}$ + and $y\in\left[\left(0,b\right)\right]$ for some + $b\in\mathbb{N}$. As $xb$.* + + *We have that* + + *$$\begin{align*} + -x=-1*x=-1*\left(0,a\right)&=\left(a,0\right)\\ + -y=-1*y=-1*\left(0,b\right)&=\left(b,0\right) + \end{align*}$$* + + *Applying the definition of $>$ to $-x$ and $-y$ gives* + + *$$\begin{equation*} + -x>-y \iff a>b + \end{equation*}$$* + + *Which we know to be true. Hence $-x>-y$.* + + *This shows part 1.* + +2. *If $x\leq y$ then $-x\geq -y$:* + + *If $x-y$ from which it follows + that $-x\geq -y$ by definition. It is left to check when $x=y$. This + is clear however as $x=y\implies -x=-y$ and so $-x\geq -y$.* + +3. *If $x>y$ then $-x<-y$:* + + *The proof of this part is similar to part 1. As in part 1. there + are three cases to consider* + + 1. *$x\geq 0$ and $y\geq 0$* + + 2. *$x\geq 0$ and $< 0$* + + 3. *$x<0$ and $y<0$* + + + + 1. *$x\geq 0$ and $y\geq 0$:* + + *Suppose that $x\geq 0$ and $y\geq 0$ then + $x\in\left[\left(a,0\right)\right]$ for some $a\in\mathbb{N}$ + and $y\in\left[\left(b,0\right)\right]$ for some + $b\in\mathbb{N}$. As $x>y$ then we must have + $a+0>b+0\Rightarrow a>b$.* + + *We have that* + + *$$\begin{align*} + -x=-1*x=-1*\left(a,0\right)&=\left(0,a\right)\\ + -y=-1*y=-1*\left(b,0\right)&=\left(0,b\right) + \end{align*}$$* + + *Now, by proposition + [48](#prop:InequalityNaturalNumbers){reference-type="ref" + reference="prop:InequalityNaturalNumbers"} part 2. we know that + $a>b$ is the same as $b 0$. We + conclude that $0y$ then we have + that $0+b>a+0\Rightarrow b>a$ which is the same as $ay$ we apply part 3. So instead suppose $x=y$ but then + $x=y\Rightarrow -x=y$ and so by definition we have $-x\leq -y$.* + +*The result is shown. $\qed$* +::: + +This proposition will play a big role in the following proposition that +extends the results for the rules of inequalities to the integers. + +::: {#prop:InequalityIntegerNumbers .proposition} +**Proposition 71**. *Properties of inequalities for the integers* + +*Let $x,y,z,c\in\mathbb{Z}$. We have the following properties for +inequalities* + +1. *$xx$:* + +2. *$x\leq y$ is the same as $y\geq x$:* + +3. *If $xy$ and $y>z$ then $x>z$:* + +8. *If $x\geq y$ and $y>z$ then $x>z$:* + +9. *If $x>y$ and $y\geq z$ then $x>z$:* + +10. *If $x\geq y$ and $y\geq z$ then $x\geq z$:* + +11. *If $xy$ then $x+z>y+z$:* + +14. *If $x\geq y$ then $x+z\geq y+z$:* + +15. *If $xyz$:* + +17. *If $x\leq y$ and $z\geq 0$ then $xz\leq yz$:* + +18. *If $x\leq y$ and $z<0$ then $xz\geq yz$:* + +19. *If $x>y$ and $z\geq 0$ then $xz>yz$:* + +20. *If $x>y$ and $z< 0$ then $xzx$:* + + *Let $x,y\in\mathbb{Z}$ with $x + + 1. *$x\geq 0$ and $y\geq 0$:* + + *Suppose $x\geq 0$ and $y\geq 0$ then + $x\in\left[\left(a,0\right)\right]$ and + $y\in\left[\left(b,0\right)\right]$ for some $a,b\in\mathbb{N}$. + We have that $x< y$ only holds if $aa$ by proposition + [48](#prop:InequalityNaturalNumbers){reference-type="ref" + reference="prop:InequalityNaturalNumbers"}. But by definition of + $>$ for integers, we have that* + + *$$\begin{equation*} + b>a \iff y>x + \end{equation*}$$* + + 2. *$x<0$ and $y\geq 0$:* + + *Suppose that $x<0$ and $y\geq 0$, then + $x\in\left[\left(0,a\right)\right]$ and + $y\in\left[\left(b,0\right)\right]$ for some $a,b\in\mathbb{N}$. + By definition of $<$ we have that* + + *$$\begin{equation*} + xx\iff a+b > 0 + \end{equation*}$$* + + *Now, $x<0\implies a>0$ and so we have that $a+b\geq a > 0$ and + so $y>x$.* + + 3. *$x<0$ and $y<0$:* + + *Now suppose that $x<0$ and $y<0$, it follows that + $x\in\left[\left(0,a\right)\right]$ and + $y\in\left[\left(0,b\right)\right]$ for some $a,b\in\mathbb{N}$. + By definition of $<$ we have that* + + *$$\begin{equation*} + xx \iff a>b + \end{equation*}$$* + + *Hence, as $bb$ and so $y>x$.* + +2. *$x\leq y$ is the same as $y\geq x$:* + + *If $x + + 1. *$x\geq 0$, $y\geq 0$ and $z\geq 0$:* + + *Suppose that $x\geq 0$, $y\geq 0$ and $z\geq 0$ then the result + follows immediately by proposition + [48](#prop:InequalityNaturalNumbers){reference-type="ref" + reference="prop:InequalityNaturalNumbers"} part 6. as $x\geq 0$, + $y\geq 0$ and $z\geq 0$ gives + $x\in\left[\left(a,0\right)\right]$, + $y\in\left[\left(b,0\right)\right]$ and + $z\in\left[\left(c,0\right)\right]$ for some + $a,b,c\in\mathbb{N}$ and therefore $x,y,z\in\mathbb{N}$.* + + 2. *$x<0$, $y\geq 0$ and $z\geq 0$:* + + *Now suppose that $x<0$, $y\geq 0$ and $z\geq 0$. We have that + $x\in\left[\left(0,a\right)\right]$, + $y\in\left[\left(b,0\right)\right]$ and + $z\in\left[\left(c,0\right)\right]$ for some for some + $a,b,c\in\mathbb{N}$. Now we have that* + + *$$\begin{align*} + x<0 &\iff a>0 \\ + y\geq 0 &\iff b\geq 0\\ + z\geq 0&\iff c\geq 0 + \end{align*}$$* + + *By assumption $x0 \\ + y<0 &\iff b>0\\ + z\geq 0&\iff c\geq 0 + \end{align*}$$* + + *By assumption $x0 \\ + y<0 &\iff b>0\\ + z<0 &\iff c>0 + \end{align*}$$* + + *As $xy$ and $y>z$ then $x>z$:* + + *By part 1. of the proposition we have that this is equivalent to + $yz$ then $x>z$:* + + *Applying part 2 to $x\geq y$ and part 1. to $y>z$ and $x>z$ gives + the equivalent statement of $y\leq x$ and $zy$ and $y\geq z$ then $x>z$:* + + *As with part 8. Applying parts 2. and 1. gives the equivalent + statement of $yy$ then $x+z>y+z$:* + + *As has been the case so far, applying part 1. gives us the + statement $yyz$:* + + *Suppose that $xyz$, by definition we have* + + *$$\begin{align*} + xz>yz \iff be+ce>ae+de\iff e\underbrace{\left(b+c\right)}_{=m}y$ and $z\geq 0$ then $xz>yz$:* + + *Let $z\geq 0$ and by applying part 1. we get the equivalent + statement of $yy$ and $z< 0$ then $xz0$ which is to say + $-\left(x-y\right)\in\mathbb{N}$* + +*The result has been shown. $\qed$* +::: + +In light of the definition of the distance function, we can define the +so-called absolute value function. This will give us a notion of the +magnitude of an integer. + +::: definition +**Definition 113**. *Absolute value function* + +*Let $x\in\mathbb{Z}$ we define the absolute value function, denoted by +$\left|x\right|$ by the function* + +*$$\begin{equation*} + \left|x\right|=d\left(x,0\right)=\begin{cases} + x,\ \text{If } x\geq 0\\ + -x,\ \text{If } x< 0 + \end{cases} +\end{equation*}$$* +::: + +With this definition, we have generalised the idea of "size" to the +integers. That is the size of an integer is its distance from $0$. We +have the basic properties of the absolute value + +::: proposition +**Proposition 73**. *Properties of the absolute value* + +*Let $x,y,z\in\mathbb{Z}$. We have that the absolute value function has +the following properties* + +1. *$\left|x\right|\geq 0$ for all $x\in\mathbb{Z}$* + +2. *$\left|x\right|=0\iff x=0$* + +3. *$\left|x-y\right|=0\iff x=y$* + +4. *$\left|xy\right|=\left|x\right|\left|y\right|$* + +5. *$\left|\left|x\right|\right|=\left|x\right|$* + +6. *$\left|-x\right|=\left|x\right|$* + +7. *$\left|x\right|\leq y \iff -y\leq x\leq y$* + +8. *$\left|x\right|\geq y\iff x\leq -y$ or $x\geq y$* + +9. *$\left|x+y\right|\leq \left|x\right|+\left|y\right|$* + +10. *$\left|x-y\right|\leq\left|x-z\right|+\left|z-y\right|$* + +11. *$\left|x-y\right|\geq \left|\left|x\right|-\left|y\right|\right|$* + +12. *$\left|\cdot\right|$ is not injective* + +13. *$\left|\cdot\right|$ is not surjective* + +*Proof:* + +1. *$\left|x\right|\geq 0$ for all $x\in\mathbb{Z}$:* + + *This follows by proposition + [72](#prop:IntegerDistanceFuncWellDefined){reference-type="ref" + reference="prop:IntegerDistanceFuncWellDefined"}.* + +2. *$\left|x\right|=0\iff x=0$:* + + *We have by definition that $\left|x\right|=0$, if and only if + $x=0$.* + +3. *$\left|x-y\right|=0\iff x=y$:* + + *$\left(\Rightarrow\right)$: Suppose that $\left|x-y\right|=0$. + There are two cases to consider.* + + *Firstly if $x\geq y$, then by definition we have that + $\left|x-y\right|=x-y=0$ from which we clearly have $x=y$. The other + case is $x + + 1. *$x\geq 0$ and $y\geq 0$:* + + *If $x\geq 0$ and $y\geq 0$ then $xy\geq 0$ and so + $\left|xy\right|=xy$. Likewise $\left|x\right|=x$ and + $\left|y\right|=y$. Hence + $\left|xy\right|=\left|x\right|\left|y\right|$.* + + 2. *$x\geq 0$ and $y<0$:* + + *If $x\geq 0$ then $\left|x\right|=x$ by definition, and if + $y<0$ then $\left|y\right|=-y$. Now $\left|xy\right|=-xy$ as + $y<0$. Moreover, we have that* + + *$$\begin{equation*} + -xy=\left(-1\right)\left(x\right)\left(y\right)=\left(x\right)\left(-1\right)\left(y\right)=\left(x\right)\left(-y\right)=\left|x\right|\left|y\right| + \end{equation*}$$* + + *Hence we get $\left|xy\right|=\left|x\right|\left|y\right|$* + + 3. *$x<0$ and $y\geq 0$:* + + *This is similar to the above but swapping the roles of $x$ and + $y$.* + + 4. *$x<0$ and $y<0$:* + + *Suppose that $x<0$ and $y<0$, then we have that + $\left|x\right|=-x$ and $\left|y\right|=-y$ by definition. + Moreover, we have that $-x*-y = xy$. Hence + $\left|xy\right|=xy=\left(-x\right)\left(-y\right)=\left|x\right|\left|y\right|$* + +5. *$\left|\left|x\right|\right|=\left|x\right|$:* + + *We have that $\left|x\right|=x$ if $x\geq 0$ and $-x$ if $x<0$.* + + *So if $x\geq 0$, we have* + + *$$\begin{equation*} + \left|\left|x\right|\right|=\left|x\right|=x=\left|x\right| + \end{equation*}$$* + + *Now if $x<0$ then* + + *$$\begin{equation*} + \left|\left|x\right|\right|=\left|-x\right|=\underbrace{-x}_{\text{As }-x>0}=\left|x\right| + \end{equation*}$$* + +6. *$\left|-x\right|=\left|x\right|$:* + + *As $-x=-1 *x$ we have by part 4 that* + + *$$\begin{equation*} + \left|-x\right|=\left|-1*x\right|=\left|-1\right|\left|x\right|=1*\left|x\right|=\left|x\right| + \end{equation*}$$* + +7. *$\left|x\right|\leq y \iff -y\leq x\leq y$:* + + *$\left(\Rightarrow\right)$: Suppose that $\left|x\right|\leq y$. If + $x\geq 0$ then we get that $\left|x\right|=x\leq y$. From this, it + is clear that $-y\leq x\leq y$ as $x\geq 0$ and + $x\leq y \Rightarrow y \geq 0$.* + + *Now if $x<0$, then $\left|x\right|=-x\leq y$. Clearly $x\leq -x$ as + $x<0$ hence we conclude that $x\leq -x\leq y$. Now by part 18 of + proposition + [71](#prop:InequalityIntegerNumbers){reference-type="ref" + reference="prop:InequalityIntegerNumbers"} we have we have* + + *$$\begin{equation*} + \left(-1\right)*\left(-x\right)\geq \left(-1\right)\left(y\right) \iff x\geq -y + \end{equation*}$$* + + *Now $x\geq -y$ is the same as $-y\leq x$ and so we have + $-y\leq x\leq -x \leq y$.* + + *Hence $-y\leq x\leq y$.* + + *$\left(\Leftarrow\right)$: Suppose that $-y\leq x\leq y$. There are + two cases to consider.* + + 1. *$x\geq 0$* + + 2. *$x<0$* + + + + 1. *$x\geq 0$:* + + *Suppose $x\geq 0$, then clearly as $x\leq y$ then + $\left|x\right|\leq \left|y\right|=y$. Moreover, we have that + $-y\leq x$ is the same $x\geq -y$ and by part 22. of proposition + [71](#prop:InequalityIntegerNumbers){reference-type="ref" + reference="prop:InequalityIntegerNumbers"} when applied to + $x\geq -y$ gives* + + *$$\begin{equation*} + \left(-1\right)*\left(x\right)\leq \left(-1\right)\left(-y\right) \iff -x\leq y + \end{equation*}$$* + + *We have that $\left|-x\right|=\left|x\right|$ by part 6. Hence + $\left|-x\right|=\left|x\right|\leq \left|y\right|=y$.* + + 2. *$x<0$:* + + *Suppose $x<0$. By assumption $x\leq y$ so either $y\geq 0$ or + $y< 0$. We can't have $y<0$ as for example take $x=-4$ and + $y=-2$ then we would have $2\leq -4\leq -2$ a contradiction.* + + *So suppose that $y\geq 0$ then as $x\leq y$ we have + $\left|x\right|\leq\left|y\right|=y$. Now as $-y\leq x$ by + assumption we have that $x\geq -y$ and so part 22. of + proposition + [71](#prop:InequalityIntegerNumbers){reference-type="ref" + reference="prop:InequalityIntegerNumbers"} gives* + + *$$\begin{equation*} + \left(-1\right)*\left(x\right)\leq \left(-1\right)\left(-y\right) \iff -x\leq y + \end{equation*}$$* + + *Hence part 6. applies and we get that $\left|x\right|\leq y$* + +8. *$\left|x\right|\geq y\iff x\leq -y$ or $x\geq y$:* + + *$\left(\Rightarrow\right)$: Suppose that $\left|x\right|\geq y$. If + $x\geq 0$ then $\left|x\right|=x\geq y$. So suppose that $x<0$ then + by definition we have that $\left|x\right|=-x$ and so $-x\geq y$ and + the result follows when applying part 22. of proposition + [71](#prop:InequalityIntegerNumbers){reference-type="ref" + reference="prop:InequalityIntegerNumbers"}.* + + *$\left(\Leftarrow\right)$: Suppose that either $x\leq -y$ or + $x\geq y$. We have three cases to consider.* + + 1. *$x\leq -y$* + + 2. *$x\geq y$* + + 3. *$x\leq -y$ and $x\geq y$* + + + + 1. *$x\leq -y$:* + + *Suppose that $x\leq -y$ holds. If $x\geq 0$ then we have that + $-y\geq 0$, Hence $y<0$. Moreover, we have that by part 18. of + proposition + [71](#prop:InequalityIntegerNumbers){reference-type="ref" + reference="prop:InequalityIntegerNumbers"} that* + + *$$\begin{equation*} + \left(-1\right)*\left(x\right)\geq \left(-1\right)\left(-y\right) \iff -x\geq y + \end{equation*}$$* + + *Now part 6. applies and we see that + $\left|-x\right|=\left|x\right|\geq\left|y\right|=y$. This is to + say $\left|x\right|\geq y$.* + + *Now suppose that $x<0$. Then as $x\leq -y$ we have that either + $-y\geq 0$ or $-y<0$. In the former case $-y\geq 0$ gives $y<0$. + Hence by part 18. of proposition + [71](#prop:InequalityIntegerNumbers){reference-type="ref" + reference="prop:InequalityIntegerNumbers"} we conclude that* + + *$$\begin{equation*} + \left(-1\right)*\left(x\right)\geq \left(-1\right)\left(y\right) \iff -x\geq y + \end{equation*}$$* + + *As $x<0$ then $-x\geq 0$. The result follows when taking the + absolute value.* + + *Now suppose that $-y<0$ then $y\geq 0$. Following similar logic + to the previous case, we see that* + + *$$\begin{equation*} + \left(-1\right)*\left(x\right)\geq \left(-1\right)\left(y\right) \iff -x\geq y + \end{equation*}$$* + + *The result again follows after taking the absolute value.* + + 2. *$x\geq y$:* + + *This case is trivial.* + + 3. *$x\leq -y$ and $x\geq y$:* + + *Suppose that $x\leq -y$ and $x\geq y$ are both true. We know by + the first case that $x\leq -y$ gives $\left|x\right|\geq y$ and + $x\leq y$ also implies $\left|x\right|\geq y$ by the second + case. Hence both inequalities being true at the same time + implies the result $\left|x\right|\geq y$.* + +9. *$\left|x+y\right|\leq \left|x\right|+\left|y\right|$:* + + *Let $x,y\in\mathbb{Z}$. There are four cases to consider.* + + 1. *$x\geq 0$ and $y\geq 0$* + + 2. *$x\geq 0$ and $y\leq 0$* + + 3. *$x\leq 0$ and $y\geq 0$* + + 4. *$x\leq 0$ and $y\leq 0$* + + + + 1. *$x\geq 0$ and $y\geq 0$:* + + *Suppose $x\geq 0$ and $y\geq 0$, then we have that* + + *$$\begin{equation*} + \left|x+y\right|=x+y=\left|x\right|+\left|y\right|\Rightarrow \left|x+y\right|\leq\left|x\right|+\left|y\right| + \end{equation*}$$* + + 2. *$x\geq 0$ and $y\leq 0$* + + *By assumption we have that $\left|x\right|=x$ and + $\left|y\right|=-y$. We have two cases based on the absolute + value, $\left|x\right|\leq\left|y\right|$ and + $\left|x\right|\geq\left|y\right|$.* + + *So suppose that $\left|x\right|\leq\left|y\right|$ then by + definition $x\leq -y$ and so by part 12. of proposition + [71](#prop:InequalityIntegerNumbers){reference-type="ref" + reference="prop:InequalityIntegerNumbers"} we have that* + + *$$\begin{equation*} + x\leq -y \Rightarrow x+y\leq 0 + \end{equation*}$$* + + *Moreover, as $x\geq 0$ then $y\leq x+y\leq 0$. Hence we have by + the definition of the absolute value that* + + *$$\begin{equation*} + \left|x+y\right|=-\left(x+y\right)\leq -y=\left|y\right| + \end{equation*}$$ As $-y>0$.* + + *In the case $\left|x\right|\geq\left|y\right|$ we have by + definition that $x\geq -y$ and so $x+y\geq 0$. Additionally it + is clear that $x\geq x+y$ as $y\leq 0$ and + $\left|x\right|\geq\left|y\right|$. Hence by definition of the + absolute value we have that* + + *$$\begin{equation*} + \left|x+y\right|=x+y\leq x=\left|x\right| + \end{equation*}$$* + + *Now, it is clear to see that + $\left|x\right|\leq \left|x\right|+\left|y\right|$ and likewise + $\left|y\right|\leq \left|x\right|+\left|y\right|$.* + + *We have hence shown that + $\left|x+y\right|leq\left|x\right|+\left|y\right|$.* + + 3. *$x\leq 0$ and $y\geq 0$:* + + *This is similar to above, interchanging the roles of $x$ and + $y$.* + + 4. *$x\leq 0$ and $y\leq 0$:* + + *Suppose that $x\leq 0$ and $y\leq 0$ then by definition we have + that $\left|x+y\right|=-\left(x+y\right)=-x-y$. As $x\leq 0$ and + $y\leq 0$ then we have that and $\left|y\right|=-y$ which shows + $\left|x+y\right|=\left|x\right|+\left|y\right|\leq\left|x\right|+\left|y\right|$* + +10. *$\left|x-y\right|\leq\left|x-z\right|+\left|z-y\right|$:* + + *We have that* + + *$$\begin{align*} + \left|x-y\right|&=\left|x-\left(z-z\right)-y\right|\\ + &=\left|x-z+z-y\right|\\ + &\leq \left|x-z\right|+\left|z-y\right| + \end{align*}$$* + +11. *$\left|x-y\right|\geq \left|\left|x\right|-\left|y\right|\right|$:* + + *We have that* + + *$$\begin{align*} + \left|x\right|&=\left|\left(x-y\right)+y\right|\leq \left|x-y\right|+\left|y\right| \Rightarrow \left|x\right|-\left|y\right|\leq \left|x-y\right|\\ + \left|y\right|&=\left|\left(y-x\right)+x\right|\leq \left|x-y\right|+\left|x\right| \Rightarrow \left|y\right|-\left|x\right|\leq \left|x-y\right|\\ + \end{align*}$$* + + *Hence we have* + + *$$\begin{align*} + \left|x\right|-\left|y\right|\leq \left|x-y\right| &\Rightarrow \left|\left|x\right|-\left|y\right|\right|\leq \left|x-y\right|\\ + \left|y\right|-\left|x\right|=\left(-1\right)\left(\left|x\right|-\left|y\right|\right)\leq \left|x-y\right| &\Rightarrow \left|\left|x\right|-\left|y\right|\right|\leq \left|x-y\right|\\ + \end{align*}$$* + + *Hence we have the result.* + +12. *$\left|\cdot\right|$ is not injective:* + + *To see that the absolute value function is not injective consider + $\left|3\right|=\left|-3\right|$. We have that $\left|3\right|=3$ + and $\left|-3\right|=3$ but $3\neq -3$.* + +13. *$\left|\cdot\right|$ is not surjective:* + + *We have that the absolute value function as there are no + $x\in\mathbb{Z}$ so that $\left|x\right|=-1$ for example.* + +*This ends the proposition. $\qed$* +::: + +#### Extending exponentiation to the integers + +We can extend the idea of exponentiation to include integers. We are now +able to consider negative bases. In other words, expressions of the form +$\displaystyle x^n$ for $x\in\mathbb{Z}$ with $x<0$. This extension is +somewhat trivial and extends naturally from the definition of the +naturals. We first look at the case where $n\geq 0$ + +::: definition +**Definition 114**. *Exponentiation of integer numbers* + +*Let $\mathbb{Z}^+=\left\{x\in\mathbb{Z}:x\geq 0\right\}$. Let +$\left(x,n\right)\in\mathbb{Z}\times\mathbb{Z}$ with $n\geq 0$ and let +$\wedge:\mathbb{Z}\times\mathbb{Z}\rightarrow\mathbb{Z}$. We define the +exponentiation of $x$ by $n$ to be $x$ multiplied by itself $n-1$ times* + +*$$\begin{align*} + \wedge:\mathbb{Z}\times\mathbb{Z}^+&\rightarrow\mathbb{Z}\\ + \left(x,n\right)&\mapsto \wedge\left(x,n\right)=\begin{cases} + 1,\ \text{If } x=0\text{ and } n=0\\ + 1,\ \text{If } n=0\\ + \displaystyle \prod_{i=1}^y x ,\ \text{If }x\neq 0\text{ and } n \geq 0\\ + \end{cases} +\end{align*}$$* + +*We will write $\wedge\left(x,n\right)$ as $x^n$. We say that $x$ is the +base and $n$ is the exponent. We sometimes say that $x$ has been raised +to the power of $n$. In the case that $x=0$ and $m=0$ we have a vacuous +product and so an empty product which by definition has a value of $1$.* +::: + +We will explore this definition by first considering $x=-1$ + +$$\begin{align*} + x*x=x^1&=-1=-1\\ + x*x=x^2&=-1*-1=1\\ + x*x*x=x^3&=-1*-1*-1=-1\\ + x*x*x*x=x^4&=-1*-1*-1*-1=1\\ +\end{align*}$$ + +This leads to the following proposition. + +::: proposition +**Proposition 74**. *Negative one to power of 2n is 1 Let +$n\in\mathbb{N}$. We have that* + +*$$\begin{equation*} + \left(-1\right)^{2n} = 1 +\end{equation*}$$* + +*Proof:* + +*We argue by induction on $n$. The base case is $n=0$ and by definition, +we have that* + +*$$\begin{equation*} + \left(-1\right)^{2*0}=\left(-1\right)^{0}=1=1 +\end{equation*}$$* + +*Now suppose the result holds for some $n=k$, that is* + +*$$\begin{equation*} + \left(-1\right)^{2k}=1 +\end{equation*}$$* + +*We show that* + +*$$\begin{equation*} + \left(-1\right)^{2*\left(k+1\right)}=1 +\end{equation*}$$* + +*We have* + +*$$\begin{align*} + \left(-1\right)^{2\left(k+1\right)}&=\left(-1\right)^{2k+2}\\ + &=\prod_{i=1}^{2k+2} \left(-1\right)\\ + &=\prod_{i=1}^{2k} \left(-1\right) *\prod_{i=2k+1}^{2k+2} \left(-1\right)\\ + &= 1 * \left(\left(-1\right)\left(-1\right)\right) + &=1*\left(1\right)=1 +\end{align*}$$* + +*Which shows the result. $\qed$* +::: + +This result generalises for any negative integer. + +::: proposition +**Proposition 75**. *Negative integer to the power of 2n is positive* + +*Let $x\in\mathbb{Z}$ with $x<0$. Let $n\in\mathbb{N}$. We have that* + +*$$\begin{equation*} + x^{2n} > 1 +\end{equation*}$$* + +*Proof:* + +*By definition we have* + +*$$\begin{align*} + x^{2n}&=\prod_{i=1}^{2n} x\\ + &=\prod_{i=1}^{2n} \left(-1*-x\right)\\ + &=\prod_{i=1}^{2n} \left(-1\right) *\prod_{i=}^{2n}\left(-x\right)\\ + &=1*\underbrace{\prod_{i=}^{2n}\left(-x\right)}_{\geq 1} \geq 1\\ +\end{align*}$$* + +*As $-x>0$ because $x<0$. $\qed$* +::: + +We also note that exponentiation is neither commutative nor associative +as they were not for the naturals. However, the following results do +extend. + +::: {#prop:IntegerExponentiationPowerLaw .proposition} +**Proposition 76**. *Power law of exponentiation for positive exponents* + +*Let $x\in\mathbb{Z}$ and let $n,m\mathbb{N}$ with $n\geq 0$ and +$m\geq 0$. We have that* + +*$$\begin{equation*} + \left(x^n\right)^m = x^{nm} +\end{equation*}$$* + +*Proof:* + +*By the definition of exponentiation, we have that* + +*$$\begin{equation*} + \left(x^n\right)^m=\prod_{i=1}^m x^n =\prod_{i=1}^m\left(\prod_{j=1}^n x\right) +\end{equation*}$$* + +*Hence we have* + +*$$\begin{align*} + \left(x^n\right)^m&=\underbrace{\prod_{j=1}^n x * \prod_{j=1}^n x *\dots * \prod_{j=1}^n x}_{n\text{ times}}\\ + &=\underbrace{\underbrace{x*x*\dots*x}_{n\text{ times}}*\underbrace{x*x*\dots*x}_{n\text{ times}}*\dots*\underbrace{x*x*\dots*x}_{n\text{ times}}}_{m\text{ times}}\\ +\end{align*}$$* + +*Therefore, there are $n*m$ total multiplications of $x$ with itself. +Which is to say* + +*$$\begin{equation*} + \left(x^n\right)^m = \underbrace{x*x*x*\dots*x}_{n*m\text{ times}} = \prod_{i=1}^{nm} x = x^{nm} +\end{equation*}$$* + +*As promised. $\qed$* +::: + +::: {#prop:IntegerExponentiationOfSameBaseAddsPowers .proposition} +**Proposition 77**. *Multiplying exponents of the same base adds the +powers* + +*Let $x\in\mathbb{Z}$ be a fixed integer and let $n,m\in\mathbb{N}$. We +have that* + +*$$\begin{equation*} + x^n *x^m = x^{n+m} +\end{equation*}$$* + +*Proof:* + +*Let $x\in\mathbb{Z}$ and $n,m\in\mathbb{N}$ If $n=0$ or $m=0$ or both +then the result is trivial. Likewise if $n=0$ and $m\geq 0$ or $n\geq 0$ +and $m=0$ again the result is trivial. So suppose that $n>0$ and $m>0$. +We have by definition of exponentiation that* + +*$$\begin{equation*} + x^n*x^m=\prod_{i=1}^n x * \prod_{i=1}^m x = \underbrace{x*x*\dots *x}_{n\text{ times}} * \underbrace{x*x*\dots *x}_{m\text{ times}}=\underbrace{x*x*\dots *x}_{n+m \text{ times}}=x^{n+m} +\end{equation*}$$* + +*As expected. $\qed$* +::: + +::: {#prop:IntegerExponentiationPowerOfProductIsProductOfPowers .proposition} +**Proposition 78**. *Power of product is product of powers* + +*Let $x,y\in\mathbb{Z}$ and $n\in\mathbb{N}$. Then* + +*$$\begin{equation*} + \left(x*y\right)^n=x^n*y^n +\end{equation*}$$* + +*Proof:* + +*If $n=0$ then $\left(x*y\right)^n=1$ and clearly $x^0*y^0=1$. So let +$n>0$ then we have* + +*$$\begin{align*} + \left(x*y\right)^n=\prod_{i=1}^n xy &=\underbrace{xy*xy*\dots *xy}_{n\text{ times}}\\ + &= \left(\underbrace{x*x*\dots *x}_{n\text{ times}}\right)*\left(\underbrace{y*y*\dots *y}_{n\text{ times}}\right),\ \text{ By commutativity of multiplication}\\ + &=x^n*y^n +\end{align*}$$* + +*Showing the proposition. $\qed$* +::: + +The awake reader may have noticed how we have only dealt with positive +exponents so far in our extension of exponentiation to the integers. +What about negative exponents? We can, loosely, justify why we can't yet +consider negative exponents by considering proposition +[77](#prop:IntegerExponentiationOfSameBaseAddsPowers){reference-type="ref" +reference="prop:IntegerExponentiationOfSameBaseAddsPowers"}. For a +second suppose that instead of $n.m\in\mathbb{N}$ we consider +$n,m\in\mathbb{Z}$. In particular $n=1$ and $m=-1$, then we have that + +$$\begin{equation*} + x^1*x^{-1}=x^{1+-1}=x^0=1 +\end{equation*}$$ + +Hence we have that when $x^1$ is multiplied by $x^{-1}$ we get back +to 1. Hence in a sense $x^{-1}$ cancels with $x$. If we let $x=2$ we +have $x^1=2$ and so $x^1*x^{-1}=1$ gives us the equation $2*x^{-1}=1$. +We intuitively know that $\displaystyle x^{-1}=\frac{1}{2}$ which we +know is not an integer. Hence if +[77](#prop:IntegerExponentiationOfSameBaseAddsPowers){reference-type="ref" +reference="prop:IntegerExponentiationOfSameBaseAddsPowers"} held for all +integer powers we have the implied existence of a new type of object. +This object has the potential that when an integer is multiplied by the +appropriate member of this new type of object, assuming such an object +even exists, then integer multiplication is undone. + +### Construction of the Rationals + +::: epigraph +A man is like a fraction whose numerator is what he is and whose +denominator is what he thinks of himself. The larger the denominator, +the smaller the fraction. + +*Leo Tolstoy* +::: + +We have now built a theory of integer numbers. One main reason for doing +this was to be able to always undo subtraction. We still have a glaring +issue at hand, however. How do we undo multiplication? For example, we +are unable to express in mathematical language how many times one +quantity goes into another. If we have $6$ pints and $3$ friends we know +that each friend should get $2$ pints as $3*2=6$. In a sense we have +that $2$ goes into $6$ a total of $3$ times and $3$ goes into $6$ a +total of $2$ times. The integers don't have a concept of how many times +one integer can go into another. This is what we call division and we +write $\displaystyle\frac{6}{2}=3$ and $\displaystyle\frac{6}{3}=2$ for +each situation respectively. + +Thankfully the method used to construct the integers can be used again +on the integers themselves to construct an even richer theory. As with +the integers, we should consider what we want to do. We seek a way to +undo the multiplication of integers. Consider $a,b,c,d\in\mathbb{Z}$ +$a=6,b=3,c=12$ and $d=6$, with these values we intuitively know that +$\displaystyle\frac{6}{3}=2$ and $\displaystyle\frac{12}{6}=2$. We also +note that $6*6=36$ and $3*12=36$. This gives us a clue on how to +proceed. We have that $\displaystyle\frac{6}{3}$ and +$\displaystyle\frac{12}{6}$ are hence similar. If we temporarily use the +language of relations we have that +$\left(a,b\right)\sim\left(c,d\right)$. + +#### Defining the Rationals + +We proceed by defining division as an ordered tuple on integers + +::: definition +**Definition 115**. *Division as an ordered tuple* + +*Let $a,b\in\mathbb{Z}$. We define division as an ordered tuple +$\left(a,b\right)\in\mathbb{Z}^2$ to mean $\displaystyle\frac{a}{b}$. We +will call $x\in\mathbb{Z}^2$ a division tuple in this context.* +::: + +Hence we can define the relation we considered above. + +::: definition +**Definition 116**. *Relation for division* + +*Let $\left(a,b\right),\left(c,d\right)\in\mathbb{Z}^2$ be division +tuples. We define the relation $\sim$ such that +$\left(a,b\right)\sim\left(c,d\right)$ if and only if $ad=bc$* +::: + +With this definition there is something we need to consider that we have +heard since school, you can't divide by zero, that is for any integer +$a$ we have $\displaystyle\frac{a}{0}$ is not defined. + +Suppose that $\left(a,0\right)\sim\left(c,d\right)$ for some +$a,c,d\in\mathbb{Z}$. We have by definition of the relation that + +$$\begin{equation*} + \left(a,0\right)\sim\left(c,d\right)\iff ad=0*c = 0 +\end{equation*}$$ + +By proposition +[69](#prop:IntegersHaveNoZeroDivisors){reference-type="ref" +reference="prop:IntegersHaveNoZeroDivisors"} we have that either $a=0$ +or $d=0$ or both. + +If $a=0$ then we have +$\left(0,0\right)\sim\left(c,d\right)\Rightarrow 0=0$ for all +$c,d\in\mathbb{Z}$. This means that every division tuple in +$\mathbb{Z}^2$ would be equivalent to $\left(0,0\right)$. Likewise if +$d=0$ we get $\left(a,0\right)\sim\left(c,0\right)\Rightarrow 0=0$ again +meaning for all division tuples in $\mathbb{Z}^2$ would be equivalent. +Finally if both $a=0$ and $d=0$ then +$\left(0,0\right)\sim\left(c,0\right)$ and so $0=0*c=0$ and again every +division tuple would be equivalent. + +This is a problem as this relation would imply that all elements are +essentially the same[^9]. This is not a useful definition to be using so +we will avoid this by not allowing $b=0$ in +$\left(a,b\right)\in\mathbb{Z}^2$. We revise the definition + +::: definition +**Definition 117**. *Division as an ordered tuple* + +*Let $a,b\in\mathbb{Z}$ with $b\neq 0$. We define division as an ordered +tuple $\left(a,b\right)\in\mathbb{Z}^2$ to mean +$\displaystyle\frac{a}{b}$. We will call $x\in\mathbb{Z}^2$ a division +tuple in this context.* +::: + +::: definition +**Definition 118**. *Relation for division* + +*Let $\left(a,b\right),\left(c,d\right)\in\mathbb{Z}^2$ be division +tuples where $b\neq 0$ and $d\neq 0$. We define the relation $\sim$ such +that $\left(a,b\right)\sim\left(c,d\right)$ if and only if $ad=bc$* +::: + +We can show that this revised definition is an equivalence relation. + +::: proposition +**Proposition 79**. *Relation for division ordered tuple is an +equivalence relation* + +*Let $x,y,z\in\mathbb{Z}^2$ be division tuples and defined the relation +$x\sim y$ as above. We have that $\sim$ is an equivalence relation.* + +*Proof:* + +*Let $x,y,z\in\mathbb{Z}^2$ be division tuples such that +$x=\left(a,b\right),y=\left(c,d\right)$ and $z=\left(e,f\right)$. We +show that $\sim$ is an equivalence relation, in other words.* + +1. *$\sim$ is reflexive* + +2. *$\sim$ is symmetric* + +3. *$\sim$ is transitive* + + + +1. *$\sim$ is reflexive:* + + *We have that for $x=\left(a,b\right)$ that $x\sim x$ as $x\sim x$ + if and only if $ab=ab$.* + +2. *$\sim$ is symmetric:* + + *Suppose that $x=\left(a,b\right)$ and $y=\left(c,d\right)$. Suppose + that $x\sim y$ then we have that $ad=bc$. Hence + $bc=ad \Rightarrow cb=ad$ and so + $\left(c,d\right)\sim\left(a,b\right)$ and so $y\sim x$.* + +3. *$\sim$ is transitive:* + + *Suppose that $x\sim y$ and $y\sim z$ then by definition we have + that $ad=bc$ and $cf=de$. We have that* + + *$$\begin{align*} + ad&=bc\\ + adf&=bcf\\ + adf&=bde\\ + af&=be + \end{align*}$$* + + *Hence $\left(a,b\right)\sim\left(e,f\right)$ and so $x\sim z$.* + +*It follows that $\sim$ is an equivalence relation. $\qed$* +::: + +We can now turn our attention to the set +$\mathbb{Z}^2/\sim=\left\{\left[x\right]_\sim:x\in\mathbb{Z}^2\right\}$. + +::: definition +**Definition 119**. *Rationals* + +*Let $\mathbb{Z}^2$ have the equivalence relation $\sim$ defined by +$\left(a,b\right)\sim\left(c,d\right)$ if and only if $ad=bc$. We define +the set of rational numbers, denoted $\mathbb{Q}$, as the quotient set +$\mathbb{Z}^2/\sim$. The set has the form* + +*$$\begin{equation} + \mathbb{Q}=\left\{\dots,-\frac{2}{3},-\frac{1}{3},-\frac{1}{2},0,\frac{1}{2},\frac{1}{3},\frac{2}{3},\dots\right\} +\end{equation}$$* +::: + +#### Extending equality to the rationals + +As with the integers, it is easy to extend equality. + +::: definition +**Definition 120**. *Equality of rationals* + +*Let $x,y\in\mathbb{Q}$ be two rational numbers. We define that two +rationals are equal, denoted $x=y$ if and only if $x\sim y$. That is $x$ +and $y$ are in the same equivalence class. If $x\not\sim y$ then we say +that $x$ is not equal to $y$ and write $x\neq y$.* +::: + +#### Extending inequality operators to the rationals + +The inequality operators can be extended to the rationals in a natural +way. + +::: definition +**Definition 121**. *Less than operator* + +*Let $x,y\in\mathbb{Q}$ where $x\in\left[a,b\right]$ and +$y\in\left[c,d\right]$ for some $a,b,c,d\in\mathbb{Z}$. The less than +operator, denoted by $xy$ is defined by the logical proposition* + +*$$\begin{equation*} + >\left(x,y\right)=\begin{cases} + 1,\ \text{If } ad>bc\\ + 0,\ \text{Otherwise} + \end{cases} +\end{equation*}$$* + +*This can equivalently be express as* + +*$$\begin{equation*} + x>y \iff ad>bc +\end{equation*}$$* +::: + +::: definition +**Definition 124**. *Greater than or equal to operator* + +*Let $x,y\in\mathbb{Q}$ where $x\in\left[a,b\right]$ and +$y\in\left[c,d\right]$ for some $a,b,c,d\in\mathbb{Z}$. The greater than +or equal to operator, denoted by $x\geq y$ is defined by the logical +proposition* + +*$$\begin{equation*} + \geq\left(x,y\right)=\begin{cases} + 1,\ \text{If } ad\geq bc\\ + 0,\ \text{Otherwise} + \end{cases} +\end{equation*}$$* + +*This can equivalently be express as* + +*$$\begin{equation*} + x\geq y \iff ad\geq bc +\end{equation*}$$* +::: + +#### Extending addition to the rationals + +We can extend addition to the rationals. To do so we need to consider +how integers are represented in the rationals. As we know an element +$\left(a,b\right)\in\mathbb{Q}$ is going to represent +$\displaystyle\frac{a}{b}$. So we can start by considering what an +integer will look like. We know by the definition of the equivalence +relation that for $\left(a,b\right),\left(c,d\right)\in\mathbb{Z}^2$ +that + +$$\begin{equation*} + \left(a,b\right)\sim\left(c,d\right)\iff ad=bc +\end{equation*}$$ + +Hence if we have for $b=d=1$ that + +$$\begin{equation*} + \left(a,1\right)\sim\left(c,1\right)\iff a=c +\end{equation*}$$ + +Hence an integer can be represented in the rationals by an element of +the form $\left(k,1\right)$ for all $k\in\mathbb{Z}$. Therefore if +$x,y\in\mathbb{Z}$ they will have the representation +$x=\left(x_1,1\right)$ and $y=\left(y_1,1\right)$ for some +$x_1,y_1\in\mathbb{Z}$. Hence by integer addition, we have that + +$$\begin{equation*} + x+y=\left(x_1,1\right)+\left(y_1,1\right)=\left(x_1+y_1,1\right) +\end{equation*}$$ + +Now what happens if $a=c=1$? From the definition of the equivalence +relation we have that + +$$\begin{equation*} + \left(1,b\right)\sim\left(1,d\right)\iff d=b +\end{equation*}$$ + +So we see that $\left(1,b\right)\sim\left(1,d\right)$ means that +intuitively $\displaystyle\frac{1}{b}=\frac{1}{d}$. The question now +becomes what is $\displaystyle\frac{1}{b}+\frac{1}{b}$? + +For example consider $\displaystyle\frac{1}{2}+\frac{1}{2}=1$, or +$\displaystyle\frac{1}{3}+\frac{1}{3}=\frac{2}{3}$. It seems the result +we need is that $\displaystyle\frac{1}{b}+\frac{1}{b}=\frac{2}{b}$. We +hence have that + +$$\begin{equation*} + \left(1,b\right)+\left(1,b\right)=\left(2,b\right) +\end{equation*}$$ + +Hence more generally we have that + +$$\begin{equation*} + \left(a,b\right)+\left(c,b\right)=\left(a+c,b\right) +\end{equation*}$$ + +Now, from intuition, we know that for example +$\displaystyle\frac{1}{3}=\frac{2}{6}=\frac{1*2}{3*2}$. In the language +of the relation we have defined, we have that + +$$\begin{equation*} + \left(a,b\right)\sim\left(ad,bd\right) +\end{equation*}$$ + +With these facts, we have enough to recover the definition of the +addition of rational numbers we were told in school. + +We have that + +$$\begin{align*} + \left(a,b\right)+\left(c,d\right)&\sim\left(ad,bd\right)+\left(bc,bd\right)\\ + &\sim\left(ad+bc,bd\right) +\end{align*}$$ + +Indeed, we have for example + +$$\begin{equation*} + \frac{1}{2}+\frac{1}{3}=\frac{3*1+2*1}{3*2}=\frac{5}{6} +\end{equation*}$$ + +We make the required definition. + +::: definition +**Definition 125**. *Addition on the Rationals* + +*Let $x,y\in\mathbb{Q}$ with $x\in\left[a,b\right]$ and +$y=\left[c,d\right]$ so that $b\neq 0$ and $d\neq 0$. We define addition +on the rationals by* + +*$$\begin{equation} + x+y=\left[a,b\right]+\left[c,d\right]=\left[ad+bc,bd\right] +\end{equation}$$* +::: + +#### Extending multiplication to the rationals + +We can extend multiplication to the rationals as well. As with extending +addition, we should consider how integers are represented in the +rationals. As before an integer in the rationals is of the form +$\left(a,1\right)$ and given the definition from the integers we know we +must have + +$$\begin{equation*} + \left(a,1\right)*\left(b,1\right)=\left(ab,1\right) +\end{equation*}$$ + +Now we need to answer the question of +$\left(1,b\right)*\left(1,d\right)$. Taking a similar approach as to +addition we will consider some examples. We intuitively know that +$\displaystyle 1*\frac{1}{2}=\frac{1}{2}$. This is to say that + +$$\begin{equation*} + \left(1,1\right)*\left(1,2\right)=\left(1,2\right) +\end{equation*}$$ + +We also knot that $2*2=4$ and so we know $\displaystyle\frac{4}{2}=2$. +In other words we must have that + +$$\begin{equation*} + \left(4,1\right)*\left(1,2\right)=\left(2,1\right)\sim\left(4,2\right) +\end{equation*}$$ + +Now, suppose we have $\displaystyle\frac{3}{2}=1.5$, what is +$\displaystyle\frac{3}{2}*\frac{1}{3}$? Again we know intuitively that +$0.5+0.5+0.5=3(0.5)=1.5$, hence we can write + +$$\begin{equation*} + \left(3,2\right)*\left(1,3\right)=\left(1,2\right)\sim\left(3,6\right) +\end{equation*}$$ + +We can now see how to handle $\left(1,b\right)*\left(1,d\right)$ and +more generally $\left(a,b\right)*\left(c,d\right)$. We make the +definition. + +::: definition +**Definition 126**. *Multiplication on the Rationals* + +*Let $x,y\in\mathbb{Q}$ with $x\in\left[a,b\right]$ and +$y=\left[c,d\right]$ so that $b\neq 0$ and $d\neq 0$. We define +multiplication on the rationals by* + +*$$\begin{equation} + x*y=\left[a,b\right]*\left[c,d\right]=\left[ac,bd\right] +\end{equation}$$* +::: + +#### Closure properties of addition and multiplication + +As with the natural numbers and integers we need to show that the +operations of addition and multiplication on the rationals are closed +and well-defined. + +::: theorem +**Theorem 26**. *Addition and multiplication on the rational are +well-defined operators and closed* + +*We have that $\forall x,y\in\mathbb{Q}$ that* + +1. *$x+y\in\mathbb{Q}$* + +2. *$x*y\in\mathbb{Q}$* + +*Proof:* + +1. *$x+y\in\mathbb{Q}$:* + + *We must show that if $\left(a,b\right)\sim\left(a',b'\right)$ and + $\left(c,d\right)\sim\left(c',d'\right)$ then we have* + + *$$\begin{equation*} + \left(ad+bc,bd\right)\sim\left(a'd'+b'c',b'd'\right) + \end{equation*}$$* + + *By definition we have that $\left(a,b\right)\sim\left(a',b'\right)$ + holds if and only if $ab'=ba'$, likewise + $\left(c,d\right)\sim\left(c',d'\right)$ holds if and only if + $cd'=c'd$. It is left to show + $\left(ad+bc,bd\right)\sim\left(a'd'+b'c',b'd'\right)$. By + definition of the equivalence relation we have that* + + *$$\begin{equation*} + \left(ad+bc,bd\right)\sim\left(a'd'+b'c',b'd'\right) \iff \left(ad+bc\right)b'd'=bd\left(a'd'+b'c'\right) + \end{equation*}$$* + + *We have that* + + *$$\begin{align*} + \left(ad+bc\right)b'd'&=adb'd'+bcb'd'\, \text{ As integer multiplication distributes over the addition}\\ + &=\left(ab'\right)\left(dd'\right)+\left(cd'\right)\left(bb'\right)\, \text{ By commutativity}\\ + &=\left(ba'\right)\left(dd'\right)+\left(dc'\right)\left(bb'\right)\, \text{ By the equivalence relation}\\ + &=\left(bd\right)\left(a'd'\right)+\left(bd\right)\left(b'c'\right)\, \text{ By commutativity}\\ + &=bd\left(a'd'+b'c' ,\right)\, \text{ As integer multiplication distributes over the addition} + \end{align*}$$* + + *Which is what we wished to show. Hence addition is well-defined. It + is left to show closure. Let $x,y\in\mathbb{Q}$ with + $x=\left(a,b\right)$ and $y=\left(c,d\right)$ so that $b\neq 0$ and + $d\neq 0$. By definition of addition we have that* + + *$$\begin{equation*} + \left(a,b\right)+\left(c,d\right)=\left(ad+bc,bd\right) + \end{equation*}$$* + + *As $ad+bc\in\mathbb{Z}$ and $bd\in\mathbb{Z}$ then + $\left(ad+bc,bd\right)\in\left[ad+bc,bd\right]$ and so + $x+y\in\mathbb{Q}$.* + +2. *$x*y\in\mathbb{Q}$:* + + *As with addition we need to show that if + $\left(a,b\right)\sim\left(a',b'\right)$ and + $\left(c,d\right)\sim\left(c',d'\right)$ that* + + *$$\begin{equation*} + \left(ac,bd\right)\sim\left(a'c',b'd'\right) + \end{equation*}$$* + + *As $\left(a,b\right)\sim\left(a',b'\right)$ holds if and only if + $ab'=ba'$, likewise $\left(c,d\right)\sim\left(c',d'\right)$ holds + if and only if $cd'=c'd$. It is left to show + $\left(ac,bd\right)\sim\left(a'c',b'd'\right)$, that is* + + *$$\begin{equation*} + \left(ac,bd\right)\sim\left(a'c',b'd'\right)\iff acb'd'=bda'c' + \end{equation*}$$* + + *We have* + + *$$\begin{align*} + acb'd'&=\left(ab'\right)\left(cd'\right)\, \text{By commutativity}\\ + &=\left(ba'\right)\left(c'd\right),\ \text{By the equivalence relation}\\ + &=bda'c',\ \text{By commutativity} + \end{align*}$$* + + *Showing that multiplication is well-defined. To show closure let + $x,y\in\mathbb{Q}$ with $x=\left(a,b\right)$ and + $y=\left(c,d\right)$ so that $b\neq 0$ and $d\neq 0$ then by + definition we have that* + + *$$\begin{equation*} + \left(a,b\right)*\left(c,d\right)=\left(ac,bd\right) + \end{equation*}$$* + + *From which it is clear that $ac,bd\in\mathbb{Z}$ so + $x*y\in\mathbb{Q}$* + +*The result is shown. $\qed$* +::: + +#### Associativity of rational addition and multiplication + +The associativity of addition and multiplication extends to the +rationals. + +::: theorem +**Theorem 27**. *Let $x,y,z\in\mathbb{Q}$. We have that* + +1. *$x+\left(y+z\right)=\left(x+y\right)+z$* + +2. *$x\left(yz\right)=\left(xy\right)z$* + +*Proof:* + +1. *$x+\left(y+z\right)=\left(x+y\right)+z$:* + + *Let $x,y,z\in\mathbb{Q}$ be such that + $x=\left(a,b\right), y=\left(c,d\right)$ and $z=\left(e,f\right)$ + where $a,b,c,d,e,f\in\mathbb{N}$ and we have that + $\left(a,b\right)\in\left[a,b\right], \left(c,d\right)\in\left[c,d\right]$ + and $\left(e,f\right)\in\left[e,f\right]$. We have that* + + *$$\begin{align*} + x+\left(y+z\right)&=\left(a,b\right)+\left(\left(c,d\right)+\left(e,f\right)\right)\\ + &=\left(a,b\right)+\left(cf+de,df\right)\\ + &=\left(adf+b\left(cf+de\right),bdf\right)\\ + &=\left(adf+bcf+bde,bdf\right)\\ + &=\left(\left(ad+bc\right)f+bde,bdf\right)\\ + &=\left(\left(ad+bc\right)f+ebd,bdf\right),\text{ By associativity of addition for integer numbers}\\ + &=\left(ad+bc,bd\right)+\left(e,f\right)\\ + &=\left(\left(a,b\right)+\left(c,d\right)\right)+\left(e,f\right)\\ + &=\left(x+y\right)+z + \end{align*}$$* + + *Which shows associativity of addition.* + +2. *$x\left(yz\right)=\left(xy\right)z$:* + + *As with addition, let $x,y,z\in\mathbb{Q}$ be such that + $x=\left(a,b\right), y=\left(c,d\right)$ and $z=\left(e,f\right)$ + where $a,b,c,d,e,f\in\mathbb{Z}$ and we have that + $\left(a,b\right)\in\left[a,b\right], \left(c,d\right)\in\left[c,d\right]$ + and $\left(e,f\right)\in\left[e,f\right]$. We then have that* + + *$$\begin{align*} + x\left(yz\right)&=\left(a,b\right)*\left(\left(c,d\right)\left(e,f\right)\right)\\ + &=\left(a,b\right)*\left(ce,df\right)\\ + &=\left(ace,bdf\right)\\ + &=\left(ac,bd\right)*\left(e,f\right)\\ + &=\left(\left(a,b\right)*\left(c,d\right)\right)*\left(e,f\right) + \end{align*}$$* + + *Showing associativity of multiplication.* + +*The result follows. $\qed$* +::: + +#### Commutativity of rational addition and multiplication + +As with the naturals and integers, addition and multiplication in the +rationals both satisfy commutativity. + +::: theorem +**Theorem 28**. *Addition and multiplication are commutative* + +*For all $x,y\in\mathbb{Q}$ we have that* + +1. *$x+y=y+x$* + +2. *$xy=yx$* + +*Proof:* + +1. *$x+y=y+x$:* + + *Let $x,y\in\mathbb{Q}$. By definition we have that + $x\in\left[a,b\right]$ and $y\in\left[c,d\right]$ for some + $a,b,c,d\in\mathbb{Z}$. Let $x=\left(a,b\right)$ and + $y=\left(c,d\right)$. We then have by definition of addition that* + + *$$\begin{align*} + x+y&=\left(a,b\right)+\left(c,d\right)\\ + &=\left(ad+bc,bd\right)\\ + &=\left(bc+ad,bd\right),\ \text{By associativity of addition for the integers}\\ + &=\left(cb+da,db\right),\ \text{By commutativity of addition for the integers}\\ + &= \left(c,d\right)+\left(a,b\right) + &=y+x + \end{align*}$$* + + *Showing commutativity holds for addition in the integers.* + +2. *$xy=yx$:* + + *Let $x,y\in\mathbb{Q}$ by definition we have that + $x\in\left[a,b\right]$ and $y\in\left[c,d\right]$ for some + $a,b,c,d\in\mathbb{Z}$. So let $x=\left(a,b\right)$ and + $y=\left(c,d\right)$. By definition of multiplication we have* + + *$$\begin{align*} + xy&=\left(a,b\right)*\left(c,d\right)\\ + &=\left(ac,bd\right)\\ + &=\left(ca,db\right), \text{By commutativity of multiplication of the integers}\\ + &=\left(c,d\right)*\left(a,b\right)\\ + &=yx + \end{align*}$$* + + *Showing commutativity for integer multiplication.* + +*The result has been shown. $\qed$* +::: + +#### The Zero and Identity laws + +The zero and identity laws from both the naturals and integers extend to +the rationals. But first, we show the following result. + +::: lemma +**Lemma 7**. *Representation of zero in the rationals* + +*We have that $0=\left[0,a\right]$ for all $a\in\mathbb{Z}$ with +$a\neq 0$* + +*Proof:* + +*Let $x,y\in\left[0,a\right]$ with $x=\left(0,a_1\right)$ and +$y=\left(0,a_2\right)$. We hence have that$x\sim y$ and* + +*Where the final $0=0$ is the zero of the integers, from which the +result is clear. $\qed$* +::: + +We take the natural representation of $0$ for the rationals. + +::: theorem +**Theorem 29**. *The zero and Identity laws* + +*Let $x\in\mathbb{Q}$. We have that* + +1. *$x+0=x=0+x$* + +2. *$1*x=x=x*1$* + +*Proof:* + +*Let $x\in\mathbb{Q}$ then we have that $x=\left(a,b\right)$ for some +$a,b\in\mathbb{Z}$* + +1. *$x+0=x=0+x$:* + + *We have that $0\in\left[0,1\right]$. Hence we have that* + + *$$\begin{align*} + x+0&=\left(a,b\right)+\left(0,1\right)\\ + &=\left(a*1+b*0,b*1\right)\\ + &=\left(a,b\right)=x\\ + &=\left(1*a+0*b.1*b\right)\\ + &=\left(0,1\right)*\left(a,b\right)\\ + &=0+x + \end{align*}$$* + +2. *$x*1=x=1*x$:* + + *As $1\in\left[1,1\right]$ then* + + *$$\begin{align*} + x*1&=\left(a,b\right)*\left(1,1\right)\\ + &=\left(a*1,b*1\right)\\ + &=\left(a,b\right)\\ + &=\left(a,b\right)=x\\ + &=\left(1*a,1*b\right)\\ + &=\left(1,0\right)\left(a,b\right)\\ + &=1*x + \end{align*}$$* + +*The result follows. $\qed$* +::: + +#### Multiplication distributes over addition + +Yet another result that extends to the rationals is that multiplication +distributes over addition. + +::: theorem +**Theorem 30**. *Multiplication distributes over addition* + +*For all $x,y,z\in\mathbb{Q}$ we have that* + +1. *$x\left(y+z\right)=xy+xz$* + +2. *$\left(y+z\right)x=yx+zx=xy+xz$* + +*Proof:* + +*Let $x,y,z\in\mathbb{Q}$ then +$x\in\left[a,b\right],y\in\left[c,d\right]$ and $z\in\left[e,f\right]$ +for some $a,b,c,d,e,f\in\mathbb{Z}$.* + +*Let $x=\left(a,b\right), y=\left(c,d\right)$ and $z=\left(e,f\right)$.* + +1. *$x\left(y+z\right)=xy+xz$:* + + *We have that* + + *$$\begin{align*} + x\left(y+z\right)&=\left(a,b\right)\left(\left(c,d\right)+\left(e,f\right)\right)\\ + &=\left(a\left(cf+ed\right),bdf\right)\\ + &=\left(acf+aed,bdf\right),\ \text{By multiplication distributes over addition for the integers}\\ + &=\left(acf+aed,bdf\right)*\left(1,1\right),\ \text{By the identity law for the rationals}\\ + &=\left(acf+aed,bdf\right)*\left(b,b\right),\ \text{As }\left(1,1\right)\sim\left(b.b\right)\\ + &=\left(\left(acf+aed\right)b,bdfb\right)\\ + &=\left(acfb+aedb,bdfb\right),\ \text{By multiplication distributes over addition for the integers}\\ + &=\left(acbf+aebd,bdbf\right),\ \text{By commutativity of integer multiplication}\\ + &=\left(ac,bd\right)+\left(ae,bf\right)\\ + &=\left(a,b\right)\left(c,d\right)+\left(a,b\right)\left(e,f\right)\\ + &=xy+xz + \end{align*}$$* + +2. *$\left(y+z\right)x=yx+zx=xy+xz$:* + + *Invoking the previous part of the proof we have that* + + *$$\begin{align*} + \left(y+z\right)x&=x\left(y+z\right), \text{By commutativity of multiplication}\\ + &=xy+xz, \text{By part }1.\\ + &=yx+zx, \text{By commutativity of multiplication} + \end{align*}$$* + +*As required. $\qed$* +::: + +#### Extending subtraction to the rationals + +We can extend subtraction from the integers to the rationals. Recall +that subtraction was defined for $x,y\in\mathbb{Z}$ by + +$$\begin{equation*} + x-y=x+\left(-y\right)=x+\left(-1*y\right) +\end{equation*}$$ + +That is to say subtraction was defined by adding the negation of $y$ to +$x$. We will use a similar idea to define subtraction on the rationals. +Firstly we need to consider what it means to negate a rational number. +To do so we need to define what it means for a rational number to be +\"positive\" or \"negative\". + +We know that any integer $x$ can be expressed as a rational by +$\left(x,1\right)$ and so in this case $\left(x,1\right)$ is positive if +$x$ is positive and $\left(x,1\right)$ is negative if $x$ is negative. +Hence a general rational number $\left(a,b\right)$ being positive or +negative will depend on $a$ and $b$ being positive or negative. There +are a few cases to consider. + +1. Suppose that $a$ is positive and $b$ is positive. We have that for + $\left(a,b\right)\sim\left(c,d\right)$ for some $c,d\in\mathbb{Z}$ + that + + $$\begin{equation*} + ad=cb + \end{equation*}$$ + + As $a$ and $b$ are positive then we are forced to conclude that $c$ + and $d$ are also positive for if not then one side of this equation + would have a different sign. + +2. Suppose that $a$ is positive and $b$ is negative. Then as before we + have that for $\left(a,b\right)\sim\left(c,d\right)$ to be true that + + $$\begin{equation*} + ad=cb + \end{equation*}$$ + + As $b$ was negative then we have that $cb$ is either positive or + negative depending on $c$. If $c$ is positive then $cb$ is negative + and so $d$ must also be negative. Likewise if $c$ is negative then + $cb$ is positive and $d$ must be positive. + +The cases for when $a$ is negative and $b$ is either positive or +negative are similar. We can use this to make a definition for a +positive and negative rational number. + +::: definition +**Definition 127**. *Positive and negative rational number* + +*Let $x\in\mathbb{Q}$ so that $x=\left(a,b\right)$ for some +$a,b\in\mathbb{Z}$. We say that $x$ is a positive rational number if and +only if $a$ is positive and $b$ is positive. That is to say +$x\in\mathbb{Q}$ is positive if and only if +$\mathop{\mathrm{sgn}}\left(a\right)=\mathop{\mathrm{sgn}}\left(b\right)$ +with $\mathop{\mathrm{sgn}}\left(a\right)\neq 0$ and +$\mathop{\mathrm{sgn}}\left(b\right)\neq 0$ where +$\mathop{\mathrm{sgn}}$ denotes the sign function of an integer.* + +*If +$\mathop{\mathrm{sgn}}\left(a\right)\neq\mathop{\mathrm{sgn}}\left(b\right)$ +and $\mathop{\mathrm{sgn}}\left(a\right)\neq 0$ and +$\mathop{\mathrm{sgn}}\left(b\right)\neq 0$ then we have that $x$ is a +negative rational number.* + +*Finally if $\mathop{\mathrm{sgn}}\left(a\right)= 0$ and +$\mathop{\mathrm{sgn}}\left(b\right)\neq 0$ then we say that $x$ is +neither positive or negative.* +::: + +We can summarise this definition using $\mathop{\mathrm{sgn}}$ just like +we did for the integers. + +::: definition +**Definition 128**. *Sign of a rational number* + +*Let $x\in\mathbb{Q}$ where $x=\left(a,b\right)$ with $a,b\in\mathbb{Z}$ +and $b\neq 0$. We define the sign of $x$, denoted by +$\mathop{\mathrm{sgn}}\left(x\right)$ to be the following function* + +*$$\begin{align*} + \mathop{\mathrm{sgn}}:\mathbb{Q}&\rightarrow\left\{-1,0,1\right\}\\ + x&\mapsto\mathop{\mathrm{sgn}}\left(x\right)=\begin{cases} + 1,\ \text{If } x\text{ is a positive rational number}\\ + -1,\ \text{If } x\text{ is a negative rational number}\\ + 0,\ \text{If } \mathop{\mathrm{sgn}}\left(a\right)=0 + \end{cases} +\end{align*}$$* +::: + +Now that we have defined the notion of a positive and negative rational +number we can consider what it means to negate a rational number. The +definition follows immediately from the representation of $-1$ in +$\mathbb{Q}$ being $\left(-1,1\right)$. Indeed for any $x\in\mathbb{Q}$ +with $x=\left(a,b\right)$ we have + +$$\begin{equation*} + -x=-1*x=\left(-1,1\right)*\left(a,b\right)=\left(-a,b\right) +\end{equation*}$$ + +We make the formal definition. + +::: definition +**Definition 129**. *Negation of a rational number* + +*Let $x\in\mathbb{Q}$. We define the negation of $x$, denoted $-x$ by* + +*$$\begin{equation*} + -x=-1*x=\left(-1,1\right)*x +\end{equation*}$$* + +*where $\left(-1,1\right)\in\left[\left(-1,1\right)\right]$. That is +$\left(-1,1\right)$ is an element of the equivalence class +$\left[\left(-1,1\right)\right]$ which represents all possible elements +that are $-1$.* +::: + +We can now make a definition for subtraction for the rational numbers + +::: definition +**Definition 130**. *Rational number subtraction* + +*Let $x,y\in\mathbb{Q}$. We define the subtraction of $y$ from $x$, +denoted $x-y$ by* + +*$$\begin{equation*} + x-y=x+\left(-y\right)=x+\left(-1*y\right) +\end{equation*}$$* +::: + +We immediately get that subtraction is closed, from the fact that both +addition and multiplication is closed. We do not have associativity of +subtraction in general. + +::: proposition +**Proposition 80**. *Rational number subtraction is not associative* + +*Let $x,y,z\in\mathbb{Q}$. We have that* + +*$$\begin{equation*} + x-\left(y-z\right)\neq \left(x-y\right)-z +\end{equation*}$$* + +*Proof:* + +*Let $\displaystyle x=\frac{1}{2}, y=\frac{1}{4}$ and +$\displaystyle z=\frac{1}{6}$, we have +$x\in\left[\left(1,2\right)\right], y\in\left[\left(1,4\right)\right]$ +and $z\in\left[\left(1,6\right)\right]$ so +$x=\left(1,2\right), y=\left(1,4\right)$ and $z=\left(1,6\right)$ . We +have that* + +*$$\begin{align*} + x-\left(y-z\right)&=\left(1,2\right)+\left(\left(1,4\right)-\left(1,6\right)\right)\\ + &=\left(1,2\right)-\left(\left(1,4\right)+\left(-1*\left(1,6\right)\right)\right)\\ + &=\left(1,2\right)-\left(\left(1,4\right)+\left(-1,6\right)\right)\\ + &=\left(1,2\right)-\left(\left(1*6+4*-1,4*6\right)\right)\\ + &=\left(1,2\right)-\left(\left(2,24\right)\right)\\ + &=\left(1,2\right)+\left(-1*\left(\left(2,24\right)\right)\right)\\ + &=\left(1,2\right)+\left(-2,24\right)\\ + &=\left(1*24+2*-1,2*24\right)\\ + &=\left(22,48\right)\\ +\end{align*}$$* + +*On the other hand we have* + +*$$\begin{align*} + \left(x-y\right)-z&=\left(1,2\right)-\left(\left(1,4\right)-\left(1,6\right)\right)\\ + &=\left(\left(1,2\right)+\left(-1*\left(1,4\right)\right)\right)-\left(1,6\right)\\ + &=\left(\left(1,2\right)+\left(-1,4\right)\right)-\left(1,6\right)\\ + &=\left(1*4+2*-1,2*4\right)-\left(1,6\right)\\ + &=\left(2,8\right)-\left(1,6\right)\\ + &=\left(2,8\right)+\left(-1*\left(1,6\right)\right)\\ + &=\left(2,8\right)+\left(-1,6\right)\\ + &=\left(2*6+8*-1,8*6\right)\\ + &=\left(4,48\right) +\end{align*}$$* + +*It is left to show that $\left(22,48\right)\neq\left(4,48\right)$. +Indeed to have $\left(22,48\right)=\left(4,48\right)$ we need +$\left(22,48\right)\sim\left(4,48\right)$ which occurs if and only if +$22*48=48*8$. However one the left hand side $48$ is multiplied by $22$ +and on the right-hand side $48$ is multiplied by $8$ so they clearly can +not be equal.* + +*The result is shown. $\qed$* +::: + +As with subtraction with integers, we can now show that formally, +subtraction is an inverse to addition. + +::: {#prop:RationalAdditiveInverse .proposition} +**Proposition 81**. *Subtracting an integer from itself gives zero* + +*Let $x\in\mathbb{Q}$. We have that* + +*$$\begin{equation*} + x-x=0 +\end{equation*}$$* + +*Proof:* + +*Let $x\in\mathbb{Q}$ where $x\in\left[\left(a,b\right)\right]$ for some +$a,b\in\mathbb{Z}$ and $b\neq 0$. We have* + +*$$\begin{align*} + x-x&=\left(a,b\right)-\left(a,b\right)\\ + &=\left(a,b\right)+\left(-a,b\right)\\ + &=\left(ab+b*-a,b*b\right)\\ + &=\left(ab-ba,b*b\right)\\ + &=\left(ab-ab,b*b\right)\\ + &=\left(0,b*b\right) +\end{align*}$$* + +*It is left to show that $\left(0,b*b\right)\sim\left(0,1\right)$. +Indeed* + +*$$\begin{equation*} + 0*1=b*b*0 \Rightarrow 0=0 +\end{equation*}$$* + +*The result is shown. $\qed$* +::: + +#### The cancellation laws + +We can now deduce that the cancellation laws extend to the rational +numbers. + +::: {#thm:CancellationLawsForRationals .theorem} +**Theorem 31**. *The cancellation laws* + +*Let $x,y,z\in\mathbb{Q}$.* + +1. *If $x+y=x+z$ then we have $y=z$.* + +2. *For $x\neq 0$, if $xy=xz$ then we have that $y=z$* + +*Proof:* + +1. *If $x+y=x+z$ then we have $y=z$:* + + *Let $x,y,z\in\mathbb{Q}$. We have that* + + *$$\begin{align*} + x+y&=x+z\\ + \Rightarrow -x+x+y&=-x+x+z,\ \text{Adding the negative of } x \text{ to both sides}\\ + \Rightarrow \left(-x+x\right)+y*&=\left(-x+x\right)+z,\ \text{Associativity of the rationals}\\ + \Rightarrow 0+y&=0+z,\ \text{By proposition \ref{prop:RationalAdditiveInverse}}\\ + \Rightarrow y&=z + \end{align*}$$* + +2. *For $x\neq 0$, if $xy=xz$ then we have that $y=z$:* + + *Let $x,y,z\in\mathbb{Q}$ where $x\neq 0$. Suppose that + $x\in\left[\left(a,b\right)\right], y\in\left[\left(c,d\right)\right]$ + and $z\in\left[\left(e,f\right)\right]$. We have* + + *$$\begin{align*} + xy&=\left(a,b\right)\left(c,d\right)=\left(ac,bd\right)\\ + xz&=\left(a,b\right)\left(e,f\right)=\left(ae,bf\right) + \end{align*}$$* + + *Now suppose that $xy=xz$ then we have that + $\left(ac,bd\right)\sim\left(ae,bf\right)$ which is to say* + + *$$\begin{equation*} + acbf=aebd + \end{equation*}$$* + + *Observer that $$\begin{align*} + &acbf=aebd\\ + &a\left(cbf\right)=a\left(ebd\right)\\ + &cbf=ebd,\ \text{By the cancellation laws for the integers}\\ + &bcf=bed,\ \text{By commutativity of the integers}\\ + &b\left(cf\right)=b\left(ed\right)\\ + &cf=ed,\ \text{By the cancellation laws for the integers}\\ + \Rightarrow&\left(c,d\right)\sim\left(e,f\right),\ \text{By definition of the equivalence relation} + \end{align*}$$* + + *It hence follows that as $\left(c,d\right)\sim\left(e,f\right)$ + then $y=z$* + +*The result is shown. $\qed$* +::: + +#### Defining multiplicative inverses and division + +When we extended the naturals to the integers we were able to extend the +notion of subtraction in such a way that we could undo any addition +operation. We were not able to do the same for multiplication in +general. For example if we have $x*2=1$ where $1,2,x\in\mathbb{Z}$ then +there is no integer $x$ that when multiplied by $2$ gives $1$. + +What happens if we consider instead the situation where we have +$1,2,x\in\mathbb{Q}$? Let $x=\left(a,b\right)$ for some +$a,b\in\mathbb{Z}$ with $b\neq 0$ and taking the natural representations +for $1$ and $2$ of $1=\left(1,1\right)$ and $2=\left(2,1\right)$. We +have that + +$$\begin{align*} + x*2&=1\\ + \left(a,b\right)\left(2,1\right)&=\left(1,1\right)\\ + \left(2a,b\right)&=\left(1,1\right)\\ + \Rightarrow\left(2a,b\right)&\sim\left(1,1\right)\iff 2a=b +\end{align*}$$ + +We don't seem to be in a better position then when we asked this +question for $\mathbb{Z}$. However as $a,b$ were arbitrary, of course +with $b\neq 0$, we are free to vary them. For example $a=1$ gives us +$b=2$, $a=2$ gives $b=4$, $a=3$ yields $b=6$ and so on. We hence have +that there is a family of possible value for $x$ which satisfies $x*2=1$ +over the rational numbers, in particular we have $x=\left(a,2a\right)$ +for $a\in\mathbb{Z}$ and $a\geq 0$. Moreover we clearly have + +$$\begin{equation*} + \left(a,2a\right)\sim\left(1,2\right)\iff 2a=2a +\end{equation*}$$ + +Hence we have that $\left(a,2a\right)$ somehow undoes multiplication by +$2$. Indeed consider $45*2=90$. We have that + +$$\begin{equation*} + 90*\left(a,2a\right)=\left(90,1\right)*\left(a,2a\right)=\left(90a,2a\right) +\end{equation*}$$ + +Where we have $\left(90a,2a\right)\sim\left(45,1\right)$ as +$90a*1=45*2a \iff 90a = 90a$. We can generalise this to $x*y=1$ for any +$y\in\mathbb{Q}$. Indeed let $x=\left(a,b\right)$ and +$y=\left(c,d\right)$ where $a,b,c,d\in\mathbb{Z}$ and $c\neq 0$ and +$d\neq 0$ then we have + +$$\begin{align*} + x*y&=\left(a,b\right)*\left(c,d\right)\\ + &=\left(ac,bd\right)=\left(1,1\right)\\ + \Rightarrow\left(ac,bd\right)&\sim\left(1,1\right)\iff bd=ac +\end{align*}$$ + +This is a somewhat unsatisfactory conclusion as it doesn't tell us what +$a$ or $b$ should actually be equal to in order for $x*y=1$, likewise, +it doesn't tell us what $c$ or $d$ should be either. + +Perhaps then we should consider a more simple setup. Suppose that +$x\in\mathbb{Z}$ then is there $y\in\mathbb{Q}$ where +$y=\left(c,d\right)$ with $d\neq 0$, such that $x*y=1$? We have + +$$\begin{equation*} + x*y=\left(x,1\right)*\left(c,d\right)=\left(xc,d\right)=\left(1,1\right) +\end{equation*}$$ + +Hence + +$$\begin{equation*} + \left(xc,d\right)\sim\left(1,1\right)\iff xc=d +\end{equation*}$$ + +Hence $y=\left(c,xc\right)$ satisfies this relation. However we can see +that $\left(c,cx\right)\sim\left(1,x\right)$. Hence for any integer +$x\neq 0$ we have a solution to $x*y=1$ with $y\in\mathbb{Q}$. We call +$y$ a multiplicative inverse of $x$ and $x$ a multiplicative inverse of +$y$. + +::: definition +**Definition 131**. *Multiplicative inverse of an integer* + +*Let $x\in\mathbb{Z}$ be such that $x\neq 0$. Then there is a +$y\in\mathbb{Q}$ such that* + +*$$\begin{equation*} + x*y=1=y*x +\end{equation*}$$* + +*where $y=\left(1,x\right)$. We can write this as +$\displaystyle y=\frac{1}{x}$ or $y=x^{-1}$. We sometimes say that +$x{-1}$ is a reciprocal of $x$ or a multiplicative inverse of $x$.* +::: + +In light of this, we have the immediate result + +::: {#prop:MultiplicativeInverseOfIntegerTimesInverseIsOriginalNumber .proposition} +**Proposition 82**. *Multiplicative inverse of an integer times its +multiplicative inverse is the original number* + +*Let $x\in\mathbb{Z}$ so that $x^{-1}\in\mathbb{Q}$ where +$\displaystyle x^{-1}=\frac{1}{x}$ is the multiplicative inverse to $x$ +in the rationals. The following result holds.* + +*$$\begin{equation*} + x*x^{-1}*x = x +\end{equation*}$$* + +*Proof:* + +*By definition of a multiplicative inverse we have that* + +*$$\begin{equation*} + x*x^{-1}=x*\frac{1}{x}=\left(x,1\right)*\left(1,x\right)=\left(x,x\right)\sim\left(1,1\right)=1 +\end{equation*}$$* + +*Hence as $x^{-1}$ is a multiplicative inverse for $x$ it follows that +$x$ is a multiplicative inverse for $x^{-1}$ and so* + +*$$\begin{equation*} + x*x^{-1}*x=1*x=x +\end{equation*}$$* + +*As required. $\qed$* +::: + +Armed with this definition we can answer the original question. In order +to find an $x$ so that $x*y=1$ we have that we need to find a +multiplicative inverse for $c$ and a multiplicative inverse for +$\displaystyle d^{-1}=\frac{1}{d}$. Clearly we have that +$\displaystyle c^{-1}=\frac{1}{c}$ and a multiplicative inverse for +$d^{-1}$ is simply $d$. Hence a candidate for $x$ is given by +$x=\left(d,c\right)$. Indeed we have that + +$$\begin{equation*} + x*y=\left(d,c\right)*\left(c,d\right)=\left(cd,cd\right)\sim\left(1,1\right)=1 +\end{equation*}$$ + +We can hence extend the idea of multiplicative inverses to the +rationals. + +::: definition +**Definition 132**. *Multiplicative inverse of a rational number* + +*Let $x\in\mathbb{Q}$ such that $x=\left(a,b\right)$ with +$a,b\in\mathbb{Z}$ and $b\neq 0$. Then there is a $y\in\mathbb{Q}$ such +that* + +*$$\begin{equation*} + x*y=1=y*x +\end{equation*}$$* + +*where $y=\left(b,a\right)$. Hence we must also have $a\neq 0$. We write +this as $\displaystyle y=\frac{b}{a}$ or as +$\displaystyle x^{-1}=y=\frac{b}{a}$. We sometimes say that $x{-1}$ is a +reciprocal of $x$ or a multiplicative inverse of $x$.* +::: + +A similar result holds as for proposition +[82](#prop:MultiplicativeInverseOfIntegerTimesInverseIsOriginalNumber){reference-type="ref" +reference="prop:MultiplicativeInverseOfIntegerTimesInverseIsOriginalNumber"} + +::: {#prop:MultiplicativeInverseOfRationalTimesInverseIsOriginalNumber .proposition} +**Proposition 83**. *Multiplicative inverse of a rational number times +its multiplicative inverse is the original number* + +*Let $x\in\mathbb{Q}$ with $x=\left(a,b\right)$ and $a,b\in\mathbb{Z}$ +so that $a\neq 0$ and $b\neq 0$. Let $x^{-1}$ denote the multiplicative +inverse of $x$. The following result holds.* + +*$$\begin{equation*} + x*x^{-1}*x = x +\end{equation*}$$* + +*Proof:* + +*By definition of a multiplicative inverse we have that* + +*$$\begin{equation*} + x*x^{-1}=1 +\end{equation*}$$* + +*Hence* + +*$$\begin{equation*} + x*x^{-1}*x=1*x=x +\end{equation*}$$* + +*As required. $\qed$* +::: + +We now have a solid grasp of undoing multiplication in the rational +numbers. In fact we are now in a position to define the operation of +division. However we are already done due to the work we have just done, +and our original motivation for defining the rational numbers in the +first place. We use the idea of multiplicative inverses! + +::: definition +**Definition 133**. *Division* + +*Let $a,b\in\mathbb{Z}$ so that $b\neq 0$. We define the division of $a$ +by $b$, denoted $\displaystyle\frac{a}{b}$ by* + +*$$\begin{equation*} + \frac{a}{b}=a*b^{-1}=\left(a,1\right)*\left(1,b\right)=\left(a,b\right) +\end{equation*}$$* +::: + +We can extend the notion of division even further by considering +$a,b\in\mathbb{Q}$ rather than $\mathbb{Z}$. At first is appears we have +a problem, we defined the rationals using integers and division in terms +of integers, so how could we possibly assign any meaning to an +expression like $\displaystyle\frac{1}{\frac{1}{2}}$? + +Consider for example the following + +$$\begin{equation*} + \frac{1}{\frac{1}{2}}*\frac{1}{2} +\end{equation*}$$ + +If we were suppose the rule for multiplication that we defined extends +to this situation then we get + +$$\begin{equation*} + \frac{1}{\frac{1}{2}}*\frac{1}{2}=\frac{1*1}{\frac{1}{2}*2}=\frac{1}{1}=1 +\end{equation*}$$ + +In the context of the work we have just done we have that +$\displaystyle \frac{1}{\frac{1}{2}}$ is a multiplicative inverse of +$\frac{1}{2}$. However we know that $\displaystyle \frac{1}{2}$ has a +multiplicative inverse of $2$. Does this mean that +$\displaystyle \frac{1}{\frac{1}{2}}=2$? A deeper analysis of +expressions of the form $\displaystyle \frac{1}{\frac{1}{a}}$. + +We know from before that $\displaystyle\frac{1}{a}=a^{-1}$ for some +non-zero $a\in\mathbb{Z}$. Hence we have that by definition +$a^{-1}\in\mathbb{Z}$. Hence we are considering the expression + +$$\begin{equation*} + \frac{1}{\frac{1}{a}}=\frac{1}{a^{-1}} +\end{equation*}$$ + +Therefore we know from the definition of the multiplicative inverse of a +rational number that there is some $y\in\mathbb{Q}$ so that + +$$\begin{equation*} + \frac{1}{a^{-1}}*y=1 +\end{equation*}$$ + +By the definition we also know what $y$ must be +$\displaystyle \frac{a^{-1}}{1}=a^{-1}=\frac{1}{a}$. Hence we can +justify our "temporary" assumption of extending the multiplication rule. +Hence hence make the following deduction + +::: {#prop:OneDividedByMultiplicativeInverseOfInteger .proposition} +**Proposition 84**. *One divided by multiplicative inverse of an integer +is the integer itself* + +*Let $x\in\mathbb{Q}$ so that $\displaystyle x=\frac{1}{\frac{1}{a}}$ +for some $a\in\mathbb{Z}$ with $a\neq 0$. we have that* + +*$$\begin{equation*} + \frac{1}{\frac{1}{a}}=a +\end{equation*}$$* + +*Proof:* + +*Let $x\in\mathbb{Q}$ be such that +$\displaystyle x=\frac{1}{\frac{1}{a}}$ for some non-zero +$a\in\mathbb{Z}$. We know by definition that* + +*$$\begin{equation*} + x=\frac{1}{a}=a^{-1} +\end{equation*}$$* + +*where $a^{-1}\in\mathbb{Z}$ and therefore +$\displaystyle x = \frac{1}{a^{-1}}$. Moreover this is still a rational +number by definition and so there exists some rational $y$ so that* + +*$$\begin{equation*} + x*y=1 +\end{equation*}$$* + +*where $\displaystyle y=\frac{a^{-1}}{1}=a^{-1}$. It follows that +$\displaystyle y=\frac{1}{a}$. Again by definition there is some +$z\in\mathbb{Q}$ so that $y*z=1$ where $\displaystyle z=\frac{a}{1}=a$ +that is to say $z$ is a multiplicative inverse of $y$.* + +*We therefore have that* + +*$$\begin{equation*} + x*y=1=y*z +\end{equation*}$$* + +*Hence by theorem +[31](#thm:CancellationLawsForRationals){reference-type="ref" +reference="thm:CancellationLawsForRationals"} we have that $x=z$ which +is to say* + +*$$\begin{equation*} + \frac{1}{\frac{1}{a}}=a +\end{equation*}$$* + +*As required. $\qed$* +::: + +We hence get an immediate corollary + +::: corollary +**Corollary 4**. *One divided by rational number* + +*Let $x\in\mathbb{Q}$ be such that $\displaystyle x=\frac{a}{b}$. We +have that* + +*$$\begin{equation*} + \frac{1}{x}=\frac{1}{\frac{a}{b}}=\frac{b}{a} +\end{equation*}$$* + +*Proof:* + +*We have* + +*$$\begin{equation*} + \frac{1}{x}=\frac{1}{\frac{a}{b}}=\frac{1}{a\frac{1}{b}}=\frac{1}{a b^{-1}}=\frac{1}{a}*\frac{1}{b^{-1}}=\frac{1}{a}b=\frac{b}{a} +\end{equation*}$$* + +*As required. $\qed$* +::: + +#### Extending the summation and product notations to the rationals + +Summation and product notation has been defined on the naturals as well +as the integers. We can extend the notation to include the rational +numbers. + +Let $q\in\mathbb{Q}^{n+m+1}$ be an ordered $n+m+1$ tuple of rational +numbers where + +$$\begin{equation*} + q=\left(q_{-m},q_{-m+1},\dots,q_{-1},q_0,q_1,\dots, q_n\right) +\end{equation*}$$ + +Define +$\mathbb{Z}_m^n=\left\{-m,-m+1,-m+2,\dots,-1,0,1,\dots,n-1,n\right\}$ to +be a set of indices and define $f:\mathbb{Z}_m^n\rightarrow\mathbb{Q}$ +by + +$$\begin{align*} + f:\mathbb{Z}_m^n&\rightarrow \mathbb{Q}\\ + i&\mapsto f\left(i\right)=q_i +\end{align*}$$ + +::: definition +**Definition 134**. *Summation notation for rational numbers* + +*Let $z\in\mathbb{Q}^{n+m+1}$ be ordered $n+m+1$ tuple of integers where +$q=\left(q_{-m},q_{-m+1},\dots,q_{-1},q_0,q_1,\dots, q_n\right)$. Define +$\mathbb{Z}_m^n$ by +$\mathbb{Z}_m^n=\left\{-m,-m+1,-m+2,\dots,-1,0,1,\dots,n-1,n\right\}$. +Let $f:\mathbb{Z}^{n+m+1}:\mathbb{Q}$ defined by* + +*$$\begin{align*} + f:\mathbb{Z}^{m+n+1}&\rightarrow\mathbb{Q}\\ + i&\mapsto f\left(i\right)=q_i +\end{align*}$$* + +*We define the summation notation for the rational numbers by* + +*$$\begin{equation*} + \sum_{i=-m}^n f\left(i\right)=f\left(-m\right)+f\left(-m+1\right)+\dots+f\left(-1\right)+f\left(0\right)+f\left(1\right)+\dots+f\left(n\right) +\end{equation*}$$* + +*Alternatively this is written* + +*$$\begin{equation*} + \sum_{i=-m}^n q_i = q_{-m}+q_{-m+1}+\dots+q_{-1}+q_0+q_1+\dots+q_n +\end{equation*}$$* + +*We have that $i$ is called the index of summation and that $i=-m$ is +the starting index of the summation, and $n$ the ending index of the +summation. If $q\in\emptyset$ then we define the summation to be $0$ and +call the summation an empty sum.* + +*We can also define the summation of some subset of $\mathbb{Z}_m^n$ +which allows for starting a summation at some starting point other than +$i=-m$. Let $T\subseteq\mathbb{Z}_m^n$. We define the summation over the +set $T$ by* + +*$$\begin{equation*} + \sum_{i\in T} z_i +\end{equation*}$$* + +*If we have a mapping $g:\mathbb{Q}\rightarrow\mathbb{Q}$ we can define +a summation over $g$ by* + +*$$\begin{equation*} + \sum_{i\in T} g\left(z_i\right) +\end{equation*}$$* + +*Finally we can define a summation over a predicate $P\left(i\right)$ +for $i\in T$ by* + +*$$\begin{equation*} + \sum_{P\left(i\right)}g\left(z_i\right) +\end{equation*}$$* + +*where we take the sum of the $g\left(z_i\right)$ for the $i$ that +satisfy the predicate $P$. We note that if we have $k>n$ for some +$k\in\mathbb{N}$ then the sum* + +*$$\begin{equation*} + \sum_{i=k}^n z_i=0 +\end{equation*}$$* +::: + +The usual proprieties shown for summations with integer numbers also +extend to the rational number version. + +::: proposition +**Proposition 85**. *Properties of summation notation* + +*Let $n,m\in\mathbb{Z}$ such that $mn$ for some +$k\in\mathbb{N}$ then the product* + +*$$\begin{equation*} + \prod_{i=k}^n z_i=1 +\end{equation*}$$* +::: + +::: proposition +**Proposition 86**. *Properties of product notation* + +*Let $n,m\in\mathbb{Z}$ such that $m-y$* + +2. *If $x\leq y$ then $-x\geq -y$* + +3. *If $x>y$ then $-x<-y$* + +4. *If $x\geq y$ then $-x\leq-y$* + +*Proof:* + +*Let $x,y\in\mathbb{Q}$ so that $\displaystyle x=\frac{a}{b}$ and +$\displaystyle y=\frac{c}{d}$ where $b\neq 0$ and $d\neq 0$.* + +1. *If $x-y$:* + + *Let $x,y\in\mathbb{Q}$ so that $x-bc + \end{equation*}$$* + + *Hence $-x>-y$.* + +2. *If $x\leq y$ then $-x\geq -y$:* + + *Let $x,y\in\mathbb{Q}$ so that $x\leq y$. Applying the definition + of $\leq$ for the rationals gives* + + *$$\begin{equation*} + x\leq y\iff ad\leq bc + \end{equation*}$$* + + *Proposition + [70](#prop:MultiplicationByNegativeOneFlipsInequalitySign){reference-type="ref" + reference="prop:MultiplicationByNegativeOneFlipsInequalitySign"} + gives* + + *$$\begin{equation*} + ad\leq bc\Rightarrow -ad\geq -bc + \end{equation*}$$ Hence $-x\geq y$.* + +3. *If $x>y$ then $-x<-y$:* + + *Let $x,y\in\mathbb{Q}$ so that $x>y$. By definition of $>$ for the + rationals we have that* + + *$$\begin{equation*} + x>y\iff ad>bc + \end{equation*}$$* + + *Proposition + [70](#prop:MultiplicationByNegativeOneFlipsInequalitySign){reference-type="ref" + reference="prop:MultiplicationByNegativeOneFlipsInequalitySign"} + shows us that* + + *$$\begin{equation*} + ad>bc\Rightarrow -ad<-bc + \end{equation*}$$* + + *Hence $-x<-y$.* + +4. *If $x\geq y$ then $-x\leq-y$:* + + *Let $x,y\in\mathbb{Q}$ so that $x>y$. By definition of $\geq$ for + the rationals, we have that* + + *$$\begin{equation*} + x\geq y\iff ad\geq bc + \end{equation*}$$* + + *Proposition + [70](#prop:MultiplicationByNegativeOneFlipsInequalitySign){reference-type="ref" + reference="prop:MultiplicationByNegativeOneFlipsInequalitySign"} we + have* + + *$$\begin{equation*} + ad\geq bc\Rightarrow -ad\leq-bc + \end{equation*}$$* + + *Hence $-x\leq -y$.* + +*The result is shown. $\qed$* +::: + +There is another useful lemma that will be useful for extending the +rules of inequalities to the rationals. + +::: {#lem:LargerRatMinusSmallIsPositive .lemma} +**Lemma 8**. *Strictly larger rational minus a smaller is positive* + +*Let $x,y\in\mathbb{Q}$. We have that $x0$* + +*Proof:* + +*$\left(\Rightarrow\right)$: Let $x,y\in\mathbb{Q}$, then +$\displaystyle x=\frac{a}{b}$ and $\displaystyle y=\frac{c}{d}$ for some +$a,b,c,d\in\mathbb{Z}$ and $b\neq 0$ and $d\neq 0$. As $x0$. By definition of +greater than, and the fact that $0\in\left[\left(0,1\right)\right]$ we +would have that* + +*$$\begin{equation*} + \left(cb-ad\right)*1>0*\left(bd\right) \Rightarrow bc-ad>0 +\end{equation*}$$* + +*Which is true as $ad0$* + +*$\left(\Leftarrow\right)$: Suppose that $y-x>0$ where +$x,y\in\mathbb{Q}$, with $\displaystyle x=\frac{a}{b}$ and +$\displaystyle y=\frac{c}{d}$ for some $a,b,c,d\in\mathbb{Z}$ and +$b\neq 0$ and $d\neq 0$. We have that* + +*$$\begin{equation*} + y-x = \left(cb-ad,bd\right) +\end{equation*}$$* + +*Moreover, $y-x>0$ implies that* + +*$$\begin{equation*} + \left(cb-ad\right)*1>0*\left(bd\right) \Rightarrow bc-ad>0 +\end{equation*}$$* + +*This is to say $bc>ad$, which by part 1 of proposition +[71](#prop:InequalityIntegerNumbers){reference-type="ref" +reference="prop:InequalityIntegerNumbers"} is the same as $ad0$ or $y=x$.* + +*Proof:* + +*$\left(\Rightarrow\right)$: Suppose $x\leq y$. If $x 0$ or $y=x$ holds. +In the first case, $y-x>0$ implies $xx$* + +2. *$x\leq y$ is the same as $y\geq x$* + +3. *If $xy$ and $y>z$ then $x>z$* + +8. *If $x\geq y$ and $y>z$ then $x>z$* + +9. *If $x>y$ and $y\geq z$ then $x>z$* + +10. *If $x\geq y$ and $y\geq z$ then $x\geq z$* + +11. *If $xy$ then $x+z>y+z$* + +14. *If $x\geq y$ then $x+z\geq y+z$* + +15. *If $xyz$* + +17. *If $x\leq y$ and $z\geq 0$ then $xz\leq yz$* + +18. *If $x\leq y$ and $z<0$ then $xz\geq yz$* + +19. *If $x>y$ and $z\geq 0$ then $xz>yz$* + +20. *If $x>y$ and $z< 0$ then $xz0$ then $\displaystyle\frac{x}{z}<\frac{y}{z}$* + +24. *If $x\leq y$ and $z>0$ then + $\displaystyle\frac{x}{z}\leq\frac{y}{z}$* + +25. *If $x>y$ and $z>0$ then $\displaystyle\frac{x}{z}>\frac{y}{z}$* + +26. *If $x\geq y$ and $z>0$ then + $\displaystyle\frac{x}{z}\geq\frac{y}{z}$* + +27. *If $x\frac{y}{z}$* + +28. *If $x\leq y$ and $z<0$ then + $\displaystyle\frac{x}{z}\geq\frac{y}{z}$* + +29. *If $x>y$ and $z<0$ then $\displaystyle\frac{x}{z}<\frac{y}{z}$* + +30. *If $x\geq y$ and $z<0$ then + $\displaystyle\frac{x}{z}\leq\frac{y}{z}$* + +31. *If $x0$ and $y>0$ then + $\displaystyle \frac{1}{x}>\frac{1}{y}$* + +32. *If $x\frac{1}{y}$* + +33. *If $x\leq y$ and $x>0$ and $y>0$ then + $\displaystyle \frac{1}{x}\geq \frac{1}{y}$* + +34. *If $x\leq y$ and $x<0$ and $y<0$ then + $\displaystyle \frac{1}{x}\geq \frac{1}{y}$* + +35. *If $x>y$ and $x>0$ and $y>0$ then + $\displaystyle \frac{1}{x}<\frac{1}{y}$* + +36. *If $x>y$ and $x<0$ and $y<0$ then + $\displaystyle \frac{1}{x}<\frac{1}{y}$* + +37. *If $x\geq y$ and $x>0$ and $y>0$ then + $\displaystyle \frac{1}{x}\leq \frac{1}{y}$* + +38. *If $x\geq y$ and $x<0$ and $y<0$ then + $\displaystyle \frac{1}{x}\leq \frac{1}{y}$* + +*Proof:* + +*Let $x,y,z\in\mathbb{Q}$. Let $\displaystyle x=\frac{a}{b}$, +$\displaystyle y=\frac{c}{d}$, $\displaystyle z=\frac{e}{f}$ for +$a,b,e,f,g,h\in\mathbb{Z}$ and $b\neq 0$, $d\neq 0$, $f\neq 0$.* + +1. *$xx$:* + + *Suppose that $xaf$ and so + $y>x$.* + +2. *$x\leq y$ is the same as $y\geq x$:* + + *If $x0$ and $z-y>0$ by lemma + [8](#lem:LargerRatMinusSmallIsPositive){reference-type="ref" + reference="lem:LargerRatMinusSmallIsPositive"}. Now we have that* + + *$$\begin{equation*} + \left(y-x\right)+\left(z-y\right)=z-x>0 + \end{equation*}$$* + + *As $y-x$ and $z-y$ are both greater than 0. Hence as $z-x>0$ then + $x0$, + likewise by corollary + [5](#cor:LargerOrEqualRatMinusSmallIsPositive){reference-type="ref" + reference="cor:LargerOrEqualRatMinusSmallIsPositive"} we have that + $y\leq z$ means either $z-y>0$ or $y=z$.* + + *If $z-y>0$ then the result is the same as part 3. So suppose $y=z$ + then clearly $xy$ and $y>z$ then $x>z$:* + + *By part 1. this is equivalent to $yz$ then $x>z$:* + + *Using parts 1. and 2. gives us the equivalent expression $y\leq x$ + and $zy$ and $y\geq z$ then $x>z$:* + + *As with the previous part, applying parts 1. and 2. gives the + statement $y0$ by lemma + [8](#lem:LargerRatMinusSmallIsPositive){reference-type="ref" + reference="lem:LargerRatMinusSmallIsPositive"}. Observer that* + + *$$\begin{align*} + y-x&=y-\left(z-z\right)-x\\ + &=\left(y-z\right)+\left(z-x\right)\\ + &=\left(y-z\right)-\left(x-z\right)>0 + \end{align*}$$* + + *So $\left(y-z\right)-\left(x-z\right)>0$ and so by the same lemma + we conclude that $x+zy$ then $x+z>y+z$:* + + *Applying part 1. and then part 11. gives the equivalent result + $y0$ by lemma + [8](#lem:LargerRatMinusSmallIsPositive){reference-type="ref" + reference="lem:LargerRatMinusSmallIsPositive"}. Hence, by + distributivity, we have $z\left(y-x\right)>0$ as $z\geq 0$. Hence* + + *$$\begin{equation*} + z\left(y-x\right)=zy-zx=yz-xz \Rightarrow xzyz$:* + + *Suppose $x0$, then applying part 15. + with $-z$ gives $-xz<-yz$. Finally by proposition + [88](#prop:MultiplicationByNegativeOneFlipsInequalitySignRational){reference-type="ref" + reference="prop:MultiplicationByNegativeOneFlipsInequalitySignRational"} + part 1 yields $xz>yz$.* + +17. *If $x\leq y$ and $z\geq 0$ then $xz\leq yz$:* + + *If $x\leq y$ there are two cases to consider. If $xy$ and $z\geq 0$ then $xz>yz$:* + + *We have $x>y$ is the same as $y0$. By + distributivity, we have that $z\left(x-y\right)>0$. Therefore we + have $zx-zy=xz-yz>0$ and so $yzyz$ by + part 1.* + +20. *If $x>y$ and $z< 0$ then $xzy$. Additionally, $z<0\Rightarrow -z>0$ so applying + part 19. gives $-xz>-yz$ and so by part 1. we conclude $xzy$ then we apply part 19. + Otherwise $x=y$ and $xz=yz$ so that $xz\geq yz$.* + +22. *If $x\geq y$ and $z<0$ then $xz\leq yz$:* + + *Again there are two cases to consider. If $x>y$ then the result + holds by part 20. Otherwise $x=y$ and so $xz=yz$ to give the result + $xz\leq yz$.* + +23. *If $x0$ then $\displaystyle\frac{x}{z}<\frac{y}{z}$:* + + *This follows by part 15.* + +24. *If $x\leq y$ and $z>0$ then + $\displaystyle\frac{x}{z}\leq\frac{y}{z}$:* + + *This follows by part 17.* + +25. *If $x>y$ and $z>0$ then $\displaystyle\frac{x}{z}>\frac{y}{z}$:* + + *This follows by part 19.* + +26. *If $x\geq y$ and $z>0$ then + $\displaystyle\frac{x}{z}\geq\frac{y}{z}$:* + + *This follows by part 21.* + +27. *If $x\frac{y}{z}$:* + + *This follows by part 16.* + +28. *If $x\leq y$ and $z<0$ then + $\displaystyle\frac{x}{z}\geq\frac{y}{z}$:* + + *This follows by part 18.* + +29. *If $x>y$ and $z<0$ then $\displaystyle\frac{x}{z}<\frac{y}{z}$:* + + *This follows by part 20.* + +30. *If $x\geq y$ and $z<0$ then + $\displaystyle\frac{x}{z}\leq\frac{y}{z}$:* + + *This follows by part 22.* + +31. *If $x0$ and $y>0$ then + $\displaystyle \frac{1}{x}>\frac{1}{y}$:* + + *Suppose that $x0$ that either + $a>0$ and $b>0$ or $a<0$ and $b<0$. Likewise as $y>0$ then either + $c>0$ and $d>0$ or $c<0$ and $d<0$. Hence there are four cases to + consider.* + + 1. *$a>0$ and $b>0$ and $c>0$ and $d>0$* + + 2. *$a>0$ and $b>0$ and $c<0$ and $d<0$* + + 3. *$a<0$ and $b<0$ and $c>0$ and $d>0$* + + 4. *$a<0$ and $b<0$ and $c<0$ and $d<0$* + + + + 1. *$a>0$ and $b>0$ and $c>0$ and $d>0$:* + + *Observe that* + + *$$\begin{align*} + ad&0\\ + d&0\\ + dc^{-1}&\frac{d}{c}$, which is + to say $\displaystyle\frac{1}{x}>\frac{1}{y}$.* + + 2. *$a>0$ and $b>0$ and $c<0$ and $d<0$:* + + *We have that as $c<0$ and $d<0$ then $ad<0$ and $bc<0$ and + $ada^{-1}bc,\ \text{By part 16. as } a^{-1}<0\\ + d&>a^{-1}bc\\ + dc^{-1}&\frac{1}{y}$.* + + 3. *$a<0$ and $b<0$ and $c>0$ and $d>0$:* + + *The argument is similar to the previous one, swapping the roles + of $a,b,c$ and $d$.* + + 4. *$a<0$ and $b<0$ and $c<0$ and $d<0$:* + + *This is similar to the first part. We give the full argument. + As $a<0$, $b<0$, $c<0$ and $d<0$ then $ad>0$ and $bc>0$ and + $ada^{-1}bc,\ \text{By part 16. as } a^{-1}<0\\ + d&>a^{-1}bc\\ + dc^{-1}&\frac{1}{y}$:* + + *This is similar to the previous part. Suppose that $x0$ and $b<0$ or $a<0$ and + $b>0$. Likewise as $y<0$ then either $c>0$ and $d<0$ or $c<0$ and + $d>0$. Hence there are four cases to consider.* + + 1. *$a>0$ and $b<0$ and $c>0$ and $d<0$* + + 2. *$a>0$ and $b<0$ and $c<0$ and $d>0$* + + 3. *$a<0$ and $b>0$ and $c>0$ and $d<0$* + + 4. *$a<0$ and $b>0$ and $c<0$ and $d>0$* + + + + 1. *$a>0$ and $b<0$ and $c>0$ and $d<0$:* + + *As $a>0$ and $b<0$ and $c>0$ and $d<0$ then we have that $ad<0$ + and $bc<0$ and $ad\frac{1}{y}$* + + 2. *$a>0$ and $b<0$ and $c<0$ and $d>0$:* + + *We have $a>0$ and $b<0$ and $c<0$ and $d>0$ then we have that + $ad>0$ and $bc>0$ and $ad\frac{1}{y}$* + + 3. *$a<0$ and $b>0$ and $c>0$ and $d<0$:* + + *This time we have $a<0$ and $b>0$ and $c>0$ and $d<0$ then we + have that $ad>0$ and $bc>0$ and $adbc\\ + a^{-1}\left(-ad\right)&>a^{-1}\left(-bc\right)\\ + -d&>a^{-1}\left(-bc\right)\\ + -dc^{-1}&>a^{-1}\left(-bc\right)c^{-1}\\ + -dc^{-1}&>-a^{-1}b\\ + \frac{d}{c}&<\frac{b}{a} + \end{align*}$$* + + *Giving the result.* + + 4. *$a<0$ and $b>0$ and $c<0$ and $d>0$:* + + *Finally, $a<0$ and $b>0$ and $c<0$ and $d>0$ which gives $ad<0$ + and $bc<0$ with $ada^{-1}bc\\ + d&>a^{-1}bc\\ + dc^{-1}&0$ and $y>0$ then + $\displaystyle \frac{1}{x}\geq \frac{1}{y}$:* + + *If $xy$ and $x>0$ and $y>0$ then + $\displaystyle \frac{1}{x}<\frac{1}{y}$:* + + *Applying part 1. the equivalent statement is $y\frac{1}{y}$ so part 32. + applies.* + +36. *If $x>y$ and $x<0$ and $y<0$ then + $\displaystyle \frac{1}{x}<\frac{1}{y}$:* + + *Likewise by part 1. this is the same as $y0$ and $y>0$ + then $\displaystyle \frac{1}{y}>\frac{1}{y}$ so part 31. applies.* + +37. *If $x\geq y$ and $x>0$ and $y>0$ then + $\displaystyle \frac{1}{x}\leq \frac{1}{y}$:* + + *If $x>y$ then part 35 applies. Otherwise, $x=y$ and the result is + clear.* + +38. *If $x\geq y$ and $x<0$ and $y<0$ then + $\displaystyle \frac{1}{x}\leq \frac{1}{y}$:* + + *Finally, if $x>y$ then we apply part 36. Otherwise $x=y$ and we get + the result.* + +*The result has been shown.[^10] $\qed$* +::: + +#### Extending exponentiation to the rational numbers + +Recall the definition of exponentiation from the integers. + +$$\begin{align*} + \wedge:\mathbb{Z}\times\mathbb{Z}^+&\rightarrow\mathbb{Z}\\ + \left(x,n\right)&\mapsto \wedge\left(x,n\right)=\begin{cases} + 1,\ \text{If } x=0\text{ and } n=0\\ + 1,\ \text{If } n=0\\ + \displaystyle \prod_{i=1}^y x ,\ \text{If }x\neq 0\text{ and } n \geq 0\\ + \end{cases} +\end{align*}$$ + +where $\mathbb{Z}^+=\left\{x\in\mathbb{Z}:x\geq 0\right\}$. We noted in +the section on extending exponentiation to the integers that we were +unable to consider the case of negative exponents. By assuming that they +did we deduced that a new type of object exists that undoes integer +multiplication. As we have seen in this section, that object type is +actually a rational number. Indeed we showed that in proposition +[82](#prop:MultiplicativeInverseOfIntegerTimesInverseIsOriginalNumber){reference-type="ref" +reference="prop:MultiplicativeInverseOfIntegerTimesInverseIsOriginalNumber"} +that if $x\in\mathbb{Z}$ then there is some $*x^{-1}\in\mathbb{Q}$ so +that $x*x^{-1}=1=x^0$. This would generalise proposition +[77](#prop:IntegerExponentiationOfSameBaseAddsPowers){reference-type="ref" +reference="prop:IntegerExponentiationOfSameBaseAddsPowers"} to all +integers rather than positive exponents. We hence generalise the +definition of exponentiation and prove the results to all integer +exponents rather than the positive. + +::: definition +**Definition 136**. *Exponentiation of integer numbers* + +*Let $\left(x,y\right)\in\mathbb{Z}\times\mathbb{Z}$ and let +$\wedge:\mathbb{Z}\times\mathbb{Z}\rightarrow\mathbb{Q}$. We define the +exponentiation of $x$ by $y$ by $$\begin{align*} + \wedge:\mathbb{Z}\times\mathbb{Z}&\rightarrow\mathbb{Q}\\ + \left(x,y\right)&\mapsto \wedge\left(x,y\right)=\begin{cases} + 1,\ \text{If } x=0\text{ and } y=0\\ + 1,\ \text{If } x=0\\ + \displaystyle \prod_{i=1}^y x ,\ \text{If }x\neq 0\text{ and } n \geq 0\\ + \displaystyle \prod_{i=1}^{\left|y\right|} \frac{1}{x} ,\ \text{If }x\neq 0\text{ and } y < 0\\ + \end{cases} +\end{align*}$$* +::: + +We can now extend the results shown in the section on integer +exponentiation extension. + +::: {#prop:IntegerExtensionExponentiationPowerLaw .proposition} +**Proposition 90**. *Power law of exponentiation for positive exponents* + +*Let $x\in\mathbb{Z}$ and let $n,m\in\mathbb{Z}$. We have that* + +*$$\begin{equation*} + \left(x^n\right)^m = x^{nm} +\end{equation*}$$* + +*Proof:* + +*If $n,m\geq 0$ the result is the same as proposition +[76](#prop:IntegerExponentiationPowerLaw){reference-type="ref" +reference="prop:IntegerExponentiationPowerLaw"}. So we must consider the +following cases* + +1. *$n\geq 0$ and $m<0$* + +2. *$n< 0$ and $m\geq 0$* + +3. *$n< 0$ and $m<0$* + + + +1. *$n\geq 0$ and $m<0$:* + + *By definition of integer exponentiation, we have that + $\displaystyle x^n=\prod_{i=1}^n x$. Now applying the general + definition of integer exponentiation we see that* + + *$$\begin{align*} + \left(x^n\right)^m=&\prod_{i=1}^{\left|m\right|} \frac{1}{x^n}\\ + &=\underbrace{\left(\frac{1}{x^n}\right)*\left(\frac{1}{x^n}\right)*\left(\frac{1}{x^n}\right)*\dots*\left(\frac{1}{x^n}\right)}_{\left|m\right|\text{ times}} + \end{align*}$$* + + *Now, we know by definition of multiplication for rationals that + $\displaystyle\frac{1}{a}*\frac{1}{b}=\frac{1}{ab}$ and so.* + + *$$\begin{align*} + \left(x^n\right)^m=&\underbrace{\left(\frac{1}{x^n}\right)*\left(\frac{1}{x^n}\right)*\left(\frac{1}{x^n}\right)*\dots*\left(\frac{1}{x^n}\right)}_{\left|m\right|\text{ times}}\\ + &=\frac{1}{x^{n\left|m\right|}}\\ + &=\prod_{i=1}^{n\left|m\right|} \frac{1}{x} =x^{nm} + \end{align*}$$* + + *By definition.* + +2. *$n< 0$ and $m\geq 0$:* + + *As $n<0$ then we have that* + + *$$\begin{equation*} + x^n=\prod_{i=1}^{\left|n\right|}\frac{1}{x}=\frac{1}{x^n} + \end{equation*}$$* + + *We can now apply similar logic to the first part to conclude the + result.* + +3. *$n< 0$ and $m<0$:* + + *Using similar logic to the two previous parts deduces the result.* + +*As promised. $\qed$* +::: + +::: {#prop:IntegerExtensionExponentiationOfSameBaseAddsPowers .proposition} +**Proposition 91**. *Multiplying exponents of the same base adds the +powers* + +*Let $x\in\mathbb{Z}$ be a fixed integer and let $n,m\in\mathbb{Z}$. We +have that* + +*$$\begin{equation*} + x^n *x^m = x^{n+m} +\end{equation*}$$* + +*Proof:* + +*If $n,m\geq 0$ the result is the same as proposition +[77](#prop:IntegerExponentiationOfSameBaseAddsPowers){reference-type="ref" +reference="prop:IntegerExponentiationOfSameBaseAddsPowers"}, so we have +to consider the following three cases* + +1. *$n\geq 0$ and $m<0$* + +2. *$n< 0$ and $m\geq 0$* + +3. *$n< 0$ and $m<0$* + + + +1. *$n\geq 0$ and $m<0$:* + + *Let $m=-k$ for some $k\in\mathbb{Z}$ with $k>0$. We know that + $\displaystyle x^m=x^{-k}=\prod_{i=1}^{-k} \frac{1}{x} = x^{-k}$. + Now we have* + + *$$\begin{equation*} + x^n*x^m=x^n x^{-k}=x^{n+-k} + \end{equation*}$$* + + *Which is equivalent to $x^{n+m}$.* + +2. *$n< 0$ and $m\geq 0$:* + + *Like the previous part let $n=-k$ for some $k\in\mathbb{Z}$ with + $k>0$ then we get* + + *$$\begin{equation*} + x^n*x^m=x^{-k}*x^{m}=x^{-k+m}=x^{n+m} + \end{equation*}$$* + +3. *$n< 0$ and $m<0$:* + + *Let $n=-k$ and $m=-j$ for $k,j\in\mathbb{Z}$ with $k>0$ and $j>0$. + Then* + + *$$\begin{equation*} + x^n*x^m=x^{-k}*x^{-j}=x^{-k+-j}=x^{n+m} + \end{equation*}$$* + +*As required. $\qed$* +::: + +::: {#prop:IntegerExtensionExponentiationPowerOfProductIsProductOfPowers .proposition} +**Proposition 92**. *Power of product is product of powers* + +*Let $x,y\in\mathbb{Z}$ and $n\in\mathbb{Z}$. Then* + +*$$\begin{equation*} + \left(x*y\right)^n=x^n*y^n +\end{equation*}$$* + +*Proof:* + +*If $n=0$ then $\left(x*y\right)^n=1$ and clearly $x^0*y^0=1$. So +suppose $n>0$ then we have* + +*$$\begin{align*} + \left(x*y\right)^n=\prod_{i=1}^n xy &=\underbrace{xy*xy*\dots *xy}_{n\text{ times}}\\ + &= \left(\underbrace{x*x*\dots *x}_{n\text{ times}}\right)*\left(\underbrace{y*y*\dots *y}_{n\text{ times}}\right),\ \text{ By commutativity of multiplication}\\ + &=x^n*y^n +\end{align*}$$* + +*Finally, let $n<0$ then a similar argument shows that* + +*$$\begin{equation*} + \left(x*y\right)^n=\frac{1}{x^n*y^n} +\end{equation*}$$ Showing the proposition. $\qed$* +::: + +We have extended integer exponentiation. What can we say about rational +exponentiation? We can clearly extend the base of exponentiation to an +arbitrary rational number. We have already used special cases of this +when we considered denominators and numerators separately in the +previous proofs. We formalise this to a fully general rational number. +Firstly, we know that if $n<0$ then $\displaystyle x^n=\frac{1}{x^n}$. +Additionally if $x\in\mathbb{Z}$ then a multiplicative inverse of $x$ in +the rationals is given by $x^{-1}=\frac{1}{x}$. We combine the two into +a general definition. + +::: definition +**Definition 137**. *Exponentiation for negative indices* + +*Let $x\in\mathbb{Z}$ with $x\neq 0$. We extend exponentiation to +negative $n\in\mathbb{Z}$ by* + +*$$\begin{equation*} + x^{-n} = \left(x^{-1}\right)^n +\end{equation*}$$* + +*Clearly we have in general that $x^{-n}\in\mathbb{Q}$* +::: + +Now we can consider the more general case of +$\displaystyle\left(\frac{a}{b}\right)^n$ for $a,b,n\in\mathbb{Z}$ and +$b\neq 0$. We have the following proposition + +::: proposition +**Proposition 93**. *Rational number raised to an integer exponent* + +*Let $x\in\mathbb{Q}$ with $\displaystyle x=\frac{a}{b}$ and $b\neq 0$. +Let $n\in\mathbb{Z}$ We have that* + +*$$\begin{equation*} + \left(\frac{a}{b}\right)^n=\frac{a^n}{b^n} +\end{equation*}$$* + +*Proof:* + +*We have that* + +*$$\begin{align*} + \left(\frac{a}{b}\right)^n&=\left(a*b^{-1}\right)^n\\ + &= \underbrace{\left(a b^{-1}\right)\left(a b^{-1}\right)\dots \left(a b^{-1}\right)}_{n \text{ times}}\\ + &=\underbrace{a*a*a*\dots*a}_{n \text{ times}}*\underbrace{b^{-1}*b^{-1}*b^{-1}\dots*b^{-1}}_{n \text{ times}}\\ + &= a^n \left(b^{-1}\right)^n\\ + &=a^n*b^{-n}\\ + &=\frac{a^n}{b^n} +\end{align*}$$* + +*As required. $\qed$* +::: + +The rules of integer exponentiation extend when the base is rational. + +::: {#propRationalExponentiationPowerLaw .proposition} +**Proposition 94**. *Power law of exponentiation for positive exponents* + +*Let $x\in\mathbb{Q}$ and let $n,m\in \mathbb{Z}$. We have that* + +*$$\begin{equation*} + \left(x^n\right)^m = x^{nm} +\end{equation*}$$* + +*Proof:* + +*Let $\displaystyle x=\frac{a}{b}$ with $a,b\in\mathbb{Z}$ and +$b\neq 0$. We have that* + +*$$\begin{align*} + \left(x^n\right)^m&=\left(\left(\frac{a}{b}\right)^n\right)^m\\ + &=\left(\frac{a^n}{b^n}\right)^m\\ + &=\left(a^n*b^{-m}\right)^m\\ + &=a^{nm}*b^{-nm}\\ + &=\frac{a^{nm}}{b^{nm}}\\ + &=x^{nm} +\end{align*}$$ $\qed$* +::: + +::: {#prop:RationalExponentiationOfSameBaseAddsPowers .proposition} +**Proposition 95**. *Multiplying exponents of the same base adds the +powers* + +*Let $x\in\mathbb{Q}$ be a fixed integer and let $n,m\in\mathbb{Z}$. We +have that* + +*$$\begin{equation*} + x^n *x^m = x^{n+m} +\end{equation*}$$* + +*Proof:* + +*Let $\displaystyle x=\frac{a}{b}$ with $a,b\in\mathbb{Z}$ and +$b\neq 0$. Observe that* + +*$$\begin{align*} + x^n*x^m&=\left(\frac{a}{b}\right)^n*\left(\frac{a}{b}\right)^m\\ + &=\frac{a^n}{b^n}*\frac{a^m}{b^m}\\ + &=\frac{a^n*a^m}{b^n*b^m}\\ + &=\frac{a^{n+m}}{b^{n+m}}\\ + &=\left(\frac{a}{b}\right)^{n+m}\\ + &=x^{n+m} +\end{align*}$$* + +*As required. $\qed$* +::: + +::: {#prop:RationalExponentiationPowerOfProductIsProductOfPowers .proposition} +**Proposition 96**. *Power of product is product of powers* + +*Let $x,y\in\mathbb{Q}$ and $n\in\mathbb{Z}$. Then* + +*$$\begin{equation*} + \left(x*y\right)^n=x^n*y^n +\end{equation*}$$* + +*Proof:* + +*Let $\displaystyle x=\frac{a}{b}$ with $a,b\in\mathbb{Z}$ and $b\neq 0$ +and let $\displaystyle y=\frac{c}{d}$ with $c,d\in\mathbb{Z}$ and +$d\neq 0$. We have* + +*$$\begin{align*} + \left(x*y\right)^n&=\left(\frac{a}{b}*\frac{c}{d}\right)^n\\ + &=\left(\frac{ac}{bd}\right)^n\\ + &=\frac{\left(ac\right)^n}{\left(bd\right)^n}\\ + &=\frac{a^n c^n}{b^n d^n}\\ + &=\frac{a^n}{b^n}*\frac{c^n}{d^n}\\ + \frac{}{} + &=x^n*y^n +\end{align*}$$* +::: + +What about rational exponents? Can we assign meaning to expressions of +the form $\displaystyle \wedge\left(\frac{a}{b},\frac{c}{d}\right)$? +Using a similar argument to when we considered extending integer +exponentiation. Suppose that proposition +[95](#prop:RationalExponentiationOfSameBaseAddsPowers){reference-type="ref" +reference="prop:RationalExponentiationOfSameBaseAddsPowers"} holds for +rational exponents. In particular we have for some $x\in\mathbb{Q}$ that + +$$\begin{equation*} + x^{\frac{1}{2}}*x^{\frac{1}{2}}=x^1 +\end{equation*}$$ + +Now, suppose that $x=2$. We are hence saying that + +$$\begin{equation*} + 2^{\frac{1}{2}}*2^{\frac{1}{2}}=2 +\end{equation*}$$ + +If we suppose that $\displaystyle 2^{\frac{1}{2}}\in\mathbb{Q}$ with say +$\displaystyle y=2^{\frac{1}{2}}$ we are saying that $y^2=2$. +Unfortunately, there is no such rational $y$ that satisfies this. +Moreover, we lack the theory required to prove this at this time. This +will be corrected in part [](#part2){reference-type="ref" +reference="part2"}. + +#### Extending the absolute value function + +When we constructed the integers we recast the notion of size into that +of distance. This was achieved using the so-called absolute value +function given by + +$$\begin{equation*} + \left|x\right|=d\left(x,0\right)=\begin{cases} + x,\ \text{If } x\geq 0\\ + -x,\ \text{If } x< 0 + \end{cases} +\end{equation*}$$ + +where + +$$\begin{align*} + d:\mathbb{Z}^2&\rightarrow\mathbb{N}\\ + \left(x,y\right)&\mapsto d\left(x,y\right)=\begin{cases} + x-y,\ \text{If } x\geq y\\ + -\left(x-y\right),\ \text{If } x< y + \end{cases} +\end{align*}$$ + +Now that we have constructed the rational numbers we can consider how +this idea extends. One thing that is clear from the definition of $d$ +for integers is that the smallest possible non-zero distance that can be +achieved is $1$, for example, $d\left(2,1\right)$. However, consider + +$$\begin{equation*} + 1-\frac{1}{2}=\frac{1}{2} +\end{equation*}$$ + +If this idea of distance is to extend to the rationals we will clearly +have that distances smaller than $1$ are now possible. In other words, +the mapping for $d$ when used with rational numbers can no longer map +into $\mathbb{N}$. This is easily remedied by defining the following +set. + +::: definition +**Definition 138**. *Positive rationals* + +*We define the set of positive rationals by* + +*$$\begin{equation*} + \mathbb{Q}^+=\left\{x\in\mathbb{Q}: x>0\right\} +\end{equation*}$$* +::: + +It is clear from the definitions for the integers how to extend the +distance function and the absolute value function to the rationals. + +::: definition +**Definition 139**. *Distance function for the rationals* + +*Let $x,y\in\mathbb{Q}$. Define the function +$d:\mathbb{Q}^2\rightarrow\mathbb{Q}^+$ by* + +*$$\begin{align*} + d:\mathbb{Q}^2&\rightarrow\mathbb{Q}^+\\ + \left(x,y\right)&\mapsto d\left(x,y\right)=\begin{cases} + x-y,\ \text{If } x\geq y\\ + -\left(x-y\right),\ \text{If } x< y + \end{cases} +\end{align*}$$* +::: + +As before we prove that this distance function is well-defined. + +::: {#prop:RationalDistanceFuncWellDefined .proposition} +**Proposition 97**. *The distance function for the rationals is +well-defined* + +*Let $x,y\in\mathbb{Q}$. We have that* + +*$$\begin{equation*} + d\left(x,y\right)=\begin{cases} + x-y,\ \text{If } x\geq y\\ + -\left(x-y\right),\ \text{If } x< y + \end{cases} +\end{equation*}$$* + +*is well-defined.* + +*Proof:* + +*Let $x,y\in\mathbb{Q}$. There are two cases to consider $x\geq y$ and +$x0$ which is to say + $-\left(x-y\right)\in\mathbb{Q}^+$* + +*The result has been shown. $\qed$* +::: + +We can now generalise the absolute value function. + +::: definition +**Definition 140**. *Absolute value function* + +*Let $x\in\mathbb{Q}$ we define the absolute value function, denoted by +$\left|x\right|$ by the function* + +*$$\begin{equation*} + \left|x\right|=d\left(x,0\right)=\begin{cases} + x,\ \text{If } x\geq 0\\ + -x,\ \text{If } x< 0 + \end{cases} +\end{equation*}$$* +::: + +We have generalised the idea of "size" to the rationals. We can now also +generalise the properties of the absolute value function explored in the +construction of the integers. + +::: proposition +**Proposition 98**. *Properties of the absolute value* + +*Let $x,y,z\in\mathbb{Q}$. We have that the absolute value function has +the following properties* + +1. *$\left|x\right|\geq 0$ for all $x\in\mathbb{Q}$* + +2. *$\left|x\right|=0\iff x=0$* + +3. *$\left|x-y\right|=0\iff x=y$* + +4. *$\left|xy\right|=\left|x\right|\left|y\right|$* + +5. *$\displaystyle \left|\frac{x}{y}\right|=\frac{\left|x\right|}{\left|y\right|}$ + with $y\neq 0$* + +6. *$\left|\left|x\right|\right|=\left|x\right|$* + +7. *$\left|-x\right|=\left|x\right|$* + +8. *$\left|x\right|\leq y \iff -y\leq x\leq y$* + +9. *$\left|x\right|\geq y\iff x\leq -y$ or $x\geq y$* + +10. *$\left|x+y\right|\leq \left|x\right|+\left|y\right|$* + +11. *$\left|x-y\right|\leq\left|x-z\right|+\left|z-y\right|$* + +12. *$\left|x-y\right|\geq \left|\left|x\right|-\left|y\right|\right|$* + +13. *$\left|\cdot\right|$ is not injective* + +14. *$\left|\cdot\right|$ is not surjective* + +*Proof:* + +1. *$\left|x\right|\geq 0$ for all $x\in\mathbb{Q}$:* + + *This follows by proposition + [97](#prop:RationalDistanceFuncWellDefined){reference-type="ref" + reference="prop:RationalDistanceFuncWellDefined"}.* + +2. *$\left|x\right|=0\iff x=0$:* + + *We have by definition that $\left|x\right|=0$, if and only if + $x=0$.* + +3. *$\left|x-y\right|=0\iff x=y$:* + + *$\left(\Rightarrow\right)$: Suppose that $\left|x-y\right|=0$. + There are two cases to consider.* + + *Firstly if $x\geq y$, then by definition we have that + $\left|x-y\right|=x-y=0$ from which we clearly have $x=y$. The other + case is $x + + 1. *$x\geq 0$ and $y\geq 0$:* + + *If $x\geq 0$ and $y\geq 0$ then $xy\geq 0$ and so + $\left|xy\right|=xy$. Likewise $\left|x\right|=x$ and + $\left|y\right|=y$. Hence + $\left|xy\right|=\left|x\right|\left|y\right|$.* + + 2. *$x\geq 0$ and $y<0$:* + + *If $x\geq 0$ then $\left|x\right|=x$ by definition, and if + $y<0$ then $\left|y\right|=-y$. Now $\left|xy\right|=-xy$ as + $y<0$. Moreover, we have that* + + *$$\begin{equation*} + -xy=\left(-1\right)\left(x\right)\left(y\right)=\left(x\right)\left(-1\right)\left(y\right)=\left(x\right)\left(-y\right)=\left|x\right|\left|y\right| + \end{equation*}$$* + + *Hence we get $\left|xy\right|=\left|x\right|\left|y\right|$* + + 3. *$x<0$ and $y\geq 0$:* + + *This is similar to the above but swapping the roles of $x$ and + $y$.* + + 4. *$x<0$ and $y<0$:* + + *Suppose that $x<0$ and $y<0$, then we have that + $\left|x\right|=-x$ and $\left|y\right|=-y$ by definition. + Moreover, we have that $-x*-y = xy$. Hence + $\left|xy\right|=xy=\left(-x\right)\left(-y\right)=\left|x\right|\left|y\right|$* + +5. *$\displaystyle\left|\frac{x}{y}\right|=\frac{\left|x\right|}{\left|y\right|}$ + with $y\neq 0$:* + + *This follows by part 4.* + +6. *$\left|\left|x\right|\right|=\left|x\right|$:* + + *We have that $\left|x\right|=x$ if $x\geq 0$ and $-x$ if $x<0$.* + + *So if $x\geq 0$, we have* + + *$$\begin{equation*} + \left|\left|x\right|\right|=\left|x\right|=x=\left|x\right| + \end{equation*}$$* + + *Now if $x<0$ then* + + *$$\begin{equation*} + \left|\left|x\right|\right|=\left|-x\right|=\underbrace{-x}_{\text{As }-x>0}=\left|x\right| + \end{equation*}$$* + +7. *$\left|-x\right|=\left|x\right|$:* + + *As $-x=-1 *x$ we have by part 4 that* + + *$$\begin{equation*} + \left|-x\right|=\left|-1*x\right|=\left|-1\right|\left|x\right|=1*\left|x\right|=\left|x\right| + \end{equation*}$$* + +8. *$\left|x\right|\leq y \iff -y\leq x\leq y$:* + + *$\left(\Rightarrow\right)$: Suppose that $\left|x\right|\leq y$. If + $x\geq 0$ then we get that $\left|x\right|=x\leq y$. From this, it + is clear that $-y\leq x\leq y$ as $x\geq 0$ and + $x\leq y \Rightarrow y \geq 0$.* + + *Now if $x<0$, then $\left|x\right|=-x\leq y$. Clearly $x\leq -x$ as + $x<0$ hence we conclude that $x\leq -x\leq y$. Now by part 18 of + proposition + [89](#prop:InequalityRationalNumbers){reference-type="ref" + reference="prop:InequalityRationalNumbers"} we have we have* + + *$$\begin{equation*} + \left(-1\right)*\left(-x\right)\geq \left(-1\right)\left(y\right) \iff x\geq -y + \end{equation*}$$* + + *Now $x\geq -y$ is the same as $-y\leq x$ and so we have + $-y\leq x\leq -x \leq y$.* + + *Hence $-y\leq x\leq y$.* + + *$\left(\Leftarrow\right)$: Suppose that $-y\leq x\leq y$. There are + two cases to consider.* + + 1. *$x\geq 0$* + + 2. *$x<0$* + + + + 1. *$x\geq 0$:* + + *Suppose $x\geq 0$, then clearly as $x\leq y$ then + $\left|x\right|\leq \left|y\right|=y$. Moreover, we have that + $-y\leq x$ is the same $x\geq -y$ and by part 22. of proposition + [71](#prop:InequalityIntegerNumbers){reference-type="ref" + reference="prop:InequalityIntegerNumbers"} when applied to + $x\geq -y$ gives* + + *$$\begin{equation*} + \left(-1\right)*\left(x\right)\leq \left(-1\right)\left(-y\right) \iff -x\leq y + \end{equation*}$$* + + *We have that $\left|-x\right|=\left|x\right|$ by part 6. Hence + $\left|-x\right|=\left|x\right|\leq \left|y\right|=y$.* + + 2. *$x<0$:* + + *Suppose $x<0$. By assumption $x\leq y$ so either $y\geq 0$ or + $y< 0$. We can't have $y<0$ as for example take $x=-4$ and + $y=-2$ then we would have $2\leq -4\leq -2$ a contradiction.* + + *So suppose that $y\geq 0$ then as $x\leq y$ we have + $\left|x\right|\leq\left|y\right|=y$. Now as $-y\leq x$ by + assumption we have that $x\geq -y$ and so part 22. of + proposition + [89](#prop:InequalityRationalNumbers){reference-type="ref" + reference="prop:InequalityRationalNumbers"} gives* + + *$$\begin{equation*} + \left(-1\right)*\left(x\right)\leq \left(-1\right)\left(-y\right) \iff -x\leq y + \end{equation*}$$* + + *Hence part 6. applies and we get that $\left|x\right|\leq y$* + +9. *$\left|x\right|\geq y\iff x\leq -y$ or $x\geq y$:* + + *$\left(\Rightarrow\right)$: Suppose that $\left|x\right|\geq y$. If + $x\geq 0$ then $\left|x\right|=x\geq y$. So suppose that $x<0$ then + by definition we have that $\left|x\right|=-x$ and so $-x\geq y$ and + the result follows when applying part 22. of proposition + [89](#prop:InequalityRationalNumbers){reference-type="ref" + reference="prop:InequalityRationalNumbers"}.* + + *$\left(\Leftarrow\right)$: Suppose that either $x\leq -y$ or + $x\geq y$. We have three cases to consider.* + + 1. *$x\leq -y$* + + 2. *$x\geq y$* + + 3. *$x\leq -y$ and $x\geq y$* + + + + 1. *$x\leq -y$:* + + *Suppose that $x\leq -y$ holds. If $x\geq 0$ then we have that + $-y\geq 0$, Hence $y<0$. Moreover, we have that by part 18. of + proposition + [89](#prop:InequalityRationalNumbers){reference-type="ref" + reference="prop:InequalityRationalNumbers"} that* + + *$$\begin{equation*} + \left(-1\right)*\left(x\right)\geq \left(-1\right)\left(-y\right) \iff -x\geq y + \end{equation*}$$* + + *Now part 6. applies and we see that + $\left|-x\right|=\left|x\right|\geq\left|y\right|=y$. This is to + say $\left|x\right|\geq y$.* + + *Now suppose that $x<0$. Then as $x\leq -y$ we have that either + $-y\geq 0$ or $-y<0$. In the former case $-y\geq 0$ gives $y<0$. + Hence by part 18. of proposition + [89](#prop:InequalityRationalNumbers){reference-type="ref" + reference="prop:InequalityRationalNumbers"} we conclude that* + + *$$\begin{equation*} + \left(-1\right)*\left(x\right)\geq \left(-1\right)\left(y\right) \iff -x\geq y + \end{equation*}$$* + + *As $x<0$ then $-x\geq 0$. The result follows when taking the + absolute value.* + + *Now suppose that $-y<0$ then $y\geq 0$. Following similar logic + to the previous case, we see that* + + *$$\begin{equation*} + \left(-1\right)*\left(x\right)\geq \left(-1\right)\left(y\right) \iff -x\geq y + \end{equation*}$$* + + *The result again follows after taking the absolute value.* + + 2. *$x\geq y$:* + + *This case is trivial.* + + 3. *$x\leq -y$ and $x\geq y$:* + + *Suppose that $x\leq -y$ and $x\geq y$ are both true. We know by + the first case that $x\leq -y$ gives $\left|x\right|\geq y$ and + $x\leq y$ also implies $\left|x\right|\geq y$ by the second + case. Hence both inequalities being true at the same time + implies the result $\left|x\right|\geq y$.* + +10. *$\left|x+y\right|\leq \left|x\right|+\left|y\right|$:* + + *Let $x,y\in\mathbb{Q}$. There are four cases to consider.* + + 1. *$x\geq 0$ and $y\geq 0$* + + 2. *$x\geq 0$ and $y\leq 0$* + + 3. *$x\leq 0$ and $y\geq 0$* + + 4. *$x\leq 0$ and $y\leq 0$* + + + + 1. *$x\geq 0$ and $y\geq 0$:* + + *Suppose $x\geq 0$ and $y\geq 0$, then we have that* + + *$$\begin{equation*} + \left|x+y\right|=x+y=\left|x\right|+\left|y\right|\Rightarrow \left|x+y\right|\leq\left|x\right|+\left|y\right| + \end{equation*}$$* + + 2. *$x\geq 0$ and $y\leq 0$* + + *By assumption we have that $\left|x\right|=x$ and + $\left|y\right|=-y$. We have two cases based on the absolute + value, $\left|x\right|\leq\left|y\right|$ and + $\left|x\right|\geq\left|y\right|$.* + + *So suppose that $\left|x\right|\leq\left|y\right|$ then by + definition $x\leq -y$ and so by part 12. of proposition + [89](#prop:InequalityRationalNumbers){reference-type="ref" + reference="prop:InequalityRationalNumbers"} we have that* + + *$$\begin{equation*} + x\leq -y \Rightarrow x+y\leq 0 + \end{equation*}$$* + + *Moreover, as $x\geq 0$ then $y\leq x+y\leq 0$. Hence we have by + the definition of the absolute value that* + + *$$\begin{equation*} + \left|x+y\right|=-\left(x+y\right)\leq -y=\left|y\right| + \end{equation*}$$ As $-y>0$.* + + *In the case $\left|x\right|\geq\left|y\right|$ we have by + definition that $x\geq -y$ and so $x+y\geq 0$. Additionally it + is clear that $x\geq x+y$ as $y\leq 0$ and + $\left|x\right|\geq\left|y\right|$. Hence by definition of the + absolute value we have that* + + *$$\begin{equation*} + \left|x+y\right|=x+y\leq x=\left|x\right| + \end{equation*}$$* + + *Now, it is clear to see that + $\left|x\right|\leq \left|x\right|+\left|y\right|$ and likewise + $\left|y\right|\leq \left|x\right|+\left|y\right|$.* + + *We have hence shown that + $\left|x+y\right|leq\left|x\right|+\left|y\right|$.* + + 3. *$x\leq 0$ and $y\geq 0$:* + + *This is similar to above, interchanging the roles of $x$ and + $y$.* + + 4. *$x\leq 0$ and $y\leq 0$:* + + *Suppose that $x\leq 0$ and $y\leq 0$ then by definition we have + that $\left|x+y\right|=-\left(x+y\right)=-x-y$. As $x\leq 0$ and + $y\leq 0$ then we have that and $\left|y\right|=-y$ which shows + $\left|x+y\right|=\left|x\right|+\left|y\right|\leq\left|x\right|+\left|y\right|$* + +11. *$\left|x-y\right|\leq\left|x-z\right|+\left|z-y\right|$:* + + *We have that* + + *$$\begin{align*} + \left|x-y\right|&=\left|x-\left(z-z\right)-y\right|\\ + &=\left|x-z+z-y\right|\\ + &\leq \left|x-z\right|+\left|z-y\right| + \end{align*}$$* + +12. *$\left|x-y\right|\geq \left|\left|x\right|-\left|y\right|\right|$:* + + *We have that* + + *$$\begin{align*} + \left|x\right|&=\left|\left(x-y\right)+y\right|\leq \left|x-y\right|+\left|y\right| \Rightarrow \left|x\right|-\left|y\right|\leq \left|x-y\right|\\ + \left|y\right|&=\left|\left(y-x\right)+x\right|\leq \left|x-y\right|+\left|x\right| \Rightarrow \left|y\right|-\left|x\right|\leq \left|x-y\right|\\ + \end{align*}$$* + + *Hence we have* + + *$$\begin{align*} + \left|x\right|-\left|y\right|\leq \left|x-y\right| &\Rightarrow \left|\left|x\right|-\left|y\right|\right|\leq \left|x-y\right|\\ + \left|y\right|-\left|x\right|=\left(-1\right)\left(\left|x\right|-\left|y\right|\right)\leq \left|x-y\right| &\Rightarrow \left|\left|x\right|-\left|y\right|\right|\leq \left|x-y\right|\\ + \end{align*}$$* + + *Hence we have the result.* + +13. *$\left|\cdot\right|$ is not injective:* + + *This follows as the absolute value function was not injective for + the integers* + +14. *$\left|\cdot\right|$ is not surjective:* + + *This follows as the absolute value function was not surjective for + the integers* + +*As required. $\qed$* +::: + +# Elementary Number Theory {#part2} + +### Introduction + +::: epigraph +Mathematics is the queen of the sciences and Number Theory is the queen +of mathematics. + +*Carl Friedrich Gauss* +::: + +In the previous part, we have gone from only having the axioms of ZFC, +the rules of logic and knowledge of mappings and have built two types of +numbers, the naturals and the integers. Unfortunately, we need to make a +detour from constructing new objects. We need to start using the objects +we have constructed to provide a guide on how to proceed with building +more mathematical objects. + +We will start with Number Theory. Number Theory primarily deals with the +properties of the integers $\mathbb{Z}$ as well as mappings defined on +$\mathbb{Z}$. This includes properties about the operations on the +integers, properties about the compositions and ways of expressing +relationships between certain "types" of integers, solving equations +involving the integers and more. + +The applications of Number Theory to the modern world are numerous. One +main example of the usage of Number Theory is encryption, the art of +obfuscating information so that it can only be read by trusted +individuals[^11]. We will later consider an example of encryption called +RSA. + +Additionally, the ideas that we will develop when studying Number Theory +are key to providing crucial insights into other branches of +mathematics. We will come to see that many of the key properties of the +integers are also enjoyed by many other types of mathematical objects, +especially in an abstract setting. + +### Divisibility + +::: epigraph +Now where there are no parts, neither extension, shape, nor divisibility +is possible. And these monads are the true atoms of nature and, in a +word, the elements of things. + +*Gottfried Leibniz* +::: + +#### Definition of divisibility of integers + +Although we have a concrete construction of the integers, we haven't +even discussed some of their most basic properties! We know how to add, +subtract and multiply them, but we don't know how to divide them without +the rational numbers $\mathbb{Q}$. It is with $\mathbb{Q}$ that we can +hope to find a rule that says that +$\displaystyle\frac{a}{b}\in\mathbb{Z}$ for some $a,b\in\mathbb{Z}$. + +Recall that in $\mathbb{Q}$ we defined an equivalence relation $\sim$ so +that for $\left(a,b\right),\left(c,d\right)\in\mathbb{Z}^2$ we have that + +$$\begin{equation*} + \left(a,b\right)\sim\left(c,d\right)\iff ad=bc +\end{equation*}$$ + +where we had $b\neq 0$ and $d\neq 0$. We also saw that +$\left(x,1\right)\in\left[\left(x,1\right)\right]$ represented an +integer. Hence the question we are resolving is when does +$\left(a,b\right)\sim\left(x,1\right)$. We have that + +$$\begin{equation*} + \left(a,b\right)\sim\left(x,1\right)\iff a=bx +\end{equation*}$$ + +That is $b$ divides $a$ and gives an integer if and only if $a=bx$. We +make this our first formal definition in the field of Number Theory. + +::: {#def:NT_Int_Div_def .definition} +**Definition 141**. *Integer divisibility* + +*Let $a,b\in\mathbb{Z}$ with $b\neq 0$. We say that $a$ is divisible by +$b$, or $b$ divides $a$, written as $b\mid a$ if and only if +$\exists c\in\mathbb{Z}$ so that $a=bc$. We say that $b$ is a divisor of +$a$.* + +*If $b$ does not divide $a$ we write $b\not\nmid a$.* +::: + +::: example +**Example 90**. *We have that $3\mid 6$ as $6=3*2$.* + +*Obverse that $2\nmid 3$. Indeed there is no integer $x$ so $3=2x$.* +::: + +We make a definition based on the definition of divisibility. Namely +based on if a number can be divided into two equal parts. + +::: definition +**Definition 142**. *Even number* + +*Let $x\in\mathbb{Z}$. We say that $x$ is even if we have that +$2\mid x$.* +::: + +This immediately gives another definition. + +::: definition +**Definition 143**. *Odd number* + +*Let $x\in\mathbb{Z}$. We say that $x$ is odd if we have that +$2\nmid x$.* +::: + +We can make another definition, based on divisibility. + +::: definition +**Definition 144**. *Integer multiple* + +*Let $a,b\in\mathbb{Z}$ so that $b\mid a$. We say that $b$ is a multiple +of $a$.* +::: + +There are two results that we can derive based on an even number, an odd +number and integer multiples. + +::: {#prop:NT_even_iff_2n .proposition} +**Proposition 99**. *Integer is even if it is a multiple of 2* + +*Let $x\in\mathbb{Z}$. We have that $x$ is even if and only if $x$ is a +multiple of $2$.* + +*Proof:* + +*$\left(\Rightarrow\right):$ Suppose that $x$ is even, then by +definition we have that $2\mid x$ and so by the definition of +divisibility we have that $x=2c$ for some $c\in\mathbb{X}$. By the +definition of being an integer multiple we have that $x$ is a multiple +of $2$.* + +*$\left(\Leftarrow\right):$ Suppose that $x$ is a multiple of $2$. By +definition of being an integer multiple, we have that $x=2r$ for some +$r\in\mathbb{Z}$. Hence by the definition of divisibility, we have that +$2\mid x$ and so by definition of an even number we have that $x$ is +even. $\qed$* +::: + +We can find a similar proposition for odd numbers. Observe that by the +previous proposition that $x$ being even means that $x=2n$ for some +integer $n$. Also, we have that $2n+2=2\left(n+1\right)$ is even, so +what can we say about $2n+1$? + +::: proposition +**Proposition 100**. *Integer is odd if and only if it is not a multiple +of 2* + +*Let $x\in\mathbb{Z}$. We have that $x$ is odd if and only if $x$ is not +a multiple of 2.* + +*Proof:* + +*The proof follows by the contra-positive, that is $x$ is a multiple of +2 if and only if $x$ is even, which is the previous proposition. $\qed$* +::: + +Hence we need to determine if $2n+1$ is even or odd. We need to develop +the theory of divisibility. + +The definition of divisibility gives an immediate result. Namely that +when considering the divisibility of integers we need only concern +ourselves with positive integers, as negative integers will also be +divisors. That is if $b\mid a$ then so does $-b$. + +::: {#prop:NT_PositiveAndNegativeDivisorsForIntsExist .proposition} +**Proposition 101**. *Integer dividing another implies negative integer +also divides* + +*Let $a,b\in\mathbb{Z}$ with $b\mid a$. We also have that $-b\mid a$.* + +*Proof:* + +*Let $a,b\in\mathbb{Z}$ with $b\mid a$. By definition of divisibility, +we have that $\exists c\in\mathbb{Z}$ so that $a=bc$. We know that +$-1*1=1$ and so we have that* + +*$$\begin{equation*} + a=bc=\left(-1*-1\right)bc=-b*-c +\end{equation*}$$* + +*As $-c\in\mathbb{Z}$ then it follows by definition that $-b\mid a$. +$\qed$* +::: + +Hence by proposition +[101](#prop:NT_PositiveAndNegativeDivisorsForIntsExist){reference-type="ref" +reference="prop:NT_PositiveAndNegativeDivisorsForIntsExist"} we will +restrict our view to positive divisors only, knowing that any results +about a positive divisor will extend to negative divisors. + +One clear divisor of any integer $a$ is itself, that is $a\mid a$ as +$a=a*1$. We will find it interesting to consider the more non-trivial +divisors of some integers. Hence we make the following definition + +::: definition +**Definition 145**. *Proper divisor* + +*Let $a,b\in\mathbb{Z}$ with $b\mid a$. If we have that $00$ and $b>0$ implies that $a\leq b$.* + +6. *If $m\in\mathbb{Z}$ is such that $m\neq 0$ then $a\mid b$ is true + if and only if $ma\mid mb$.* + +7. *For all $a\in\mathbb{Z}$ with $a\neq 0$ we have $a\mid 0$* + +*Proof:* + +1. *$a\mid b \Rightarrow a\mid bc$ for any $c\in\mathbb{Z}$:* + +2. *$a\mid b$ and $b\mid c$ implies that $a\mid c$:* + + *Suppose that $a\mid b$, then by definition there exists + $d\in\mathbb{Z}$ so that $b=ad$. Hence we have that* + + *$$\begin{equation*} + bc=adc \Rightarrow a\mid bc + \end{equation*}$$* + + *as $dc\in\mathbb{Z}$.* + +3. *$a\mid b$ and $a\mid c$ implies that $a\mid\left(bx+cy\right)$ for + any $x,y\in\mathbb{Z}$:* + + *Suppose that $a\mid b$ and $b\mid c$, then by the definition of + divisibility, and by part 1., we have that $b=ax$ and $c=by$ for all + $x,y\in\mathbb{Z}$. We hence see that* + + *$$\begin{equation*} + c=axy + \end{equation*}$$* + + *Hence as $xy\in\mathbb{Z}$ then we conclude that $a\mid c$.* + +4. *$a\mid b$ and $b\mid a$ implies $a=\pm b$, that is either $a=b$ or + $a=-b$:* + + *Let $a\mid b$ and $a\mid c$, then there are $d,e\in\mathbb{Z}$ such + that $b=ad$ and $c=ae$. Now, let $x,y\in\mathbb{Z}$ then we have + that $bx=adx$ and $cy=aey$ and $bx+cy=adx+aey=a\left(dx+ey\right)$. + Hence $a\mid\left(bx+cy\right)$.* + +5. *$a\mid b$ and $a>0$ and $b>0$ implies that $a\leq b$:* + + *If $a\mid b$ then $\exists x\in\mathbb{Z}$ so that $b=ax$, likewise + if $b\mid a$ then $\exists y\in\mathbb{Z}$ so that $a=by$. It + follows that $b=byx$. We have that $b=byx$ is true if and only if + $yx=1$. Therefore either $x=y=1$ or $x=y=-1$.* + + *The result is clear after substituting $y$ into $a=by$.* + +6. *If $m\in\mathbb{Z}$ is such that $m\neq 0$ then $a\mid b$ is true + if and only if $ma\mid mb$:* + + *$\left(\Rightarrow\right)$: Let $m\in\mathbb{Z}$ be non-zero and + let $a\mid b$. By definition, there is some $c\in\mathbb{Z}$ so that + $b=ac$. Multiplying both sides by $m$ gives* + + *$$\begin{equation*} + bm=acm=amc + \end{equation*}$$* + + *and so $am\mid bm$.* + + *$\left(\Leftarrow\right):$ Suppose that $am\mid bm$, then again by + the definition of divisibility we have that there is some + $c\in\mathbb{Z}$ so that $bm=amc$. By the cancellation law, we can + cancel the $m$ to get $b=ac$ and the result follows.* + +7. *For all $a\in\mathbb{Z}$ with $a\neq 0$ we have $a\mid 0$:* + + *Let $a\in\mathbb{Z}$, where $a\neq 0$. We have that $0=ka$ has the + solution $k=0$ by part I proposition + [69](#prop:IntegersHaveNoZeroDivisors){reference-type="ref" + reference="prop:IntegersHaveNoZeroDivisors"}. Hence $a\mid 0$.* + +*As required. $\qed$* +::: + +Part 3. of the previous proposition can be generalised. We will work +with an example to see how this can be achieved. + +::: example +**Example 91**. *Let $a=2$, $b=16$ and $c=32$. Clearly we have that +$a\mid b$ as $16=4*2$ and likewise $a\mid c$ as $32=5*2$.* + +*Now part 3. states that if $a\mid b$ and $a\mid c$ then we must have +that $a\mid\left(bx+cy\right)$ for any $x,y\in\mathbb{Z}$.* + +*Indeed, for example, we can see that +$2\mid\left(-5\left(16\right)+7\left(32\right)\right)$. As +$-5\left(16\right)+7\left(32\right)=-80+224=144$. Now suppose that +$d=64$ and say $z=5$. We can see that* + +*$$\begin{equation*} + -5\left(16\right)+7\left(32\right)+5\left(64\right)=144+320=464 \Rightarrow 2\mid\left(-5\left(16\right)+7\left(32\right)+5\left(64\right)\right) +\end{equation*}$$* +::: + +We prove the general statement now. + +::: {#prop:NT_Divisor_dividing_all_in_set_divides_linear_combination .proposition} +**Proposition 103**. *Divisor that divides a set of integers divides a +combination of the set* + +*Let $a\in\mathbb{Z}$ and let $S=\left\{b_1,b_2,b_3,\dots,b_n\right\}$ +be a set of $n$ integers where $b_i\in\mathbb{Z}$ for each $b_i$. +Moreover suppose that $a\mid b_i$ for each $b_i\in S$. We have that* + +*$$\begin{equation*} + a\mid\sum_{i=1}^n b_i x_i +\end{equation*}$$* + +*for any $x_i\in\mathbb{Z}$.* + +*Proof:* + +*We argue by induction on $n$. The base case is $n=2$ which is shown in +proposition [102](#prop:NT_divisibility_properties){reference-type="ref" +reference="prop:NT_divisibility_properties"}. So suppose that the result +holds for some $k\geq 1$, which is to say that if +$S=\left\{b_1,b_2,\dots,b_k\right\}$ and we have that $a\mid b_i$ for +each $b_i\in S$ then* + +*$$\begin{equation*} + a\mid\sum_{i=1}^k b_i x_i +\end{equation*}$$* + +*We need to show that the result holds for $k+1$. That is if +$\Tilde{S}=S\cup \left\{b_{k+1}\right\}$ so that $a\mid b_i$ for each +$b_i\in\Tilde{S}$ then* + +*$$\begin{equation*} + a\mid\sum_{i=1}^{k+1} b_i x_i +\end{equation*}$$* + +*So take $\Tilde{S}=S\cup \left\{b_{k+1}\right\}$ so that $a\mid b_i$ +for each $b_i\in\Tilde{S}$. By applying part 1. of proposition +[102](#prop:NT_divisibility_properties){reference-type="ref" +reference="prop:NT_divisibility_properties"} to each $a\mid b_i$ we know +that for all $x_i\in\mathbb{Z}$ that $a\mid b_ix_i$.* + +*Now, by the induction hypothesis we know that $\forall b_i\in S$ that +$a\mid b_i$ and moreover we have that* + +*$$\begin{equation*} + a\mid\sum_{i=1}^k b_i x_i +\end{equation*}$$* + +*Let $\displaystyle d=\sum_{i=1}^k b_i x_i$. Again by part 1 of +proposition [102](#prop:NT_divisibility_properties){reference-type="ref" +reference="prop:NT_divisibility_properties"} we have that $a\mid ad$. +Additional we know that $a\mid b_{k+1}$ and so by part 3. of +[102](#prop:NT_divisibility_properties){reference-type="ref" +reference="prop:NT_divisibility_properties"}, As $d\in\mathbb{Z}$, we +have that* + +*$$\begin{align*} + a &\mid\left(1*d + b_{k+1}x_{k+1}\right)\\ + a &\mid\left(\sum_{i=1}^k b_i x_i + b_{k+1}x_{k+1}\right)\\ + a &\mid\left(\sum_{i=1}^{k+1} b_i x_i\right)\\ +\end{align*}$$* + +*Which implies the result holds for $k+1$ and hence for any +$n\in\mathbb{N}$ by induction. $\qed$* +::: + +#### The greatest common divisor and the least common multiple + +Now that we have a solid grasp of the basics of integer divisibility, we +can start looking towards some applications. One immediate question is +given a set of integers say + +$$\begin{equation*} + S=\left\{a_1,a_2,a_3,\dots,a_n\right\} +\end{equation*}$$ + +What is the largest integer which divides each $a_i\in S$. and what is +the largest integer $m$ so that $m$ has each $a_i\in S$ as a proper +divisor? An immediate use of these two ideas is very useful when doing +arithmetic with rational numbers. For example, consider trying to +simplify the fraction $\displaystyle\frac{525}{2925}$. To simplify this +we need to find the integers that multiply to make $525$ and those that +multiply to make $2925$. If there are any in common then we know from +the construction of the rationals that $\displaystyle \frac{x}{x}=1$ and +in particular we have that +$\displaystyle\frac{xy}{xz}=\frac{y}{z}*\frac{x}{x}=1$. + +Likewise suppose we wanted to add $\displaystyle\frac{1}{4}$ and +$\displaystyle\frac{1}{7}$. It is true that by definition of addition, +we would have + +$$\begin{equation*} + \frac{1}{4}+\frac{1}{7}=\frac{1*7+1*4}{7*4}=\frac{7+4}{7*4}=\frac{11}{28} +\end{equation*}$$ + +The key stage was $\displaystyle\frac{1*7+1*4}{7*4}$, breaking this down +we see that + +$$\begin{equation*} + \frac{1*7+1*4}{7*4}=\frac{1*7}{7*4}+\frac{1*4}{7*4} +\end{equation*}$$ + +In other words, we are finding a multiple in common with $7$ and $4$ to +turn the denominator into. It is therefore worthwhile to work out the +theory of working out common divisors and common multiples. + +We will start by working out common divisors, by first making a +definition. + +::: definition +**Definition 146**. *Common divisor* + +*Let $a,b,c\in\mathbb{Z}$ be non-zero integers. We say that $c$ is a +common divisor of $a$ and $b$ if $c\mid a$ and $c\mid b$.* +::: + +::: example +**Example 92**. *Consider the integers $35$ and $25$. The divisors of +$35$ are $1$, $5$ and $7$ and $35$, likewise the divisors of $25$ are +$1$ and $5$ and $25$. The largest common divisor is therefore 5.* +::: + +::: example +**Example 93**. *Consider the integers $24$ and $54$. Doing the same as +before, we can see that the divisors of $24$ are $1$, $2$, $3$, $4$, +$6$, $8$, $12$ and $24$. Looking at the divisors of $54$ we see that +they are $1$, $2$, $3$, $6$, $9$, $18$, $27$ and $54$.* + +*The common divisors of $24$ and $54$ are therefore $1$, $2$, $3$ and +$6$,* +::: + +::: example +**Example 94**. *Consider the common divisors of $3$ and $5$. The +divisors of $3$ are simply $1$ and $3$, likewise the divisors of $5$ are +$1$ and $5$. The only common divisor is $1$.* +::: + +We can see from the previous examples that there was a largest, or +greatest common divisor between the pairs of integers in each case. We +can show that for any two integers, there is always a greatest common +divisor. + +::: {#thm:NT_gcd_exists .theorem} +**Theorem 32**. *The greatest common divisor of two integers exists* + +*Let $a,b\in\mathbb{Z}$ so that $a\neq 0$ or $b\neq 0$. Then there +exists $d\in\mathbb{Z}$ so that $d$ is the largest possible common +divisor, that is there is no $g\in\mathbb{Z}$ with $g>d$ so that +$g\mid a$ and $g\mid b$.* + +*Proof:* + +*Firstly, we note that as $1\mid a$ and $1\mid b$, the largest possible +common divisor is at least 1, proving existence. To show that there is +the largest possible common divisor we must show that this divisor can't +exceed some integer, say $M$, where $M$ depends on $a$ and $b$. Moreover +by proposition +[101](#prop:NT_PositiveAndNegativeDivisorsForIntsExist){reference-type="ref" +reference="prop:NT_PositiveAndNegativeDivisorsForIntsExist"} we only +need to consider the case where $a\geq 0$ and $b\geq 0$.* + +*So. suppose that $c\mid a$ and $c\mid b$ for some $c\geq 1$. By part 5. +of proposition +[102](#prop:NT_divisibility_properties){reference-type="ref" +reference="prop:NT_divisibility_properties"} we have that as $c\mid a$ +then $c\leq a$, likewise as $c\mid b$ then $c\leq b$. There are three +possibilities to consider* + +1. *$a=b$* + +2. *Without any loss of generality we have $a + +1. *$a=b$:* + + *In this case we easily take $M$ to be the largest divisor of $a$, + or equivalently $b$, then $c\leq M$* + +2. *Without any loss of generality we have $aa$ a contradiction to the fact that $c\leq a$ as $c\mid a$.* + +3. *One of $a=0$ or $b=0$ but not both at the same time:* + + *Suppose that $a=0$ and $b\neq 0$, then we have that for all + $M\in\mathbb{Z}$ that $M\mid a$, but as $c\mid b$ then $c\leq b$ and + so we take $M=b$ as $b\mid b$. Likewise if we assume $b=0$ and + $a\neq 0$.* + +*In each case we found a $M$ so that if we take $c\leq M$ then $c\mid a$ +and $c\mid b$.* +::: + +We have shown that the for any two integers a greatest common divisor +always exists. We can make a formal definition. + +::: definition +**Definition 147**. *Greatest common divisor* + +*Let $a,b\in\mathbb{Z}$ so that $a\neq 0$ and $b\neq 0$. Let +$d\in\mathbb{Z}$ be such that $d\mid a$ and $d\mid b$. We say the +largest value of $d$ where $d\mid a$ and $d\mid b$ is the greatest +common divisor of $a$ and $b$, denoted +$d=\mathop{\mathrm{GCD}}\left(a,b\right)$, sometimes written +$\gcd\left(a,b\right)$ and in some texts simply by $\left(a,b\right)$.* + +*As $a\mid 0$ for any integer $a$. We define +$\mathop{\mathrm{GCD}}\left(a,0\right)=a$, similarly +$\mathop{\mathrm{GCD}}\left(0,b\right)=b$.* +::: + +We will use the notation $\mathop{\mathrm{GCD}}$ in this text and we +will usually abbreviate saying the greatest common divisor to +$\mathop{\mathrm{GCD}}$. Although we have proved that the greatest +common divisor exists, we do not yet actually have a method of +calculating what it is other than trying through trial and error. To see +how we can attempt to construct a method of finding +$\mathop{\mathrm{GCD}}$ we should look to cases where integer division +does not fail and to cases where it does fail. + +::: example +**Example 95**. *It is clear that $2\nmid 3$ as there is no integer $x$ +so that $3=2x$. If we take $x=1$ we get the false equality of $3=2$, if +we take $x=2$ we get another false equality of $3=4$. We observe however +that $3=2*1+1$.* +::: + +::: example +**Example 96**. *Let $a=25$ and $b=7$. It is clear that $b\nmid a$. The +first couple multiples of $7$ are $7=7*1$, $14=7*2$, $21=7*3$, $28=7*4$ +and so on. However, we can see that $25=7*3+4$.* +::: + +::: example +**Example 97**. *Let $a=36$ and $b=12$. Clearly that $b\mid a$ as +$36=12*3$. The first couple multiples of $7$ are $7=7*1$, $14=7*2$, +$21=7*3$, $28=7*4$ and so on.* +::: + +::: example +**Example 98**. *This time, let $a=8$ and $b=2$. Then we have that +$2\mid 8$ as $8=2*4$. In a similar way to the previous examples we see +that $8=2*4+0$* +::: + +If we let $a,b\in\mathbb{Z}$ so that $b\nmid a$ then, in the previous +examples it seems that we can always find a multiple of $b$ so that +$bx\leq a$ for some $x\in\mathbb{Z}$ and in particular we have that + +$$\begin{equation*} + a=bx+\left(a-bx\right) +\end{equation*}$$ + +In the case that $b\mid a$ then $a-bx=0$. Interpreting what $a-bx$ +means, when $b\nmid a$ then $a-bx\neq 0$ and when $b\mid a$ we had that +$a-bx=0$. Hence $a-bx\neq 0$ is a measure of how far off we are from +having $b\mid a$. This is to say that if $a-bx>0$ then we are a little +short of making a multiple of $a$ from $b$ and if $a-bx<0$ we are a +little over of making a multiple of $a$ from $b$. + +In general, we can see that any integer division can be viewed in this +way, that is if $a,b\in\mathbb{Z}$ we can see the result of $a$ divided +by $b$ in the form $a=qb+r$ for some $q,r\in\mathbb{Z}$. + +::: {#thm:NT_divAlg .theorem} +**Theorem 33**. *The division algorithm* + +*Let $a,b\in\mathbb{Z}$ so that $b> 0$, then there exist +$q,r\in\mathbb{Z}$ with $q,r$ being unique so that* + +*$$\begin{equation*} + a=bq+r +\end{equation*}$$* + +*where $0\leq r < b$* + +*Proof:* + +*There are three cases to consider* + +1. *$a=b$* + +2. *$ab$* + + + +1. *$a=b$:* + + *If $a=b$ then $b\mid a$ holds trivially and we see that $a=1*b+0$ + where $q=1$ and $r=0$.* + +2. *$ab$:* + + *This case is the meat of the theorem. To prove the division theorem + we will argue by induction on $a$. The base case is $a=1$ where we + either have $a=b$ or $a1$. Likewise in the base + case, we only need to consider the case of $k+1>b$, or equivalently + $b0$ then $q_2-q_1=0$ +giving $q_2=q_1$. $\qed$* +::: + +Based on this theorem we make a definition. + +::: definition +**Definition 148**. *Quotient and remainder* + +*Let $a,b\in\mathbb{Z}$ so that $b>0$. We have by the division algorithm +that* + +*$$\begin{equation*} + a=qb+r +\end{equation*}$$* + +*where $q,r\in\mathbb{Z}$ and $0\leq r < b$. We say that $q$ is the +quotient of the division and that $r$ is the remainder.* +::: + +In the theorem, we assumed that $b>0$. However by proposition +[101](#prop:NT_PositiveAndNegativeDivisorsForIntsExist){reference-type="ref" +reference="prop:NT_PositiveAndNegativeDivisorsForIntsExist"} we know +that negative divisors are also valid. To resolve this we reformulate +theorem [33](#thm:NT_divAlg){reference-type="ref" +reference="thm:NT_divAlg"} so that $0\leq r <\left|b\right|$. + +::: {#thm:NT_divAlg_ext .theorem} +**Theorem 34**. *The division algorithm (Extended)* + +*Let $a,b\in\mathbb{Z}$ so that $b\neq 0$, then there exist +$q,r\in\mathbb{Z}$ with $q,r$ being unique so that* + +*$$\begin{equation*} + a=bq+r +\end{equation*}$$* + +*where $0\leq r < \left|b\right|$* + +*Proof:* + +*By the division algorithm, theorem +[33](#thm:NT_divAlg){reference-type="ref" reference="thm:NT_divAlg"} we +have for $\left|a\right|$ and $\left|b\right|$ that there exist unique +$q,r\in\mathbb{Z}$ so that* + +*$$\begin{equation*} + \left|a\right|=q\left|b\right|+r +\end{equation*}$$* + +*where $0\leq r<\left|b\right|$. There are a few cases to consider.* + +1. *$r=0$* + +2. *$r>0$ and $a\geq 0$* + +3. *$r>0$ and $a<0$* + + + +1. *$r=0$:* + + *If $r=0$, then $\left|a\right|=q\left|b\right|$ and so by the + properties of the absolute value we have that $a=\pm qb$, hence + $a=b\left(\pm q\right)$ and we have the result.* + +2. *$r>0$ and $a\geq 0$:* + + *Now suppose $r>0$ and $a\geq 0$. We hence have that + $a=q\left|b\right|+r$ which gives* + + *$$\begin{align*} + a&=bq+r,\ \text{If } b>0\\ + a&=\left(-b\right)q+r,\ \text{If } b<0\\ + \end{align*}$$* + + *The first is simply the first version of the division algorithm and + the second can be written as $a=b\left(-q\right)+r$ which gives the + result.* + +3. *$r>0$ and $a<0$:* + + *Finally if $r>0$ and $a<0$ then we have* + + *$$\begin{equation*} + -a=\left|b\right|q+r \Rightarrow a=-\left|b\right|q-r + \end{equation*}$$* + + *This is a problem as it would give a negative remainder. We can + employ a trick that doesn't change the value of $a$ but allows us to + express $a=-\left|b\right|q-r$ in a more suitable form.* + + *$$\begin{align*} + a&=-\left|b\right|q-r\\ + a&=-\left|b\right|q+\left(\left|b\right|-\left|b\right|\right)-r\\ + a&=-\left|b\right|q+\left|b\right|+\left(\left|b\right|r\right)\\ + a&=\left|b\right|\left(-1-q\right)+\left(\left|b\right|r\right)\\ + \end{align*}$$* + + *By assumption we have that $00$ and for $b<0$ + we write $q'=1+q$.* + +*This completes the proof. $\qed$* +::: + +We can now go back to a problem from the first section, namely showing +that $2n+1$ must be odd + +::: {#prop:NT_Odd_iff_2n+1 .proposition} +**Proposition 104**. *Integer is odd if and only if it is a multiple of +$2n+1$* + +*Let $x\in\mathbb{Z}$. We have that $x$ is odd if and only if it is a +multiple of $2n+1$ where $x=2n+1$ for $n\in\mathbb{Z}$. Then $n$ is +odd.* + +*Proof:* + +*Suppose $x\in\mathbb{Z}$, then by the division algorithm we have that* + +*$$\begin{equation*} + x=2q+r +\end{equation*}$$* + +*where $0\leq r< \left|2\right|$. Hence the only remainders possible are +$r=0$ or $r=1$. Hence either $x=2q$ or $x=2q+1$. In the first case we +have $x=2q$ is even by definition. In the case $x=2q+1$ we have that +$2\nmid 2n+1$ and so $x$ can't be even by definition. It follows that +$x$ is odd. $\qed$* +::: + +With this proposition and proposition +[99](#prop:NT_even_iff_2n){reference-type="ref" +reference="prop:NT_even_iff_2n"} we can derive the evenness or oddness +when adding or multiplying even or odd integers. + +::: proposition +**Proposition 105**. *Even and oddness for addition and multiplication* + +*Let $x,y\in\mathbb{Z}$. We have that* + +1. *If $x$ is even and $y$ is even then $x+y$ is even and $xy$ is + even.* + +2. *If $x$ is even and $y$ is odd then $x+y$ is odd and $xy$ is even.* + +3. *If $x$ is odd and $y$ is even then $x+y$ is odd and $xy$ is even.* + +4. *If $x$ is odd and $y$ is odd then $x+y$ is even and $xy$ is odd.* + +*Proof:* + +1. *If $x$ is even and $y$ is even then $x+y$ is even and $xy$ is + even:* + + *Suppose that $x$ and $y$ are even, then by proposition + [99](#prop:NT_even_iff_2n){reference-type="ref" + reference="prop:NT_even_iff_2n"} we have $x=2n$ for some + $n\in\mathbb{Z}$ and $y=2m$ for some $m\in\mathbb{Z}$. We have that + $x+y=2n+2m=2\left(n+m\right)$ hence $x+y$ is even by proposition + [99](#prop:NT_even_iff_2n){reference-type="ref" + reference="prop:NT_even_iff_2n"}. Likewise, we have that + $xy=2n*2m=2\left(n*m\right)$ and therefore even.* + +2. *If $x$ is even and $y$ is odd then $x+y$ is odd and $xy$ is odd:* + + *Suppose that $x$ is even and $y$ is odd. By we have that $x=2n$ for + some $n\in\mathbb{Z}$ by + [99](#prop:NT_even_iff_2n){reference-type="ref" + reference="prop:NT_even_iff_2n"} and by proposition + [104](#prop:NT_Odd_iff_2n+1){reference-type="ref" + reference="prop:NT_Odd_iff_2n+1"} we have that $y=2m+1$ for some + $m\in\mathbb{Z}$.* + + *We have $x+y=2n+2m+1-2\left(n+m\right)+1$ and so $x+y$ is odd by + proposition [104](#prop:NT_Odd_iff_2n+1){reference-type="ref" + reference="prop:NT_Odd_iff_2n+1"}. Additionally, + $xy=2n\left(2m+1\right)=2\left(2mn+n\right)$ and so by proposition + [99](#prop:NT_even_iff_2n){reference-type="ref" + reference="prop:NT_even_iff_2n"} we have that $xy$ is even.* + +3. *If $x$ is odd and $y$ is even then $x+y$ is odd and $xy$ is even:* + + *Similar to above, swapping the roles of $x$ and $y$.* + +4. *If $x$ is odd and $y$ is odd then $x+y$ is even and $xy$ is odd:* + + *By proposition [104](#prop:NT_Odd_iff_2n+1){reference-type="ref" + reference="prop:NT_Odd_iff_2n+1"} we have that $x=2n+1$ for some + $n\in\mathbb{Z}$ and $y=2m+1$ for some $m\in\mathbb{Z}$.* + + *Now, + $x+y=\left(2n+1\right)+\left(2m+1\right)=2\left(n+m\right)+2=2\left(\left(n+m\right)+1\right)$. + So by proposition [99](#prop:NT_even_iff_2n){reference-type="ref" + reference="prop:NT_even_iff_2n"} we have $x+y$ is even.* + + *Finally, + $xy=\left(2n+1\right)\left(2m+1\right)=4nm+2n+2m+1=2\left(2nm+\left(n+m\right)\right)+1$ + and so by proposition + [104](#prop:NT_Odd_iff_2n+1){reference-type="ref" + reference="prop:NT_Odd_iff_2n+1"} is odd.* + +*As required. $\qed$* +::: + +Continuing with our quest to find a method to compute the greatest +common divisor. At first, it might seem that we haven't made much +progress in finding a way to calculate the $\mathop{\mathrm{GCD}}$. +However, consider the following examples. + +::: example +**Example 99**. *Consider $a=56$ and $b=24$. By the division algorithm, +we have that $56=2*24+8$. Now what about $a=24$ and $b=8$? Again, by the +division algorithm, we have that $24=3*8+0$.* + +*Now, the divisors of $56$ are $1$, $2$, $4$, $7$, $8$, $14$, $28$ and +$56$, the divisors of $24$ are $1$, $2$, $3$, $4$, $6$, $8$, $12$ and +$24$. The largest common divisor was $8$, which was the remainder after +the first use of the division algorithm. Likewise, it was the quotient +in the second application of the division algorithm.* +::: + +::: example +**Example 100**. *Consider $a=4947$ and $b=1552$. By the division +algorithm, we have that $4974=3*1552+291$. Applying the division +algorithm to $a=1552$ and $b=291$ gives $1552=5*291+97$. A third +application of the division algorithm to $a=291$ and $b=97$ gives +$291=3*97+0$.* + +*Unlike with the previous example, there may be potentially too many +divisors for $4947$ to list them out by trying each integer +$00$. Moreover, we clearly have $0\in S$ as we can take $x=0$ and +$y=0$.* + +*Now consider the set $\Tilde{S}$ given by* + +*$$\begin{equation*} + \Tilde{S}=\left\{s\in S: s>0\right\} +\end{equation*}$$* + +*We have by definition of $\Tilde{S}$ that $\forall s \in \Tilde{S}$ +that $s>0$ and so $\Tilde{S}\subset\mathbb{N}$. Hence by the +well-ordering principle, theorem [18](#thm:WOP){reference-type="ref" +reference="thm:WOP"}, there is a smallest element, say $\Bar{s}$. By +definition of being an element of $\Tilde{S}$ we have that +$\Bar{s}=ax_0+by_0$ for some $x_0,y_0\in\mathbb{Z}$, where $x_0,y_0$ +each have a fixed value.* + +*We show that $\Bar{s}\mid a$ and $\Bar{s}\mid b$. Suppose instead that +$\Bar{s}\nmid a$, then by the division algorithm we have that +$a=q\Bar{s}+r$ where $00$, then + $\mathop{\mathrm{GCD}}\left(am,bm\right)=m*\mathop{\mathrm{GCD}}\left(a,b\right)$* + +6. *If $d\mid a$ and $d\mid b$ where $d\in\mathbb{Z}$ and $d>0$ then + $\displaystyle\mathop{\mathrm{GCD}}\left(\frac{a}{d},\frac{b}{d}\right)=\frac{1}{d}\mathop{\mathrm{GCD}}\left(a,b\right)$* + +7. *If $\mathop{\mathrm{GCD}}\left(a,b\right)=d$ then + $\displaystyle\mathop{\mathrm{GCD}}\left(\frac{a}{d},\frac{b}{d}\right)=1$* + +*Proof:* + +1. *$\mathop{\mathrm{GCD}}\left(a,a\right)=a$:* + + *Clearly, we have that $a\mid a$. Now by proposition + [102](#prop:NT_divisibility_properties){reference-type="ref" + reference="prop:NT_divisibility_properties"} part 5. We have that if + $a\mid a$ with $a>0$ then $a\leq a$. Hence $a$ is the largest such + divisor so $\mathop{\mathrm{GCD}}\left(a,a\right)=a$* + +2. *$\mathop{\mathrm{GCD}}\left(a,b\right)=\mathop{\mathrm{GCD}}\left(b,a\right)$:* + + *This is trivial. If $d=\mathop{\mathrm{GCD}}\left(a,b\right)$ then + $d$ is the largest common divisor of $a$ and $b$.* + +3. *Let $D$ be the set of all common divisors of $a$ and $b$. then + $\forall d\in D$ we have that + $d\mid\mathop{\mathrm{GCD}}\left(a,b\right)$:* + + *Let $D$ be defined as above, then* + + *$$\begin{equation*} + D=\left\{x\in\mathbb{Z}: x>0\text{ and } x\mid a \text{ and } x\mid b\right\} + \end{equation*}$$* + + *Then by definition of $D$ we have that $\forall d\in D$ that $d$ is + a common divisor of $a$ and $d$ is a common divisor of $b$. Clearly + then $d\mid\mathop{\mathrm{GCD}}\left(a,b\right)$ as + $\mathop{\mathrm{GCD}}\left(a,b\right)$ is the largest such common + divisor of $a$ and $b$ and therefore + $\mathop{\mathrm{GCD}}\left(a,b\right)\in D$.* + +4. *We have that $\mathop{\mathrm{GCD}}\left(a,b\right)$ is the + smallest such $ax+by$ where $x,y\in\mathbb{Z}$ so that + $\mathop{\mathrm{GCD}}\left(a,b\right)=ax+by$:* + + *This follows from the proof of theorem $\ref{thm:NT_bezout_id}$. + For it it were not we would have a contradiction.* + +5. *Let $m\in\mathbb{Z}$ with $m>0$, then + $\mathop{\mathrm{GCD}}\left(a,b\right)=m\mathop{\mathrm{GCD}}\left(a,b\right)$:* + + *By the previous part we have that + $\mathop{\mathrm{GCD}}\left(a,b\right)$ is the smallest such element + of the set* + + *$$\begin{equation*} + S=\left\{ax+by:x,y\in\mathbb{Z}\right\} + \end{equation*}$$* + + *Let $s\in S$ denote the smallest such $ax+by$, that is $s=ax+by$ + and $s=\mathop{\mathrm{GCD}}\left(a,b\right)$.* + + *As $s=\mathop{\mathrm{GCD}}\left(a,b\right)$ then $s\mid a$ and + $s\mid b$. As $s\mid a$ then $a=ks$ for some $k\in\mathbb{Z}$ and so + $am=k\left(ms\right)$ which is to say $ms\mid am$. Likewise as + $s\mid b$ then $b=ls$ for some $l\in\mathbb{Z}$ and hence + $bm=l\left(ms\right)$ giving $ms\mid bm$.* + + *Now as $s=ax+by$ then we have that + $ms=m\left(ax+by\right)=a\left(mx\right)+b\left(my\right)$. + Moreover, as $s\in S$ is the smallest such $ax+by$ then + $m\left(ax+by\right)$ will be the smallest such element of the set* + + *$$\begin{equation*} + \Tilde{S}=\left\{amx+bmy:x,y\in\mathbb{Z}\right\} + \end{equation*}$$* + + *Hence we have that + $amx+bmy=\mathop{\mathrm{GCD}}\left(am,bm\right)=ms=m*\mathop{\mathrm{GCD}}\left(a,b\right)$.* + +6. *If $d\mid a$ and $d\mid b$ where $d\in\mathbb{Z}$ and $d>0$ then + $\displaystyle\mathop{\mathrm{GCD}}\left(\frac{a}{d},\frac{b}{d}\right)=\frac{1}{d}\mathop{\mathrm{GCD}}\left(a,b\right)$* + + *Let $a,b,d\in\mathbb{Z}$ so that $d\mid a$ and $d\mid b$. As + $d\mid a$ then we have that $\displaystyle\frac{a}{d}\in\mathbb{Z}$, + likewise as $d\mid b$ then $\displaystyle\frac{b}{d}\in\mathbb{Z}$. + The result now follows by applying the previous part.* + +7. *If $\mathop{\mathrm{GCD}}\left(a,b\right)=d$ then + $\displaystyle\mathop{\mathrm{GCD}}\left(\frac{a}{d},\frac{b}{d}\right)=1$:* + + *This follows by the previous part.* + +*Concluding the proof. $\qed$* +::: + +We have talked a lot about the greatest common divisor but nothing about +the least common multiple. As with common divisors, we start by making a +definition of a common multiple. + +::: definition +**Definition 149**. *Common multiple* + +*Let $a,b,c\in\mathbb{Z}$ so that $a\mid m$ and $b\mid m$. We say that +$m$ is a common multiple of $a$ and $b$.* +::: + +::: example +**Example 104**. *Let $a=2$, $b=4$ and $c=8$. We have that $2\mid 8$ and +$4\mid 8$ and so $8$ is a common multiple of $2$ and $4$. In fact, $4$ +is a common multiple of $2$ and $4$.* +::: + +::: example +**Example 105**. *Let $a=4$ and $b=14$. Listing multiples of $2$ we have +$4$, $8$, $12$, $16$, $20$, $24$, $28$, $32$ and so on. Doing a similar +procedure for $14$ we see we have $14$, $28$, $42$ and so on. We see +that $28$ is a common multiple of $4$ and $14$.* +::: + +::: example +**Example 106**. *Consider $a=24$ and $b=54$. Listing the first ten +multiples of $a$ and $b$ we have* + +*$$\begin{align*} + &24,\ 48,\ 72,\ 96,\ 120,\ 144,\ 168,\ 192,\ 216,\ 240,\ \dots\\ + &54,\ 108,\ 162,\ 216,\ 270,\ 324,\ 378,\ 432,\ 486,\ 540,\ \dots\\ +\end{align*}$$* + +*The first common multiple is $216$. Interestingly, we saw that +$\mathop{\mathrm{GCD}}\left(a,b\right)$ was $6$. We have that +$216*6=1296$ and $24*54$=1296.* +::: + +::: example +**Example 107**. *We observe for any integer $a$ that $a\mid 0$ as +$0=am$ for some $m\in\mathbb{z}$ and by proposition +[69](#prop:IntegersHaveNoZeroDivisors){reference-type="ref" +reference="prop:IntegersHaveNoZeroDivisors"} we must have either $a=0$ +or $m=0$. Hence $0$ can be argued to be a common multiple of any +integers $a$ and $b$. This result is not particularly useful.* +::: + +These examples indicate that a common multiple always exists. In fact, +there is always a smallest common multiple + +::: {#thm:NT_lcm_exists .theorem} +**Theorem 37**. *The least common multiple of two integers exists* + +*Let $a,b\in\mathbb{Z}$ where $a>0$ and $b>0$. We have that +$\exists m\in\mathbb{Z}$ with $m>0$ so that $m$ is the smallest common +multiple of $a$ and $b$. That is $m$ is the smallest such integer so +that $a\mid m$ and $b\mid m$.* + +*Proof:* + +*We first prove that a non-trivial common multiple of $a$ and $b$ +exists. That is some $m\neq 0$ as $0$ can be viewed as a common divisor +of any two integers $a,b$. Clearly $ab$ is a common multiple of $a$ and +$b$ as $a\mid ab$ and $b\mid ab$. Hence a non-trivial common multiple +exists.* + +*It is left to show that there is a minimal common multiple. Let $S$ be +the set of all positive common multiples of $a$ and $b$. By the +well-ordering principle, $S$ has a smallest element as +$S\subset\mathbb{N}$. The result follows. $\qed$* +::: + +We can now make a formal definition. However, first, we can note that +the restriction of $a>0$ and $b>0$ is not needed. + +::: corollary +**Corollary 6**. *Let $a,b\in\mathbb{Z}$, where $a\neq 0$ and $b\neq 0$. +We have that $\exists m\in\mathbb{Z}$ with $m>0$ so that $m$ is the +smallest common multiple of $a$ and $b$. This is, $m$ is the smallest +such integer so that $a\mid m$ and $b\mid m$.* + +*Proof:* + +*The proof is similar to theorem +[37](#thm:NT_lcm_exists){reference-type="ref" +reference="thm:NT_lcm_exists"}. We have that $ab$ is a common multiple +of $a$ and $b$ as is $-ab$. Hence we have that one of $ab>0$ or $-ab>0$. +Let $S$ be the set of all positive common multiples of $a$ and $b$. Then +the well-ordering principle gives us that $S$ has the smallest such +element. $\qed$.* +::: + +::: definition +**Definition 150**. *Least common multiple* + +*Let $a,b\in\mathbb{Z}$ so that $a\neq 0$ and $b\neq 0$. We say that the +smallest positive value $m$ so that $a\mid m$ and $b\mid m$ is the least +common multiple of $a$ and $b$, denoted +$m=\mathop{\mathrm{LCM}}\left(a,b\right)$, sometimes written +$\mathop{\mathrm{lcm}}\left(a,b\right)$.* +::: + +It is important to note why we say that the least common multiple is +positive. If we allowed a negative least common multiple, say $-m$, then +for all $n\in\mathbb{Z}$ with $n>0$ we have that $-nm$ is a smaller +common multiple than $-m$ and so we could always find a smaller such +multiple. + +As with the greatest common divisor, we need a way to compute the least +common multiple. We should look again at the example where $a=24$ and +$b=54$. We saw that the first, smallest, common multiple was $216$, and +that the greatest common divisor was $6$. We also noted that the product +$ab=1296$ which is also the product $216*6$. We should look to more +examples to see if this holds in other cases. + +::: example +**Example 108**. *Let $a=14$ and $b=21$. Using the method of writing +multiples out we have* + +*$$\begin{align*} + &14,\ 28,\ 42,\ 56,\ \dots\\ + &21,\ 42,\ 63,\ 84,\ \dots\\ +\end{align*}$$* + +*So the smallest positive common multiple is $42$. Now, +$\mathop{\mathrm{GCD}}\left(14,21\right)=7$. Finally, $14*21=294$ and +$7*42=294$.* + +*Hence we have that +$\displaystyle \mathop{\mathrm{LCM}}\left(14,21\right)=\frac{14*21}{\mathop{\mathrm{GCD}}\left(14,21\right)}$.* + +*In general we might expect that +$\displaystyle \mathop{\mathrm{LCM}}\left(a,b\right)=\frac{a*b}{\mathop{\mathrm{GCD}}\left(a,b\right)}$* +::: + +::: example +**Example 109**. *Let $a=6$ and $b=36$. Using our expected result, we +have that +$\displaystyle \mathop{\mathrm{LCM}}\left(a,b\right)=\frac{a*b}{\mathop{\mathrm{GCD}}\left(a,b\right)}$. +So computing $\mathop{\mathrm{GCD}}\left(a,b\right)$ we see that +$\mathop{\mathrm{GCD}}\left(a,b\right)=6$ and so we suspect that +$\displaystyle\mathop{\mathrm{LCM}}\left(6,36\right)=\frac{6*36}{6}=36$. +Writing out the multiples of both $6$ and $36$* + +*$$\begin{align*} + &6,\ 12,\ 18,\ 24,\ 30,\ 36,\ 42,\ \dots\\ + &36,\ 72,\ 108,\ \dots\\ +\end{align*}$$* + +*So the smallest common multiple is indeed $36$.* +::: + +We have enough evidence to postulate and prove the following theorem. + +::: {#thm:NT_LCM_by_GCD_is_product .theorem} +**Theorem 38**. *Least common multiple by greatest common divisor equals +product* + +*Let $a,b\in\mathbb{Z}$ so that $a> 0$ and $b> 0$. We have that* + +*$$\begin{equation*} + \mathop{\mathrm{GCD}}\left(a,b\right)*\mathop{\mathrm{LCM}}\left(a,b\right)=ab +\end{equation*}$$* + +*Proof:* + +*Let $d=\mathop{\mathrm{GCD}}\left(a,b\right)$, then by definition we +have that $d\mid a$ so by proposition +[102](#prop:NT_divisibility_properties){reference-type="ref" +reference="prop:NT_divisibility_properties"} part 1. implies that +$d\mid ac$ for any $c\in\mathbb{Z}$ and in particular $d\mid ab$. Hence +by the definition of divisibility, there exists $n\in\mathbb{Z}$ so that +$ab=dn$.* + +*Now as $d\mid a$ then there is an integer $u$ so that $a=du$, likewise +as $d\mid b$ then there is an integer $v$ so that $b=dv$. Hence we have +that* + +*$$\begin{align*} + dn&=dub \Rightarrow n=ub,\ \text{By the cancellation law for the integers}\\ + dn&=adv \Rightarrow n=av,\ \text{By the cancellation law for the integers} +\end{align*}$$* + +*Hence as $n=ub$ we have that $b\mid n$ and likewise as $n=av$ we have +that $a\mid n$. Hence it follows that $n$ is a common multiple of $a$ +and $b$. We need to show that $n$ is the smallest such multiple so then +$\mathop{\mathrm{LCM}}\left(a,b\right)=n$.* + +*So, let $S$ denote the set of positive common multiples of $a$ and $b$ +and let $s\in S$ be a common multiple of $a$ and $b$. By definition of a +common multiple, we have that there exists some $k_1,k_2\in\mathbb{Z}$ +so that $s=ak_1$ and $s=bk_2$.* + +*Now, we have by Bézout's identity we have that +$\exists x,y\in\mathbb{Z}$ so that* + +*$$\begin{equation*} + \mathop{\mathrm{GCD}}\left(a,b\right)=d=ax+by +\end{equation*}$$* + +*Now, consider $sd$, we have that* + +*$$\begin{align*} + sd&=s\left(ax+by\right)\\ + &=sax+sby\\ + &=\left(bk_2\right)ax+\left(ak_1\right)by\\ + &=abk_2x+abk_1y\\ + &=ab\left(k_2x+k_1y\right)\\ + &=dn\left(k_2x+k_1y\right)\\ + s&=n\left(k_2x+k_1y\right),\ \text{By the cancellation law for the integers} +\end{align*}$$* + +*Now $\left(k_2x+k_1y\right)\in\mathbb{Z}$ and so we have that +$n\mid s$. Now by proposition +[102](#prop:NT_divisibility_properties){reference-type="ref" +reference="prop:NT_divisibility_properties"} part 5. we have that +$n\leq s$. As $s\in S$ was arbitrary we have that $n$ divides the +smallest element of $S$ by the well-ordering principle, i.e $n$ is the +smallest common divisor and so by definition +$\mathop{\mathrm{LCM}}\left(a,b\right)=n$.* + +*Hence we have that +$ab=dn=\mathop{\mathrm{GCD}}\left(a,b\right)\mathop{\mathrm{LCM}}\left(a,b\right)$. +As required. $\qed$.* +::: + +We can now justify the following corollary to compute the least common +multiple. + +::: {#cor:NT_lcm_formula .corollary} +**Corollary 7**. *Least common multiple is product divided by greatest +common divisor* + +*Let $a,b\in\mathbb{Z}$ so that $a>0$ and $b>0$. We have that* + +*$$\begin{equation*} + \mathop{\mathrm{LCM}}\left(a,b\right)=\frac{ab}{\mathop{\mathrm{GCD}}\left(a,b\right)} +\end{equation*}$$* + +*Proof:* + +*By theorem [38](#thm:NT_LCM_by_GCD_is_product){reference-type="ref" +reference="thm:NT_LCM_by_GCD_is_product"} we have that* + +*$$\begin{equation*} + \mathop{\mathrm{GCD}}\left(a,b\right)*\mathop{\mathrm{LCM}}\left(a,b\right)=ab +\end{equation*}$$* + +*Let $d=\mathop{\mathrm{GCD}}\left(a,b\right)$ then by definition we +have that $d\mid a$ and $d\mid b$ so that $d\mid ab$. Hence +$\displaystyle\frac{ab}{d}\in\mathbb{Z}$. Hence +$\mathop{\mathrm{LCM}}\left(a,b\right)\in\mathbb{Z}$. $\qed$* +::: + +We can now show some similar results to proposition +[108](#prop:NT_GCD_properties){reference-type="ref" +reference="prop:NT_GCD_properties"} + +::: {#prop:NT_LCM_properties .proposition} +**Proposition 109**. *Properties of the least common multiple* + +*Let $a,b\in\mathbb{Z}$ with $a>0$ $b> 0$. We have the following +properties of the $\mathop{\mathrm{LCM}}$ hold.* + +1. *$\mathop{\mathrm{LCM}}\left(a,a\right)=a$* + +2. *$\mathop{\mathrm{LCM}}\left(a,b\right)=\mathop{\mathrm{LCM}}\left(b,a\right)$* + +3. *Let $M$ be the set of all positive common multiples of $a$ and $b$. + then $\forall m\in M$ we have that + $\mathop{\mathrm{LCM}}\left(a,b\right)\mid m$* + +4. *We have that $\mathop{\mathrm{LCM}}\left(a,b\right)$ is the + greatest $\displaystyle \frac{ab}{ax+by}$ where + $\mathop{\mathrm{GCD}}\left(a,b\right)=ax+by$.* + +*Proof:* + +1. *$\mathop{\mathrm{LCM}}\left(a,a\right)=a$:* + + *As $\mathop{\mathrm{GCD}}\left(a,a\right)=a$ and $a*a=a^2$, we have + by corollary [7](#cor:NT_lcm_formula){reference-type="ref" + reference="cor:NT_lcm_formula"} that* + + *$$\begin{equation*} + \mathop{\mathrm{LCM}}\left(a,a\right)=\frac{a*a}{\mathop{\mathrm{GCD}}\left(a,a\right)}=\frac{a^2}{a}=a + \end{equation*}$$* + +2. *$\mathop{\mathrm{LCM}}\left(a,b\right)=\mathop{\mathrm{LCM}}\left(b,a\right)$:* + + *This follows as + $\mathop{\mathrm{GCD}}\left(a,b\right)=\mathop{\mathrm{GCD}}\left(b,a\right)$ + and integer multiplication is commutative, this is to say* + + *$$\begin{equation*} + \mathop{\mathrm{LCM}}\left(a,b\right)=\frac{a*b}{\mathop{\mathrm{GCD}}\left(a,b\right)}=\frac{b*a}{\mathop{\mathrm{GCD}}\left(b,a\right)}=\mathop{\mathrm{LCM}}\left(b,a\right) + \end{equation*}$$* + +3. *Let $M$ be the set of all positive common multiples of $a$ and $b$. + then $\forall m\in M$ we have that + $\mathop{\mathrm{LCM}}\left(a,b\right)\mid m$:* + + *Let $M$ be the set of all positive common multiples. By the + well-ordering principle, there is a smallest element $\Tilde{m}$. By + the definition of the least common multiple we have that + $\mathop{\mathrm{LCM}}\left(a,b\right)$ divides any other common + multiple, so $\mathop{\mathrm{LCM}}\left(a,b\right)\mid\Tilde{m}$. + For every $m\in M$, we have that $m\geq\Tilde{m}$ and so + $\mathop{\mathrm{LCM}}\left(a,b\right)\mid m$ for every $m\in M$.* + +4. *We have that $\mathop{\mathrm{LCM}}\left(a,b\right)$ is the + greatest $\displaystyle \frac{ab}{ax+by}$ where + $\mathop{\mathrm{GCD}}\left(a,b\right)=ax+by$:* + + *By proposition [108](#prop:NT_GCD_properties){reference-type="ref" + reference="prop:NT_GCD_properties"} part 4. we have that + $\mathop{\mathrm{GCD}}\left(a,b\right)=ax+by$ for some + $x,y\in\mathbb{Z}$ is the smallest such $ax+by$. Hence* + + *$$\begin{equation*} + \mathop{\mathrm{LCM}}\left(a,b\right)=\frac{ab}{\mathop{\mathrm{GCD}}\left(a,b\right)} + \end{equation*}$$* + + *Will be the greatest such fraction. For if not then there is either + $x_0,y_0\in\mathbb{Z}$ so that $ax_0+by_0ax+by$ then by part 35. of + proposition + [89](#prop:InequalityRationalNumbers){reference-type="ref" + reference="prop:InequalityRationalNumbers"} we have that* + + *$$\begin{equation*} + \frac{ab}{ax_1+by_1}<\frac{ab}{ax+by} + \end{equation*}$$* + +*Concluding the proof. $\qed$* +::: + +### Prime and co-prime numbers + +::: epigraph +God may not play dice with the universe, but something strange is going +on with the prime numbers. + +*Paul Erdos* +::: + +So far we have been building a theory of divisibility. This theory has +allowed us to define what it means to be an odd or an even integer. To +know when one integer divides another, and computing the largest divisor +of two integers. Where do we go from here? One question we could ask is +how many divisors does a given integer have? + +#### The divisor function + +We start with the following definition. + +::: definition +**Definition 151**. *The Divisor function* + +*Let $x\in\mathbb{Z}$. We define +$\sigma:\mathbb{Z}\rightarrow\mathbb{Z}$ by* + +*$$\begin{align*} + \sigma:\mathbb{Z}&\mathlarger{\mathlarger{\rightarrow}}\mathbb{Z}\\ + x&\mapsto \sigma\left(x\right)=\sum_{d\mid x} 1 +\end{align*}$$* + +*here we are summing over all of the divisors $d$ of $x$, where if +$d\mid x$ then we add one to the sum total.* +::: + +Rather than work with explicit examples we will provide a table of the +first 20 integers. + + $x$ 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 + ------------------------ --- --- --- --- --- --- --- --- --- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- + $\sigma\left(x\right)$ 1 2 2 3 2 4 2 4 3 4 2 6 2 4 4 5 2 6 2 6 + + : The divisor function for the integers $1\leq x\leq 20$ + +There are a few things to note from this table. Firstly the only integer +with a single divisor is $1$. Secondly, there are many examples of +integers having only $2$ divisors. These are $2$, $3$, $5$, $7$, $11$, +$13$, $17$ and $19$. As $1$ is a divisor of every integer we can +conclude the other divisors in the case of $\sigma\left(x\right)=2$ must +be $x$ itself. + +What about the case when $\sigma\left(x\right)>2$. Looking at $6$ we see +the divisors are $1$, $2$, $3$ and $6$ itself, and from the table +$\sigma\left(2\right)=\sigma\left(3\right)=2$. Moreover, we have that +$6=2*30$. + +Similarly with $12$ we have that the divisors are $1$, $2$, $3$, $4$, +$6$ and $12$. Again, we have that +$\sigma\left(2\right)=\sigma\left(3\right)=2$. Now, as $12=2*6$ and +$6=2*3$ then we have that $12=2*2*3$. In both cases, we have seen that a +number $x$ with $\sigma\left(x\right)>2$ can be written into a product +of integers with exactly $2$ divisors. We can ask does this hold in +general? To do so we need to make some definitions. + +#### Prime numbers + +With the remarks of the previous section, we give a special name to any +integer $x$ where $\sigma\left(x\right)=2$. + +::: definition +**Definition 152**. *Prime number* + +*Let $x\in\mathbb{Z}$ with $x\geq 2$. We say that $x$ is a prime number, +or simply that $x$ is prime, if and only if $\sigma\left(x\right)=2$. In +other words, we say that $x$ is prime, if and only if the only two +distinct positive divisors of $x$ are $1$ and itself. If $x$ is not +prime we say that $x$ is composite.* +::: + +We noted that there were many $x\in\mathbb{Z}$ with +$\sigma\left(x\right)=2$. A natural question that arises is are there +infinitely many such $x$, or are there only finitely so many? To answer +this we need to see how primes and divisibility interact. We first have +to make another definition based on the greatest common divisor of two +integers. We show some examples to motivate this new definition. + +::: example +**Example 110**. *Let $a=6$ and $b=35$. By the Euclidean algorithm, we +see that* + +*$$\begin{align*} + 35&=5\left(6\right)+5\\ + 6&=5+1\\ + 5&=5\left(1\right) +\end{align*}$$* + +*Hence $\mathop{\mathrm{GCD}}\left(a,b\right)=1$.* +::: + +::: example +**Example 111**. *Let $a=2$ and $b=3$. By the Euclidean algorithm, we +see that* + +*$$\begin{align*} + 3&=2+1\\ + 2&=2\left(1\right) +\end{align*}$$* + +*Hence $\mathop{\mathrm{GCD}}\left(a,b\right)=1$. We note that $a$ and +$b$ are prime.* +::: + +::: example +**Example 112**. *Let $a=4$ and $b=9$. By the Euclidean algorithm, we +see that* + +*$$\begin{align*} + 9&=2\left(4\right)+1\\ + 4&=4\left(1\right) +\end{align*}$$* + +*Hence $\mathop{\mathrm{GCD}}\left(a,b\right)=1$.* +::: + +We see that there are integers $a,b\in\mathbb{Z}$ so that +$\mathop{\mathrm{GCD}}\left(a,b\right)=1$. Meaning that they have no +common divisors other than $1$. This situation turns out to happen +enough in Number Theory to warrant a definition. + +::: definition +**Definition 153**. *Co-prime Integers* + +*Let $a,b\in\mathbb{Z}$. We say that $a$ is co-prime to $b$, or $a$ and +$b$ are co-prime, or $a$ and $b$ are relatively prime, if and only if +$\mathop{\mathrm{GCD}}\left(a,b\right)=1$.* +::: + +We have some immediate results. + +::: {#prop:NT_Bezout_coprime .proposition} +**Proposition 110**. *Bézout's Identity for co-prime integers* + +*Let $a,b\in\mathbb{Z}$ so that +$\mathop{\mathrm{GCD}}\left(a,b\right)=1$. We have that +$\exists x,y\in\mathbb{Z}$ so that* + +*$$\begin{equation*} + 1=ax+by +\end{equation*}$$* + +*Proof:* + +*This immediately follows from theorem +[36](#thm:NT_bezout_id){reference-type="ref" +reference="thm:NT_bezout_id"}. $\qed$* +::: + +::: proposition +**Proposition 111**. *Distinct prime numbers are co-prime* + +*Let $p,q\in\mathbb{Z}$ so that $p$ and $q$ are prime. We have that +$\mathop{\mathrm{GCD}}\left(p,q\right)=1$.* + +*Proof:* + +*Let $p,q\in\mathbb{Z}$ so that $p$ and $q$ are prime and $p\neq q$. As +$p$ is prime then the only positive divisors are $p$ and $1$, likewise +for $q$. Hence the largest divisor of both $p$ and $q$ is 1 so that +$\mathop{\mathrm{GCD}}\left(p,q\right)=1$ by definition. $\qed$* +::: + +::: {#cor:NT_PrimeNotDividing_Integer_implies_coprime .corollary} +**Corollary 8**. *Prime not dividing integer implies co-prime* + +*Let $a,p\in\mathbb{Z}$ where $p$ is prime. If $p\nmid a$ then +$\mathop{\mathrm{GCD}}\left(a,p\right) = 1$* + +*Proof:* + +*Let $a,p\in\mathbb{Z}$ where $p$ is prime and where $p\nmid a$. Suppose +that $\mathop{\mathrm{GCD}}\left(a,p\right)=d$ for some +$d\in\mathbb{Z}$. By definition of the greatest common divisor, we have +that $d\mid p$ and by definition of a prime, we have that either $d=1$ +or $d=p$. But if $d=p$ then $p\mid a$ by definition of the greatest +common divisor, contradicting the assumption that $p\nmid a$. Hence +$d=1$. $\qed$* +::: + +::: proposition +**Proposition 112**. *Product of co-prime integers is equal to their +least common multiple* + +*Let $a,b\in\mathbb{Z}$ so that +$\mathop{\mathrm{GCD}}\left(a,b\right)=1$. We have that +$ab=\mathop{\mathrm{LCM}}\left(a,b\right)$.* + +*Proof:* + +*Let $a,b\in\mathbb{Z}$ be as given in the proposition. We have by +corollary [7](#cor:NT_lcm_formula){reference-type="ref" +reference="cor:NT_lcm_formula"} that* + +*$$\begin{equation*} + \mathop{\mathrm{LCM}}\left(a,b\right)= + \frac{ab}{\mathop{\mathrm{GCD}}\left(a,b\right)} +\end{equation*}$$* + +*As $a$ and $b$ are co-prime, we have +$\mathop{\mathrm{GCD}}\left(a,b\right)=1$, hence the result. $\qed$* +::: + +::: {#prop:NT_Bezout_coef_coprime .proposition} +**Proposition 113**. *Coefficients in Bézout's identity are co-prime* + +*Let $a,b\in\mathbb{Z}$ with $d=\mathop{\mathrm{GCD}}\left(a,b\right)$ +so that by Bézout's identity we have $\exists x,y\in\mathbb{Z}$ so that* + +*$$\begin{equation*} + d=ax+by +\end{equation*}$$* + +*We have that $\mathop{\mathrm{GCD}}\left(x,y\right)=1$* + +*Proof:* + +*Let $a,b\in\mathbb{Z}$ with $d=\mathop{\mathrm{GCD}}\left(a,b\right)$. +By Bézout's identity we have that there exists $x,y\in\mathbb{Z}$ so +that* + +*$$\begin{equation*} + d=ax+by +\end{equation*}$$* + +*Now, dividing by $d$ gives* + +*$$\begin{equation*} + 1=\frac{a}{d}x+\frac{b}{d}y +\end{equation*}$$* + +*As $d\mid a$ and $d\mid b$. Hence we have that $1=k_1x+k_2y$ where +$\displaystyle k_1=\frac{a}{d}$ and $\displaystyle k_2=\frac{b}{d}$. +Hence $\mathop{\mathrm{GCD}}\left(x,y\right)=1$ and so by definition $x$ +and $y$ are co-prime. $\qed$* +::: + +With some basic results out of the way, we can start seeing more +meaningful consequences of defining prime and co-prime numbers. One of +the first things we should do is see how primes divide other integers. + +::: example +**Example 113**. *Let $n=10$, we have that $2\mid 10$ and +$\sigma\left(2\right)=2$, hence $2$ is prime. Moreover $10=2*5$ and +clearly $2\mid 2$.* +::: + +::: example +**Example 114**. *let $n=4$, clearly $4=2*2$ and so $2\mid 4$. Moreover, +$2\mid 2$.* +::: + +::: example +**Example 115**. *Let $n=14=2*7$. Both $2$ and $7$ are prime and so +$2\mid 14$ and $7\mid 14$.* +::: + +Then, if a prime $p$ divides $n=ab$ we seem to have that either +$p\mid a$ or $p\mid b$. + +::: {#lem:NT_Euclid .lemma} +**Lemma 9**. *Euclid's Lemma* + +*Let $a,b\in\mathbb{Z}$ and let $p\in\mathbb{Z}$ be prime. Suppose that +$p\mid ab$. We have that either $p\mid a$ or $p\mid b$.* + +*Proof:* + +*Let $p\mid ab$. Suppose that $p\nmid b$. As the only divisors of $p$ +are $1$ and itself then we have that +$\mathop{\mathrm{GCD}}\left(p,b\right)=1$ by corollary +[8](#cor:NT_PrimeNotDividing_Integer_implies_coprime){reference-type="ref" +reference="cor:NT_PrimeNotDividing_Integer_implies_coprime"}. Now by +proposition [110](#prop:NT_Bezout_coprime){reference-type="ref" +reference="prop:NT_Bezout_coprime"} we have that +$\exists x,y\in\mathbb{Z}$ so that* + +*$$\begin{equation*} + 1=px+by +\end{equation*}$$* + +*Multiplying by $a$ gives $a=apx+aby$ and as $p\mid apx$ and $p\mid ab$ +we have that $p\mid a$. Likewise if $p\nmid a$. $\qed$* +::: + +This result generalises to products of more than two integers. + +::: {#lem:NT_Euclid_general .lemma} +**Lemma 10**. *Generalised Euclid's lemma* + +*Let $p\in\mathbb{Z}$ be prime. Let $n\in\mathbb{Z}$ be such that* + +*$$\begin{equation*} + n=\prod_{i=1}^m a_i +\end{equation*}$$* + +*where $a_i\in\mathbb{Z}$ for each $i$. Suppose that $p\mid n$, then +there exists an $i\in\mathbb{N}$ so that $p\mid a_i$.* + +*Proof:* + +*We argue by induction on $m$. The base case is $m=2$ which follows by +Euclid's lemma. So suppose the result holds for some $k>2$ that is if +$n$ is such that* + +*$$\begin{equation*} + n=\prod_{i=1}^k a_i +\end{equation*}$$* + +*then there is some $i\in\mathbb{N}$ so that $p\mid a_i$. We show that +if $n$ is such that* + +*$$\begin{equation*} + n=\prod_{i=1}^{k+1} a_i +\end{equation*}$$* + +*then there is some $i\in\mathbb{N}$ so that $p\mid a_i$. So suppose +that $p\mid n$, then* + +*$$\begin{equation*} + p\mid\prod_{i=1}^{k+1} a_i +\end{equation*}$$* + +*We have that* + +*$$\begin{align*} + p\mid&\prod_{i=1}^{k+1} a_i \\ + p\mid&\left(\prod_{i=1}^{k} a_i *a_k\right) +\end{align*}$$* + +*By the induction hypothesis we have that as +$\displaystyle p\mid\prod_{i=1}^{k} a_i$ then there is some +$i\in\mathbb{N}$ so that $p\mid a_i$ where $1\leq i \leq k$. Hence we +have that either $p\mid a_i$ or $p\mid a_{k+1}$. The result now follows +by induction. $\qed$* +::: + +With Euclid's lemma, we can provide a very famous theorem. Namely, there +is no $x\in\mathbb{Q}$ so that $x^2=2$. We first need a definition, +based on co-prime integers. + +::: definition +**Definition 154**. *Reduced fraction* + +*Let $x\in\mathbb{Q}$ where $\displaystyle x=\frac{a}{b}$ and $b\neq 0$. +We say that $x$ is a reduced fraction, or a fraction in its lowest terms +if $\mathop{\mathrm{GCD}}\left(a,b\right)=1$.* +::: + +We give some examples. + +::: example +**Example 116**. *Let $\displaystyle x=\frac{1}{2}$. As +$\mathop{\mathrm{GCD}}\left(1,2\right)=1$ we have that $x$ is a reduced +fraction.* +::: + +::: example +**Example 117**. *Let $\displaystyle x=\frac{3}{6}$. We can compute that +$\mathop{\mathrm{GCD}}\left(3,6\right)=3$, hence we have that $3\mid 3$ +and $3\mid 6$. We hence can write* + +*$$\begin{equation*} + x=\frac{3}{6}=\frac{3*1}{3*2}=\frac{1}{2} +\end{equation*}$$* + +*And as $\mathop{\mathrm{GCD}}\left(1,2\right)=1$ we can conclude $x$ is +now in its lowest terms.* +::: + +We can now show the theorem. + +::: {#thm:NT_Root2Irrational .theorem} +**Theorem 39**. *No rational exists whose square is $2$* + +*We have that $\not\exists x\in\mathbb{Q}$ with $x^2=2$.* + +*Proof:* + +*Suppose instead that $x\in\mathbb{Q}$ where +$\displaystyle x=\frac{a}{b}$ with $b\neq 0$. Moreover assume that $x$ +is a reduced fraction, i.e $\mathop{\mathrm{GCD}}\left(a,b\right)=1$. We +can make this assumption as otherwise we can reduce $x$ until it is +reduced without affecting the proof.* + +*We have that* + +*$$\begin{align*} + x^2&=2\\ + \frac{a^2}{b^2}&=2\\ + a^2&=2b^2 +\end{align*}$$* + +*Hence by the definition of divisibility, we have $2\mid a^2$ and so by +Euclid's lemma we have that $2\mid a$ as $2$ is prime. So write $a=2k$ +for some $k\in\mathbb{Z}$. Then we have that* + +*$$\begin{align*} + a^2&=2b^2\\ + \left(2k\right)^2&=2b^2\\ + 4k^2&=2b^2\\ + 2k^2&=b^2\\ +\end{align*}$$* + +*Hence $2\mid b^2$ and again by Euclid's lemma we have that $2\mid b$. +We have a contradiction as $2\mid a$ and $2\mid b$ implies that +$\mathop{\mathrm{GCD}}\left(a,b\right)\geq 2$ and so $x$ can't have been +a reduced fraction. But then if $x$ was not a reduced fraction and $a$ +and $b$ can't be co-prime then we can conclude that there is no rational +$x$ so that $x^2=2$. $\qed$* +::: + +This raises the question if there is no rational $x$ whose square is $2$ +then what exactly is $x$? Unfortunately, we are not quite ready to +properly answer this question in a satisfying way, all we can is that we +have seen a hint of a new type of number. One that we can define but not +study in more detail at the moment. + +::: definition +**Definition 155**. *Irrational number* + +*If we have $x\not\in\mathbb{Q}$, then we say that $x$ is irrational. In +other words, $x$ is irrational if and only if +$\displaystyle x=\frac{a}{b}$ where $a,b\in\mathbb{Z}$ and $b\neq 0$.* +::: + +Clearly, if $S$ denotes the set of irrational numbers then by theorem +$\ref{thm:NT_Root2Irrational}$ that $S\neq\emptyset$. Perhaps then it +makes sense, for now, to consider which elements of $x\in\mathbb{Q}$ so +that $x^2=y$ where $y\in\mathbb{Z}$, or more restrictively, which +$x\in\mathbb{Z}$ are such that we have $x^2=y$ where $y\in\mathbb{Z}$. + +Before we start answering this question, we note one useful result by +generalising Euclid's lemma from the prime case to the co-prime case. + +::: {#lem:NT_Euclid_co_primes .lemma} +**Lemma 11**. *Euclid's lemma for co-primes* + +*Let $a,b,c\in\mathbb{Z}$ and suppose that $c\mid ab$ and +$\mathop{\mathrm{GCD}}\left(b,c\right)=1$. We have that $c\mid a$.* + +*Proof:* + +*Let $a,b,c\in\mathbb{Z}$ be such that $c\mid ab$ and +$\mathop{\mathrm{GCD}}\left(b,c\right)=1$. As +$\mathop{\mathrm{GCD}}\left(b,c\right)=1$, we have by proposition +[110](#prop:NT_Bezout_coprime){reference-type="ref" +reference="prop:NT_Bezout_coprime"} that there exists integers +$x,y\in\mathbb{Z}$ so that* + +*$$\begin{equation*} + bx+cy=1 +\end{equation*}$$* + +*On multiplication by $a$ we have that $abx+acy=a$. Clearly $c\mid abx$ +and $c\mid acy$ and so $c\mid a$ as required. $\qed$* +::: + +There is a useful application of this lemma. + +::: {#exam:NT_solutions_to_ax_plus_by .example} +**Example 118**. + +*Let $a,b\in\mathbb{Z}$ and let +$d=\mathop{\mathrm{GCD}}\left(a,b\right)$. We know by Bézout's identity +that $\exists x,y\in\mathbb{Z}$ so that* + +*$$\begin{equation*} + ax+by=d +\end{equation*}$$* + +*The theorem for Bézout's identity, theorem +[36](#thm:NT_bezout_id){reference-type="ref" +reference="thm:NT_bezout_id"}, doesn't state anything about there not +being another pair $x',y'$ so that* + +*$$\begin{equation*} + ax'+by'=d +\end{equation*}$$* + +*For example, consider $a=30$ and $b=105$, then +$\mathop{\mathrm{GCD}}\left(a,b\right)=15$ and we have that +$15=-3*30+1*105$, i.e $x=-3$ and $y=1$ in this case. We could have also +have $x=-10$ and $y=3$ as $-10*30+3*105=-300+315=15$.* + +*So supposing that $a,b\in\mathbb{Z}$ and +$d=\mathop{\mathrm{GCD}}\left(a,b\right)$ we know that +$\exists x,x',y, y'\in\mathbb{Z}$ with* + +*$$\begin{align*} + ax+by&=d\\ + ax'+by'&=d +\end{align*}$$* + +*Can we find a relation between the pair $x$ and $y$ and the pair $x'$ +and $y'$? As $d\mid a$ then there exists $a'\in\mathbb{Z}$ so that +$a=a'd$ and likewise as $d\mid b$ then there exists $b'\in\mathbb{Z}$ so +that $b=b'd$. Hence we see that* + +*$$\begin{align*} + ax+by&=d\\ + a'dx+b'dy&=d\\ + a'x+b'y&=1 +\end{align*}$$* + +*Now, we have that $x$ and $y$ are co-prime so we can deduce that $a'$ +and $b'$ are also co-prime. Now, we have that* + +*$$\begin{equation*} + ax+by=d=ax'+by' +\end{equation*}$$* + +*So, re-arranging we see that* + +*$$\begin{align*} + ax-ax'&=by'-by\\ + a\left(x-x'\right)&=b\left(y'-y\right) +\end{align*}$$* + +*Dividing by $d$ gives* + +*$$\begin{equation*} + a'\left(x-x'\right)=b'\left(y'-y\right) +\end{equation*}$$* + +*Now, as $a'$ and $b'$ are co-prime, we have by Euclid's lemma for +co-primes that $a'\mid\left(y'-y\right)$, We, therefore have that +$\exists k\in\mathbb{Z}$ so that* + +*$$\begin{equation*} + y'-y=a'k \Rightarrow y'=y+a'k +\end{equation*}$$* + +*But as $y'-y=a'k$ we have that* + +*$$\begin{align*} + a'\left(x-x'\right)&=b'\left(a'k\right)\\ + x-x'&=b'k\\ + x'&=x-b'k\\ +\end{align*}$$* + +*Therefore, we can conclude that* + +*$$\begin{align*} + x'&=x-\frac{b}{d}k\\ + y'&=y+\frac{a}{d}k +\end{align*}$$* + +*where $k\in\mathbb{Z}$. To check this is the case we return to the +example of $a=30$ and $b=105$ where we had that +$\mathop{\mathrm{GCD}}\left(a,b\right)=15$. We saw that $x=-3$ and +$y=1$. Using these values in the equations above we get* + +*$$\begin{align*} + x'&=-3-\frac{105}{15}k \Rightarrow x'=-3-7k\\ + y'&=1+\frac{30}{15}k \Rightarrow y'=1+2k +\end{align*}$$* + +*Using $k=1$ gives us the alternative solution we saw of $x'=-10$ and +$y'=3$.* +::: + +From Euclid's lemma for co-primes we have deduced the full set of values +where $d=\mathop{\mathrm{GCD}}\left(a,b\right)$ and $d=ax+by$. + +We now return to the problem at hand. We wish to consider the elements +of $x\in\mathbb{Z}$ so that $x^2=y$ where $y\in\mathbb{Z}$. As is the +theme of this section we will do some exploratory examples. + +::: example +**Example 119**. *Let $x\in\mathbb{Q}$ be such that +$\displaystyle x=\frac{2}{1}$, then +$\displaystyle x^2=\frac{4}{1}=4\in\mathbb{Z}$. In particular, we have +that $4=2*2=2^2$.* +::: + +::: example +**Example 120**. *Consider $\displaystyle x=\frac{10}{1}=10$. Clearly +$x^2=100\in\mathbb{Z}$. We have that* + +*$$\begin{equation*} + 100=2*50=2*2*25=2*2*5*5=2^2*5^2 +\end{equation*}$$* +::: + +::: example +**Example 121**. *We generalise the example of there being no +$x\in\mathbb{Q}$ so that $x^2=2$. We will show that for a prime +$p\in\mathbb{Z}$, there is no $x\in\mathbb{Q}$ so that $x^2=p$. So +suppose there is such an $x$, that is $\displaystyle x=\frac{a}{b}$ so +that $a,b\in\mathbb{Z}$ and $b\neq 0$ and moreover suppose that $x$ is a +reduced fraction, which is to say +$\mathop{\mathrm{GCD}}\left(a,b\right)=1$. We then have that* + +*$$\begin{equation*} + x^2=\frac{a^2}{b^2}=p \Rightarrow a^2=pb^2 +\end{equation*}$$* + +*Hence $p\mid a^2$. Hence by Euclid's lemma, we have that $p\mid a$. +Hence let $a=pk$ for some $k\in\mathbb{Z}$. We then have that* + +*$$\begin{align*} + a^2&=pb^2\\ + \left(pk\right)^2&=pb^2\\ + p^2k^2&=pb^2\\ + pk&=b^2 +\end{align*}$$* + +*Therefore $p\mid b^2$ and so by Euclid's lemma we have that $p\mid b$, +a contradiction to the assumption that $x$ was a fraction in reduced +form.* +::: + +This last example shows that for any prime $p$ there is no rational +number $x$ with $x^2=p$. We also saw an example of when $x^2=p^2$, +namely when $p=2$. Also an example of a product of primes satisfying +$x^2=p^2*q^2$ for some primes $p$ and $q$. It seems therefore that the +question of what $x\in\mathbb{Z}$ so that $x^2=y$ for some integer $y$ +is deeply connected to primes. In particular, we have seen that the +powers of the primes must be even. We need more examples before we can +make a claim. + +::: example +**Example 122**. *Consider $x=4$, we have that $x^2=8$ and $8$ is not +prime as $\sigma\left(8\right)=4$, with divisors $1$, $2$, $4$ and $8$. +However, we have that $8=2^4$ and we know that $2$ is prime.* +::: + +::: example +**Example 123**. *Let $y=3^2*5^4=5625$, a product of primes. We can see +that we can take $x=3*5^2=75$.* +::: + +With these examples, we can see that to answer the question of what +$x\in\mathbb{Z}$ are such that $x^2=y$ for some $y\in\mathbb{Z}$, it is +enough to consider the structure of the primes that make $y$. This leads +us to, perhaps, the most important theorem of elementary Number +Theory[^12]. + +::::: {#thm:NT_FTOA .theorem} +**Theorem 42**. *The fundamental theorem of arithmetic* + +*Let $n\in\mathbb{Z}$ be such that $n\geq 2$. We have that $n$ can be +expressed as a product of one or more primes. This product is uniquely +up to the order of the primes. This is to say we have that* + +*$$\begin{equation*} + n=p_1^{e_1}*p_2^{e_2}*p_3^{e_3}*\dots*p_k^{e_k} +\end{equation*}$$* + +*where $p_i$ are the primes and $e_i$ are the powers for the prime +$p_i$. Here uniquely up to the order of the primes means that, for +example, $6=2*3=3*2$ are considered the same product.* + +*Proof:* + +*There are two parts to this theorem, firstly we must show that every +integer $n\geq 2$ is expressible as a product of primes. Secondly that +this product is unique up to the ordering of the primes.* + +*As a result, we will break this theorem down into two sub-theorems.* + +::: {#thm:NT_FOTA_EveryIntIsProductOfPrimes .theorem} +***Theorem 40**. *Every integer greater than one is expressible as a +product of primes** + +Let $n\in\mathbb{Z}$ be such that $n>1$. We have that + +$$\begin{equation*} + n=p_1*p_2*p_3*\dots*p_k +\end{equation*}$$ + +where $p_i$ are the primes. + +Proof: + +We argue by induction on $n$. The base case is $n=2$ for which we have +$n=2$ which is a prime. So the base case is immediate. So suppose the +result holds for some $k>2$, that is $n=k$ can be written as a product +of primes. We show that $n=k+1$ can be written as a product of primes. + +If $k+1$ is itself prime we are done, so suppose not, then +$\sigma\left(k+1\right)>2$ and so there are some factors, say $a$ and +$b$ so that $k+1=ab$, where $2\leq a < k+1$ and $2\leq b < k+1$. +However, this means that we have $2\leq a \leq k$ and $2\leq b \leq k$ +and so by the induction hypothesis we can write $a$ and $b$ as a product +of primes. But then $ab$ will be a product of primes and so $k+1$ is a +product of primes. + +The result follows by induction. $\qed$ +::: + +::: {#thm:NT_FOTA_PrimeProdUnique .theorem} +***Theorem 41**. *The product of primes expression for an integer is +unique** + +Let $n\in\mathbb{Z}$ be such that $n\geq 2$. We have that the expression +for $n$ as a product of primes is unique. + +Proof: + +Let $n\in\mathbb{Z}$ be as given. Suppose that $n$ has two different +representations into a product of primes, that is + +$$\begin{align*} + n&=p_1p_2p_3\dots p_r\\ + n&=q_1q_2q_3\dots q_s +\end{align*}$$ + +where without loss of generality we suppose that $r\leq s$. Moreover, +Without loss of generality suppose that we have the primes in ascending +order, that is, $p_1\leq p_2\leq p_3\leq\dots\leq p_r$ and that +$q_1\leq q_2\leq q_3\leq\dots\leq q_s$. + +Now as $p_1\mid q_1q_2q_3\dots q_s$ we have by Euclid's lemma that +$p_1\mid q_i$ for some $1\leq i \leq s$. Therefore $p_1\geq q_1$ as the +primes are in ascending order. Likewise, as +$q_1\mid p_1p_2p_3\dots p_r$, then $q_1\mid p_j$ for some +$1\leq j\leq r$. Hence $q_1\geq p_1$. As $p_1\geq q_1$ and $q_1\geq p_1$ +we must have that $p_1=q_1$. Hence we have + +$$\begin{align*} + p_1p_2p_3\dots p_r&=q_1q_2q_3\dots q_s\\ + p_1p_2p_3\dots p_r&=p_1q_2q_3\dots q_s\\ + p_2p_3\dots p_r&=q_2q_3\dots q_s\\ +\end{align*}$$ + +This process can be repeated for each prime $p_j$ for the remaining +$2\leq j\leq r$. Now if $rp_n$, which would be a contradiction. So $N$ is composite and by the +fundamental theorem of arithmetic, we have that $N$ has a factorisation +into primes. Clearly, none of the $p_i$ divide $N$, but then none of the +$p_i$ divide the prime factorisation of $N$ from the fundamental theorem +of arithmetic, a contradiction. Hence $P$ can't be a finite set and the +number of primes must be infinite. $\qed$* +::: + +The fundamental theorem of arithmetic can be used to recast some +previous results for the greatest common divisor. We start with a result +for integers being co-primes. + +::: {#prop:NT_co-prime_iff_no_common_primes .proposition} +**Proposition 114**. *Greatest common divisor is 1 if and only if +no-common prime in factorisation* + +*Let $a,b\in\mathbb{Z}$ with $b\neq 0$. We have that +$\mathop{\mathrm{GCD}}\left(a,b\right)=1$ if and only if $a$ and $b$ +share no common primes in their factorisations.* + +*Proof:* + +*We have that $\mathop{\mathrm{GCD}}\left(a,b\right)=1$ if and only if +the largest divisor of both $a$ and $b$ is $1$, which occurs if and only +if there are no primes in the factorisation of $a$ and in the +factorisation of $b$ in common. $\qed$* +::: + +We can compute the greatest common divisor by considering the prime +factorisations of $a$ and $b$. To do so we need a helpful result. + +::: {#prop:NT_express_primes_in_common_basis .proposition} +**Proposition 115**. *Expression for integers as powers of same primes* + +*Let $a,b\in\mathbb{Z}$ with $a\geq 2$ and $b\geq 2$. Consider the prime +factorisations of $a$ and $b$ given by* + +*$$\begin{align*} + a&=p_1^{e_1}p_2^{e_2}p_3^{e_3}\dots p_n^{e_n}\\ + &=\prod_{\substack{p_i\mid a \\ p_i\text{ is prime}}} p_i^{e_i}\\ + b&=q_1^{f_1}q_2^{f_2}q_3^{f_3}\dots q_m^{f_m}\\ + &=\prod_{\substack{q_i\mid b \\ q_i\text{ is prime}}} q_i^{f_i}\\ +\end{align*}$$* + +*where $n$ need not be equal to $m$. We have that there exist prime +numbers* + +*$$\begin{equation*} + t_1 + + 1. *$k'=1$:* + + *Suppose for a contradiction that $k'\neq 1$. As $d>0$ and $D>0$ + then we must have that $k>0$ which means that $k'>0$. Hence as + $k'>0$ we have by the fundamental theorem of arithmetic that + $k'$ has a factorisation into primes, say* + + *$$\begin{equation*} + k'=q_1^{r_1}q_2^{r_2}q_3^{r_3}\dots q_c^{r_c} + \end{equation*}$$* + + *Moreover, no $q_j=t_i$ as $k'$ has no common primes with + $t_1^{\lambda_1}t_2^{\lambda_2}t_3^{\lambda_3}\dots t_v^{\lambda_v}$. + Pick one of the primes in $k'$ say $q=q_j$ then we have that + $q\mid d$. Moreover we have that $d\mid a$ as + $d=\mathop{\mathrm{GCD}}\left(a,b\right)$ hence we must have + that $q\mid a$. Hence we have that $q$ is one of the primes + $t_i$ a contradiction. Therefore $k'=1$.* + + 2. *$\lambda_i\leq \sigma_i$ for all $1\leq i\leq v$:* + + *Suppose for a contradiction that $\lambda_i>\sigma_i$ for all + $1\leq i\leq v$. Without loss of generality take $i=1$, for if + this is not the case re-label the primes. Now by definition of + $\sigma_1$ we have that $\sigma_1=\min\left(e_,f_1\right)$ and + so we must have that either $\sigma_1=e_1$ or $\sigma_1=f_1$. + Without loss of generality let $\sigma_1=e_1$ as the case where + $\sigma_1=f_1$ is similar.* + + *We, therefore, have that $\lambda_1>e_1$. Now, as $d$ is the + greatest common divisor of $a$ there is a $s\in\mathbb{Z}$ so + that $ds=a$ where $s>0$ as both $a$ and $d$ are. Now, comparing + the prime factorisations of $ds$ and $a$ we have that* + + *$$\begin{equation*} + s*t_1^{\lambda_1}t_2^{\lambda_2}t_3^{\lambda_3}\dots t_v^{\lambda_v}=t_1^{e_1}t_2^{e_2}t_3^{e_3}\dots t_v^{e_v} + \end{equation*}$$* + + *Dividing by $\displaystyle t_1^{e_1}$ we get that* + + *$$\begin{equation*} + s*t_1^{\lambda_1-e_1}t_2^{\lambda_2}t_3^{\lambda_3}\dots t_v^{\lambda_v}=t_1^{e_1-e_1}t_2^{e_2}t_3^{e_3}\dots t_v^{e_v} + \end{equation*}$$* + + *Where clearly $\displaystyle t_1^{e_1-e_1}=1$. So this can be + re-written as* + + *$$\begin{equation*} + s*t_1^{\lambda_1-e_1}t_2^{\lambda_2}t_3^{\lambda_3}\dots t_v^{\lambda_v}=t_2^{e_2}t_3^{e_3}\dots t_v^{e_v} + \end{equation*}$$* + + *As $\lambda_1>e_1$, we have $\lambda_1-e_1>0$. and so $t_1$ + divides the left-hand side of the equation. But by the + fundamental theorem of arithmetic if $t_1$ divides the left-hand + side it must also divide the right-hand side and so would appear + in the factorisation, but it is not in the factorisation of the + right-hand side a contradiction. So $\lambda_i\leq\sigma_i$ for + all $1\leq i\leq v$.* + + *Therefore $d\leq D$* + + *As $D\leq d$ and $d\leq D$ we must have that $d=D$. As required. + $\qed$* +::: + +Proposition +[116](#prop:NT_gcd_can_be_computed_by_primes){reference-type="ref" +reference="prop:NT_gcd_can_be_computed_by_primes"} allows us to compute +the greatest common divisor by considering the prime factorisations, +rather than using the Euclidean algorithm. Unfortunately, we now have a +new problem, how do we compute the prime factorisation of an integer? +Thankfully to answer this question we have to answer the original +question posed, what $x\in\mathbb{Z}$ are such that $x^2=y$ for some +$y\in\mathbb{Z}$? Clearly if $x\in\mathbb{Z}$ then $x$ has some prime +factorisation, say + +$$\begin{equation*} + x=p_1^{e_1}p_2^{e_2}p_3^{e_3}\dots p_k^{e_k} +\end{equation*}$$ + +So that + +$$\begin{align*} + x^2&=\left(p_1^{e_1}p_2^{e_2}p_3^{e_3}\dots p_k^{e_k}\right)\left(p_1^{e_1}p_2^{e_2}p_3^{e_3}\dots p_k^{e_k}\right)\\ + &=\left(p_1^{e_1}p_1^{e_1}\right)\left(p_2^{e_2}p_2^{e_2}\right)\left(p_3^{e_3}p_3^{e_3}\right)\dots\left(p_k^{e_k}p_k^{e_k}\right)\\ + &=p_1^{2e_1}p_2^{2e_2}p_3^{2e_3}\dots p_k^{2e_k}=y +\end{align*}$$ + +For each prime $p_i$, the power of that prime is now of the form $2e_i$ +and therefore the power is even. We make this fact a definition. + +::: {#def:NT_square_number .definition} +**Definition 158**. *Square number* + +*Let $y\in\mathbb{Z}$ where $y>0$, if there exists an $x\in\mathbb{Z}$ +so that* + +*$$\begin{equation*} + x^2=y +\end{equation*}$$* + +*Then we say that $y$ is a square number.* +::: + +In light of the above discussion, we have the following result. + +::: {#prop:NT_square_number_iff_prime_exonents_even .proposition} +**Proposition 117**. *Square number if and only if prime factorisation +has even powers* + +*Let $x\in\mathbb{Z}$. We have that $x$ is a square number if and only +if the prime factorisation of $x$ only contains even prime powers. This +is to say that each prime $p_i$ in the factorisation of $x$ has an +exponent of the form $2e_i$.* + +*Proof:* + +*$\left(\Rightarrow\right)$: Suppose that $x$ is a square number, by +definition there exists $y\in\mathbb{Z}$ so that $y^2=x$. Let the prime +factorisation of $y$ be* + +*$$\begin{equation*} + y=q_1^{f_1}q_2^{f_2}q_3^{f_3}\dots q_k^{f_k} +\end{equation*}$$* + +*We have that then* + +*$$\begin{equation*} + x=y^2=q_1^{2f_1}q_2^{2f_2}q_3^{2f_3}\dots q_k^{2f_k} +\end{equation*}$$* + +*Hence all the prime factors of $x$ have an exponent of the form $2f_i$ +making them even.* + +*$\left(\Leftarrow\right)$: Suppose that the prime factorisation of $x$ +has prime factors which only have even powers, that is* + +*$$\begin{equation*} + x=p_1^{e_1}p_2^{e_2}p_3^{e_3}\dots p_k^{e_k} +\end{equation*}$$* + +*As each $e_i$ is even we have that +$\displaystyle \frac{e_i}{2}\in\mathbb{Z}$. Define $y$ to be* + +*$$\begin{equation*} + y=p_1^{e_1/2}p_2^{e_2/2}p_3^{e_3/2}\dots p_k^{e_k/2} +\end{equation*}$$* + +*Where clearly $y\in\mathbb{Z}$. We then have that* + +*$$\begin{align*} + y^2&=\left(p_1^{e_1/2}p_2^{e_2/2}p_3^{e_3/2}\dots p_k^{e_k/2}\right)\left(p_1^{e_1/2}p_2^{e_2/2}p_3^{e_3/2}\dots p_k^{e_k/2}\right)\\ + &=\left(p_1^{e_1/2}p_1^{e_1/2}\right)\left(p_2^{e_2/2}p_2^{e_2/2}\right)\left(p_3^{e_3/2}p_1^{e_3/2}\right)\dots \left(p_k^{e_k/2}p_k^{e_k/2}\right)\\ + &=p_1^{e_1}p_2^{e_2}p_3^{e_3}\dots p_k^{e_k}=x +\end{align*}$$* + +*Hence as $x=y^2$ for some $y\in\mathbb{Z}$ we conclude that $x$ is a +square number. $\qed$* +::: + +We also have an immediate proposition. + +::: {#prop:NT_product_of_sqaure_numbers_is_sqaure_number .proposition} +**Proposition 118**. *Product of two square numbers is a square number* + +*Let $x,y\in\mathbb{Z}$ be square numbers. We have that $xy$ is a square +number.* + +*Proof:* + +*Let $x,y\in\mathbb{Z}$ be square numbers. We have by definition that +$\exists a,b\in\mathbb{Z}$ so that* + +*$$\begin{align*} + a^2&=x\\ + b^2&=y +\end{align*}$$* + +*Now, consider the product $xy$, we have* + +*$$\begin{equation*} + xy=a^2*b^2=\left(ab\right)^2 +\end{equation*}$$* + +*Hence by definition, $xy$ is a square number. $\qed$* +::: + +With proposition +[117](#prop:NT_square_number_iff_prime_exonents_even){reference-type="ref" +reference="prop:NT_square_number_iff_prime_exonents_even"} we can +finally answer the question of what $x\in\mathbb{Z}$ are such that +$x^2=y$ for some $y\in\mathbb{Z}$. It is those $x\in\mathbb{Z}$ so that +$x^2$ is a square number! At first, this doesn't seem too useful as we +can clearly take any $n\in\mathbb{Z}$ and see that $n^2\in\mathbb{Z}$. +However, the real meaning of this result is actually the converse, given +some $n\in\mathbb{Z}$ we can see if there is an $x\in\mathbb{Z}$ so that +$x^2=n$. With this, we make a definition + +::: definition +**Definition 159**. *Square root function* + +*Let $x\in\mathbb{Z}$ be a positive square number. We define the square +root function, denoted by $\sqrt{}$ as follows* + +*$$\begin{align*} + \sqrt{}:\mathbb{Z}&\rightarrow\mathbb{Z}\\ + x&\mapsto \sqrt{x}=\begin{cases} + n,\ \text{If } n^2=x\\ + \text{Undefined otherwise} + \end{cases} +\end{align*}$$* + +*That is, we define the square root of an integer $x$ to be the integer +$n$ that when squared gives $x$.* +::: + +In light of this definition, we have the following result. + +::: {#prop:NT_root_of_product_is_product_of_roots .proposition} +**Proposition 119**. *Square root of product is product of square roots* + +*Let $x,y\in\mathbb{Z}$ be square numbers. We have that* + +*$$\begin{equation*} + \sqrt{xy}=\sqrt{x}\sqrt{y} +\end{equation*}$$* + +*Proof:* + +*Let $x,y$ be as given. By proposition +[118](#prop:NT_product_of_sqaure_numbers_is_sqaure_number){reference-type="ref" +reference="prop:NT_product_of_sqaure_numbers_is_sqaure_number"} we have +that $xy$ is a square number and so $\sqrt{xy}$ is well-defined. We need +to show that* + +*$$\begin{equation*} + \sqrt{xy}=\sqrt{x}\sqrt{y} +\end{equation*}$$* + +*By definition, we suppose that $\sqrt{xy}=n$, where $n^2=xy$. +Additionally, we can suppose that $\sqrt{x}=a$ where $a^2=x$ and +$\sqrt{y}=b$ where $b^2=y$. Now, we have that* + +*$$\begin{equation*} + \left(\sqrt{x}\sqrt{y}\right)^2=\left(ab\right)^2=a^2b^2=xy=n^2=\left(\sqrt{xy}\right)^2 +\end{equation*}$$* + +*As $n^2=a^2b^2$ we have that $n=ab$. Hence we have that +$\sqrt{xy}=\sqrt{x}\sqrt{y}$ as required. $\qed$* +::: + +The idea of a square number actually generalises, meaning the question +of what $x\in\mathbb{Z}$ are such that $x^2=y$ for some $y\in\mathbb{Z}$ +can be generalised to the question what $x\in\mathbb{Z}$ are such that +$x^n=y$ for some $y\in\mathbb{Z}$ and every $n\in\mathbb{N}$. + +The generalisation works very similarly to how we got to square numbers. +As before let $x\in\mathbb{Z}$ which has a factorisation + +$$\begin{equation*} + x=p_1^{e_1}p_2^{e_2}p_3^{e_3}\dots p_k^{e_k} +\end{equation*}$$ + +Now, consider $x^n$, the factorisation is given by + +$$\begin{equation*} + x=p_1^{ne_1}p_2^{ne_2}p_3^{ne_3}\dots p_k^{ne_k} +\end{equation*}$$ + +Hence the power of each prime $p_i$ is of the form $ne_i$. This is the +defining characteristic for the next definition. + +::: definition +**Definition 160**. *$n$-th power number* + +*Let $y\in\mathbb{Z}$ with $y>0$ and let $n\in\mathbb{N}$, if there +exists an $x\in\mathbb{Z}$ so that* + +*$$\begin{equation*} + x^n=y +\end{equation*}$$* + +*We say that $y$ is the $n$-th power of $x$. We have already seen the +case of $n=2$ where $y$ is called a square number. For $n=3$ we call $y$ +a cube number. For $n>4$, there is no formal term hence the definition +using the terminology of $n$-th power.* +::: + +The next step is to prove an equivalent proposition to +[117](#prop:NT_square_number_iff_prime_exonents_even){reference-type="ref" +reference="prop:NT_square_number_iff_prime_exonents_even"}. + +::: {#prop:NT_nth_power_number_iff_prime_exonents_multiple_of_n .proposition} +**Proposition 120**. *$n$-th power number if and only if prime +factorisation has multiples of $n$ powers* + +*Let $x\in\mathbb{Z}$. We have that $x$ is a $n$-th power number if and +only if the prime factorisation of $x$ only contains prime powers that +are a multiple of $n$. this is to say that each prime $p_i$ in the +factorisation of $x$ has an exponent of the form $ne_i$.* + +*Proof:* + +*$\left(\Rightarrow\right)$: Suppose that $x$ is a $n$-th power number, +by definition there exists $y\in\mathbb{Z}$ so that $y^n=x$. Let the +prime factorisation of $y$ be* + +*$$\begin{equation*} + y=q_1^{f_1}q_2^{f_2}q_3^{f_3}\dots q_k^{f_k} +\end{equation*}$$* + +*We have that then* + +*$$\begin{equation*} + x=y^2=q_1^{nf_1}q_2^{nf_2}q_3^{nf_3}\dots q_k^{nf_k} +\end{equation*}$$* + +*Hence all the prime factors of $x$ have an exponent of the form $nf_i$, +meaning each prime power is a multiple of $n$.* + +*$\left(\Leftarrow\right)$: Suppose that the prime factorisation of $x$ +has prime factors which only have multiples of $n$, that is* + +*$$\begin{equation*} + x=p_1^{ne_1}p_2^{ne_2}p_3^{ne_3}\dots p_k^{ne_k} +\end{equation*}$$* + +*As each $e_i$ is a multiple of $n$ we have that +$\displaystyle \frac{e_i}{n}\in\mathbb{Z}$. Define $y$ to be* + +*$$\begin{equation*} + y=p_1^{e_1/n}p_2^{e_2/n}p_3^{e_3/n}\dots p_k^{e_k/n} +\end{equation*}$$* + +*Where clearly $y\in\mathbb{Z}$. We then have that* + +*$$\begin{align*} + y^n&=\prod_{i=1}^n\left(p_1^{e_1/n}p_2^{e_2/n}p_3^{e_3/n}\dots p_k^{e_k/n}\right)\\ + &=\prod_{j=1}^k\left(\prod_{i=1}^n\left(p_j^{e_j/n}\right)\right)\\ + &=\prod_{j=1}^k\left(p_j^{e_j}\right)\\ + &=p_1^{e_1}p_2^{e_2}p_3^{e_3}\dots p_k^{e_k}=x +\end{align*}$$* + +*Hence as $x=y^n$ for some $y\in\mathbb{Z}$ we conclude that $x$ is an +$n$-th power number. $\qed$* +::: + +As before, there is an immediate proposition. + +::: proposition +**Proposition 121**. *Product of two $n-th$ power numbers is an $n$-th +power number* + +*Let $x,y\in\mathbb{Z}$ be $n$-th power numbers. We have that $xy$ is an +$n$-th power number.* + +*Proof:* + +*Let $x,y\in\mathbb{Z}$ be $n$-th power numbers. By definition, we have +that $\exists a,b\in\mathbb{Z}$ so that* + +*$$\begin{align*} + a^n&=x\\ + b^n&=y +\end{align*}$$* + +*We have* + +*$$\begin{equation*} + xy=a^n*b^n=\left(ab\right)^n +\end{equation*}$$* + +*Giving the result. $\qed$* +::: + +We can now now generalise the square root function. + +::: definition +**Definition 161**. *$n$-th root function* + +*Let $x\in\mathbb{Z}$ be a positive $n$-th power number. We define the +$n$-th root function, denoted by $\displaystyle \sqrt[n]{}$ is given by* + +*$$\begin{align*} + \sqrt[n]{}:\mathbb{Z}&\rightarrow\mathbb{Z}\\ + x&\mapsto \sqrt[n]{x}=\begin{cases} + m,\ \text{If } m^n=x\\ + \text{Undefined otherwise} + \end{cases} +\end{align*}$$* + +*That is, we define the $n$-th root of an integer $x$ to be the integer +$m$ that when raised to the power of $n$ gives $x$.* +::: + +### The integers modulo n + +::: epigraph +Mathematicians call it "the arithmetic of congruences." You can think of +it as clock arithmetic + +*John Derbyshire* +::: + +So far in the study of the divisibility of integers, we have considered +what it means for an integer $a$ to divide another $b$, namely we have +that $a\mid b$ if there is some $c\in\mathbb{Z}$ such that $ac=b$. We +now explore the implications of the case where $a\nmid b$, in +particular, we look at the the remainders from the division algorithm. + +#### Remainders after division + +Recall that for $a,b\in\mathbb{Z}$ we have that $a\mid b$ if +$\exists c\in\mathbb{Z}$ so that $b=ac$. When this is not the case we +have that $a\nmid b$. By the division algorithm, when $a\nmid b$ we have +that $0a$. We have by the +division algorithm that* + +*$$\begin{equation*} + b=2q+r +\end{equation*}$$* + +*Hence $r$ can only be one of $0$ or $1$. Now if $r=0$ then we must have +that $b$ is an even number and if $r=1$ we must have that $b$ is an odd +number. Then as $b$ is an arbitrary integer we must have that dividing +any integer by $2$ will give us all of the possible remainders as an +integer $x\in\mathbb{Z}$ is either even or odd.* +::: + +::: example +**Example 129**. *Let $a=3$ and consider some $b>a$. By the division +algorithm we have that $r$ is either $0$, $1$ or $2$. Like before, if +$r=0$ then $b$ is a multiple of $3$ so that $b=3q$.* + +*Now, suppose $b$ is a multiple of $3$. We have that $b+3$ is also a +multiple of $3$ as* + +*$$\begin{equation*} + b+3=3q+3\Rightarrow 3\left(q+1\right) +\end{equation*}$$* + +*So, as $b$ is a multiple of $3$ and $b+3$ is a multiple of $3$ then +these will give a remainder $r=0$ by the division algorithm. What can we +say about $b+1$ and $b+2$? Using $b+1$ in the division algorithm with +$3$ gives* + +*$$\begin{align*} + b+1=3q+1 +\end{align*}$$* + +*as $b=3q$. Hence the remainder is $1$. Likewise using $b+2$ in the +division algorithm with $3$ gives a remainder of $2$. As $b$ was an +arbitrary integer, we can conclude that the possible remainders when +dividing an arbitrary integer by $3$ are $0, 1$ and $2$. All of the +possibilities are realised for the division of an arbitrary integer.* +::: + +::: example +**Example 130**. *Let $a=4$ and consider some $b>a$. The division +algorithm gives the possible range of remainders of $0, 1, 2$ and $3$. +Like the previous example, we see that if the remainder is $0$ then $b$ +is a multiple of $4$, so similarly $b+4$ is a multiple of $4$. Looking +at $b+1$, $b+2$ and $b+3$ we see by the division algorithm that* + +*$$\begin{align*} + b+1=4q+1\\ + b+2=4q+2\\ + b+3=4q+3\\ +\end{align*}$$* + +*So dividing an arbitrary integer by $4$ will give a remainder in the +range $0$ to $3$ inclusive.* +::: + +These examples suggest that when dividing an arbitrary integer $b$ by +some $a\in\mathbb{Z}$ with $a>0$ will always give one value $r$ with +$00$. We can prove this but first make an important +observation. + +::: {#cor:NT_integer_minus_remainder_is_divisable .corollary} +**Corollary 9**. + +*Let $a,b\in\mathbb{Z}$ with $b>a$. Consider the division algorithm for +$b$ divided by $a$, that is we have* + +*$$\begin{equation*} + b=qa+r +\end{equation*}$$* + +*for some $q,r\in\mathbb{Z}$ and $00$. If we have that $a$ and $b$ +have the same remainder when divided by $n$ we say that $a$ and $b$ a +congruent modulo $n$. This is denoted by* + +*$$\begin{equation*} + a\equiv b \ (\mathrm{mod}\ n) +\end{equation*}$$* + +*We call $b$ a residue of $a$ modulo $n$. We usually say that $a$ is +congruent to $b$ modulo $n$. We define a congruence to capture the +notion of congruent numbers and residue numbers. We call the number $n$ +the modulus of the congruence.* + +*If $a$ is not congruent to $b$, equivalently if $b$ is not a residue of +$a$ we write $a\not\equiv b\ (\mathrm{mod}\ n)$.* +::: + +We can make use of corollary +[9](#cor:NT_integer_minus_remainder_is_divisable){reference-type="ref" +reference="cor:NT_integer_minus_remainder_is_divisable"} to connect +division to congruences. + +::: {#prop:NT_congruent_iff_difference_is_divisible .proposition} +**Proposition 122**. *Congruent if and only if the difference is +divisible by modulus* + +*Let $a,b,n\in\mathbb{Z}$ and fix $n\geq 1$. We have that +$a\equiv b\ (\mathrm{mod}\ n)$ if and only if $n\mid\left(a-b\right)$.* + +*Proof:* + +*By the division algorithm, we have that* + +*$$\begin{align*} + a&=qn+r\\ + b&=q'n+r' +\end{align*}$$* + +*for some $q,q',r,r'\in\mathbb{Z}$ where $00$ and +$j-i0$ with a prime factorisation + +$$\begin{equation*} + n=p_1^{e_1}p_2^{e_2}p_3^{e_3}\dots p_k^{e_k} +\end{equation*}$$ + +then we might expect that $a\equiv b\ (\mathrm{mod}\ n)$ if and only if +$a\equiv b\ (\mathrm{mod}\ p_i^{e_i})$ where $i=1,2,\dots, k$. We can +prove this. + +::: proposition +**Proposition 128**. *Congruent if and only if congruent to each prime +in factorisation* + +*Let $n\in\mathbb{Z}$ so that $n>0$ and $n$ has a prime factorisation +given by* + +*$$\begin{equation*} + n=p_1^{e_1}p_2^{e_2}p_3^{e_3}\dots p_k^{e_k} +\end{equation*}$$* + +*Let $a,b\in\mathbb{Z}$. We have that $a\equiv b\ (\mathrm{mod}\ n)$ if +and only if $a\equiv b\ (\mathrm{mod}\ p_i^{e_i})$ for each $i$ where +$i=1,2,\dots, k$* + +*Proof:* + +*Let $n\in\mathbb{Z}$ be as given in the hypothesis. We have that* + +*$$\begin{align*} + a\equiv b\ (\mathrm{mod}\ n) &\iff n\mid\left(a-b\right)\\ + &\iff p_1^{e_1}p_2^{e_2}p_3^{e_3}\dots p_k^{e_k}\mid\left(a-b\right)\\ + &\iff p_i^{e_i}\mid\left(a-b\right),\ \text{For each } i=1,2,\dots, k\\ + &\iff a\equiv b\ (\mathrm{mod}\ p_i^{e_i}),\ \text{For each } i=1,2,\dots, k +\end{align*}$$* + +*As required. $\qed$* +::: + +Another use of congruences is in cryptography, which is a field of study +of taking messages and encoding (obfuscating) them in such a way that +only the person the message was intended for can read it. This is +especially true for the RSA[^14] encryption method. We already have some +of the mathematical machinery required to explore how this method of +cryptography works, namely prime numbers and congruences. On the other +hand, we still lack some important theory. If cryptography is the field +of encoding messages so that only the person the message was intended +for can read it, then there is some method that encodes the message and +a method that decodes the message using some information known to both +the sender and recipient. This means that using this information the +recipient will have some method of finding out the original message! We +look at this idea in more detail. + +### Diophantine equations and Polynomials + +::: epigraph +I had a Polynomial once. My Doctor removed it. + +*Micheal Grant* +::: + +We start with a definition that we have seen numerous times so far but +have not formally defined. That of an equation. + +::: definition +**Definition 165**. *Equation* + +*An equation is a mathematical statement that states that two +expressions are equal.* +::: + +This seems simple enough, but what does it mean? Unfortunately, this +depends on the situation, different situations will have a different +meaning of what a statement is. Thankfully, we have seen equations +already throughout the text so this abstract definition is familiar to +us. For example + +$$\begin{equation*} + 1+1=2 +\end{equation*}$$ + +is an equation. So is $\mathop{\mathrm{GCD}}\left(a,b\right)=d$, in a +similar vain we have from Bézout's Identity that $d=ax+by$ is also an +equation. So why define something if it is really this simple? Simply +put, we can use the idea of an equation in a more complex way. For +example, + +$$\begin{equation*} + 1+x=2 +\end{equation*}$$ + +says that $1$ plus $x$ is equal to $2$ but we don't know what $x$ is. +However, we can see that + +$$\begin{align*} + 1+x&=2\\ + x&=1 +\end{align*}$$ + +That is, we see that $x=1$, this is an equation! This is where the power +of an equation starts to show its worth. If we have a problem where we +don't know the value of some quantity of interest, we might be able to +work out what that quantity is. We have seen more complex examples of +equations, for example $x^2=2$ which we have shown has no value of +$x\in\mathbb{Q}$ where it is true. + +Hence, equations that contain a value, or maybe multiple values that we +don't know but want to know, are important. This section is focused on +looking at such equations. We make another couple of definitions for +when an equation contains a value we don't know. + +::: definition +**Definition 166**. *Variable* + +*A variable is a value that is allowed to be changed either freely or +restricted by some constraint or equation. A variable can be taken to be +any meaningful value, either inside or outside of some set $S$. The +context of the statement under study usually makes it clear where the +variable belongs.* +::: + +::: definition +**Definition 167**. *Indeterminate variable* + +*An indeterminate variable is a variable value which has not been +specified. As with a variable, it could be inside or outside of some set +$S$.* +::: + +::: definition +**Definition 168**. *An unknown variable* + +*An unknown variable, or simply an unknown, is a variable whose value is +unknown but we wish to find its value. As before, this unknown variable +is to be taken as a member of a set $S$. If a value for the unknown +variable can be found, we call it a solution to the equation.* +::: + +For example, the equation $5x+1=2$ would have $x$ as the indeterminate +variable, if we were solving for $x$ then $x$ would be the unknown +variable as well. The equation $2x+5y=6$ has two indeterminate +variables, $x$ and $y$. We can potentially have many indeterminate +variables in an equation. Moreover, in many problems, we will have a +certain type of variable whose value can vary but is not the unknown +that we are looking to solve for. We define this type of variable as +well. + +::: definition +**Definition 169**. *Coefficient* + +*A variable which can vary but is not the variable that is being solved +for is called a coefficient, or a parameter of the equation.* +::: + +So, let's start simply and consider the simplest equation possible with +one unknown variable and two coefficients. + +$$\begin{equation*} + x+a=b +\end{equation*}$$ + +This is simple to solve for the unknown $x$, simply take $a$ from both +sides to give $x=b-a$. So for example if we let $a,b\in\mathbb{Z}$ say +with $a=5$ and $b=3$, then we see that $x\in\mathbb{Z}$ with $x=3-5=-2$. +This is also true if we take $a,b\in\mathbb{Q}$. A more complex form of +the above equation is + +$$\begin{equation*} + ax+b=c +\end{equation*}$$ + +Now we hit a problem we are looking for a solution $x\in\mathbb{Z}$. +Firstly, we have that $ax=c-b$, but then a solution $x\in\mathbb{Z}$ can +occur if and only if $a\mid\left(b-c\right)$. If we look for a solution +where $x\in\mathbb{Q}$ then no such problem occurs. Therefore, the set +that we are looking for solutions in is crucial in solving equations. +With our current theory, the situation gets more hopeless the more +complicated the equation becomes. For example, if we consider the +equation + +$$\begin{equation*} + 4x^2+2x+3=0 +\end{equation*}$$ + +Does this equation have solutions in $\mathbb{Z}$? How about +$\mathbb{Q}$?. Additionally, what happens if we have more than one +equation or unknowns? For example, consider the two equations given by + +$$\begin{align*} + 4x+2y&=6\\ + -2x+5y&=7 +\end{align*}$$ + +How do we solve equations like this? This section aims to answer +questions like these. We make a final definition, a special case for +when we only seek integer solutions. + +::: definition +**Definition 170**. *Diophantine equation* + +*An equation for which the solutions have to be integers is called a +Diophantine equation[^15].* +::: + +#### Linear Diophantine equations + +##### Linear equations with two variables + +We start where the previous section left off, by looking at the simplest +type of equation that can be solved. + +::: definition +**Definition 171**. *Linear equation of a single indeterminate variable* + +*Let $S$ be a set. We say an equation is a linear equation in a single +variable $x$ if it has the form* + +*$$\begin{equation*} + ax+b=c +\end{equation*}$$* + +*for some coefficients $a,b,c\in S$ and an indeterminate variable $x$. +In particular as this equation only has one indeterminate variable we +say it is a single-variable linear equation.* +::: + +We have already seen that solutions to this equation exist in +$\mathbb{Z}$ if and only if $a\mid\left(c-b\right)$, and a solution +always exists if we want $x\in\mathbb{Q}$. Things are a bit more +interesting if we introduce a second variable. + +::: definition +**Definition 172**. *Linear equation of two indeterminate variables* + +*Let $S$ be a set. We say an equation is a linear equation in two +variables $x,y$ if it has the form* + +*$$\begin{equation*} + ax+by=c +\end{equation*}$$* + +*for some coefficients $a,b,c\in S$ and indeterminate variables $x$ and +$y$.* +::: + +We have seen this type of equation before, in Bézout's Identity (Theorem +[36](#thm:NT_bezout_id){reference-type="ref" +reference="thm:NT_bezout_id"}). In Bézout's Identity, we have that the +greatest common divisor, $d$, of two integers $a,b$ can be expressed as + +$$\begin{equation*} + ax+by=d +\end{equation*}$$ + +for some $x,y\in\mathbb{Z}$. This gives us examples of already solved +equations, but what about the other way? Given an equation of the form + +$$\begin{equation*} + ax+by=c +\end{equation*}$$ + +with $a,b,c\in\mathbb{Z}$ given, can we find integer values for $x$ and +$y$?. That is, we are considering $ax+by=c$ to be a Diophantine +equation. If the reader is sufficiently alert, they will notice that by +mentioning Bézout's Identity we are hinting that it will be crucial to +finding the solutions. + +We know of one solution, namely if +$\mathop{\mathrm{GCD}}\left(a,b\right)=d$ and $c=d$ then the solution is +found by the Euclidean algorithm. Now if $c$ were a multiple of $d$ can +we find solutions? Recall proposition +[108](#prop:NT_GCD_properties){reference-type="ref" +reference="prop:NT_GCD_properties"} part 4. We have that +$\mathop{\mathrm{GCD}}\left(a,b\right)=d$ is the smallest such so that +$ax+by=d$, given that this is the smallest such then we can show that +there exist others, namely these solutions are multiples of $d$. + +::: {#prop:NT_bezout_extension .proposition} +**Proposition 129**. *Integer has form $ax+by$ if it is a multiple of +the greatest common divisor of $a$ and $b$* + +*Let $a,b\in\mathbb{Z}$ and $d=\mathop{\mathrm{GCD}}\left(a,b\right)$. +Let $c\in\mathbb{Z}$. We have that* + +*$$\begin{equation*} + c=ax+by +\end{equation*}$$* + +*if and only if $d\mid c$. Which is to say $c$ is a multiple of $d$* + +*Proof:* + +*$\left(\Rightarrow\right)$: Clearly if $c=ax+by$ then as +$d=\mathop{\mathrm{GCD}}\left(a,b\right)$ we have by proposition +[102](#prop:NT_divisibility_properties){reference-type="ref" +reference="prop:NT_divisibility_properties"} part 3 that $d\mid c$.* + +*$\left(\Leftarrow\right)$: Suppose that $c=de$ for some +$e\in\mathbb{Z}$. By Bézout's Identity, we have that +$\exists u,v\in\mathbb{Z}$ so that* + +*$$\begin{equation*} + d=au+bv +\end{equation*}$$* + +*where $d=\mathop{\mathrm{GCD}}\left(a,b\right)$. Multiplying both sides +by $e$ we get* + +*$$\begin{equation*} + c=aue+bve=ax+by +\end{equation*}$$* + +*Hence $x=ue$ and $y=ve$.* + +*As required. $\qed$* +::: + +Armed with this proposition we can find the solutions to the Diophantine +equation $ax+by=c$. + +::: {#prop:NT_solutions_to_two_var_linear_diophantine_equation .proposition} +**Proposition 130**. *Solutions to the Diophantine equation $ax+by=c$* + +*Let $a,b,c\in\mathbb{Z}$ be such that* + +*$$\begin{equation*} + ax+by=c +\end{equation*}$$* + +*for the indeterminate variables $x,y$ and let +$d=\mathop{\mathrm{GCD}}\left(a,b\right)$. We have that there are +solutions so that $x,y\in\mathbb{Z}$ if and only if $d\mid c$.* + +*Moreover, there are infinitely many solutions where the solutions are +given by* + +*$$\begin{align*} + x&=x_0+\frac{bn}{d}\\ + y&=y_0-\frac{an}{d} +\end{align*}$$* + +*where $x_0,y_0\in\mathbb{Z}$ is one solution.* + +*Proof:* + +*The existence of a solution is given by proposition +[129](#prop:NT_bezout_extension){reference-type="ref" +reference="prop:NT_bezout_extension"}. It is left to show that the +suggested solutions $x,y$ are solutions and that there are infinitely +many solutions. This follows the argument in example +[118](#exam:NT_solutions_to_ax_plus_by){reference-type="ref" +reference="exam:NT_solutions_to_ax_plus_by"}. We give the argument again +to refresh the reader's memory.* + +*Let $x_0,y_0\in\mathbb{Z}$ be a solution, then we have that* + +*$$\begin{equation*} + ax_0+by_0=c +\end{equation*}$$* + +*For any $n\in\mathbb{Z}$ let* + +*$$\begin{align*} + x&=x_0+\frac{bn}{d}\\ + y&=y_0-\frac{an}{d} +\end{align*}$$* + +*We then have that $\displaystyle\frac{bn}{d}\in\mathbb{Z}$ as $d\mid b$ +by definition of the greatest common divisor, likewise for +$\displaystyle\frac{ab}{d}$. Hence, we have that* + +*$$\begin{align*} + ax+by&=a\left(x_0+\frac{bn}{d}\right)+b\left(y_0-\frac{an}{d}\right)\\ + &=ax_0+a\frac{bn}{d}+by_0-b\frac{an}{d}\\ + &=ax_0+\frac{abn}{d}+by_0-\frac{abn}{d}\\ + &=ax_0+by_0=c\\ +\end{align*}$$* + +*Hence $x,y$ is a solution. Moreover, as $n\in\mathbb{Z}$ is any integer +we have shown that there are infinitely many solutions. It is left to +show that these are the only solutions.* + +*Let $x,y\in\mathbb{Z}$ be any solution to $ax+by=c$, and let +$x_0,y_0\in\mathbb{Z}$ be a particular solution. Hence* + +*$$\begin{equation*} + ax+by=ax_0by_0 +\end{equation*}$$* + +*Subtracting $ax_0by_0$ from the right-hand side gives* + +*$$\begin{align*} + ax+by-ax_0by_0&=0\\ + a\left(x-x_0\right)+b\left(y-y_0\right)&=0 +\end{align*}$$* + +*Now, as $d=\mathop{\mathrm{GCD}}\left(a,b\right)$ then we have that +$d\mid a$ and $d\mid b$ so that* + +*$$\begin{align*} + \frac{a}{d}\left(x-x_0\right)+\frac{b}{d}\left(y-y_0\right)&=0\\ + \frac{a}{d}\left(x-x_0\right)&=-\frac{b}{d}\left(y-y_0\right) +\end{align*}$$* + +*If $a=b=0$, we are done so suppose not. Then one of $a$ or $b$ is +non-zero. Without loss of generality, suppose that $a\neq 0$. We have +that by proposition [108](#prop:NT_GCD_properties){reference-type="ref" +reference="prop:NT_GCD_properties"} that if +$\mathop{\mathrm{GCD}}\left(a,b\right)=d$ then +$\displaystyle\mathop{\mathrm{GCD}}\left(\frac{a}{d},\frac{b}{d}\right)=1$, +moreover by definition of co-prime integers we have that +$\displaystyle\frac{a}{d}$ and $\displaystyle\frac{b}{d}$ are co-prime.* + +*By Euclid's lemma for co-primes (lemma +[11](#lem:NT_Euclid_co_primes){reference-type="ref" +reference="lem:NT_Euclid_co_primes"}) we have that +$\displaystyle\frac{a}{d} \mid-\left(y-y_0\right)$. Hence there is some +$n\in\mathbb{Z}$ so that* + +*$$\begin{equation*} + -\left(y-y_0\right)=n\frac{a}{d} +\end{equation*}$$* + +*Which is to say* + +*$$\begin{equation*} + y=y_0-\frac{an}{d} +\end{equation*}$$* + +*Similarly, we have that* + +*$$\begin{equation*} + x=x_0+\frac{bn}{d} +\end{equation*}$$* + +*As required. $\qed$* +::: + +##### Linear equations with more than two variables + +A natural question to ask now is what happens when we have more than two +indeterminate variables? For example $ax+by+cz=e$? We can take some +inspiration from the two variable case. + +Recall that for $ax+by=c$ with $d=\mathop{\mathrm{GCD}}\left(a,b\right)$ +that there are solutions with $x,y\in\mathbb{Z}$ if and only if +$d\mid c$. More importantly, we have that if +$d=\mathop{\mathrm{GCD}}\left(a,b\right)$ then we can express $d$ by +$d=ax+by$ for some $x,y\in\mathbb{Z}$ by Bézout's Identity. Moreover by +proposition +[103](#prop:NT_Divisor_dividing_all_in_set_divides_linear_combination){reference-type="ref" +reference="prop:NT_Divisor_dividing_all_in_set_divides_linear_combination"} +we have that for a set of $n$ integers +$S=\left\{b_1,b_2,b_3,\dots,b_n\right\}$ and additionally we have that +that $a\mid b_i$ for each $b_i\in S$ then + +$$\begin{equation*} + a\mid\sum_{i=1}^n b_i x_i +\end{equation*}$$ + +This hints at an extension to Bézout's Identity, given a suitable +extension to the definition of the greatest common divisor for more than +two inputs. Hence, our goal is to build this suitable extension to the +greatest common divisor. We will start by looking at some exploratory +examples before moving on with the generalisation. + +::: example +**Example 136**. *Let $a=2, b=4$ and $c=6$. What is +$\mathop{\mathrm{GCD}}\left(a,b,c\right)$? Clearly, by inspection, we +have that $2$ is the largest divisor of $a,b$ and $c$. In particular we +have that $\mathop{\mathrm{GCD}}\left(2,4\right)=2$ and +$\mathop{\mathrm{GCD}}\left(2,6\right)=2$. In other words, we have that* + +*$$\begin{equation*} + \mathop{\mathrm{GCD}}\left(2,4,6\right)=\mathop{\mathrm{GCD}}\left(\mathop{\mathrm{GCD}}\left(2,4\right),6\right) +\end{equation*}$$* + +*Equivalently, we could have first considered +$\mathop{\mathrm{GCD}}\left(4,6\right)=2$ and then +$\mathop{\mathrm{GCD}}\left(2,2\right)=2$ so we have* + +*$$\begin{equation*} + \mathop{\mathrm{GCD}}\left(2,4,6\right)=\mathop{\mathrm{GCD}}\left(2,\mathop{\mathrm{GCD}}\left(4,6\right)\right) +\end{equation*}$$* +::: + +::: example +**Example 137**. *Let $a=3, b=6$ and $c=30$. What is +$\mathop{\mathrm{GCD}}\left(a,b,c\right)$? Breaking this problem down we +have that $\mathop{\mathrm{GCD}}\left(3,6\right)=3$, +$\mathop{\mathrm{GCD}}\left(3,30\right)=3$ and +$\mathop{\mathrm{GCD}}\left(6,30\right)=6$. As the greatest common +divisor must divide all of the numbers we must conclude that +$\mathop{\mathrm{GCD}}\left(3,6,30\right)=3$.* +::: + +::: example +**Example 138**. *Let $a=3, b=5$ and $c=7$. As $a,b$ and $c$ are all +prime we clearly see that $\mathop{\mathrm{GCD}}\left(a,b,c\right)=1$* +::: + +::: example +**Example 139**. *Let $a=14$, $b=35$, $c=7$ and $d=5$. We again break +this down. We see that* + +*$$\begin{align*} + \mathop{\mathrm{GCD}}\left(14,33\right)&=7\\ + \mathop{\mathrm{GCD}}\left(14,7\right)&=7\\ + \mathop{\mathrm{GCD}}\left(14,5\right)&=1\\ + \mathop{\mathrm{GCD}}\left(35,7\right)&=5\\ + \mathop{\mathrm{GCD}}\left(35,5\right)&=7\\ + \mathop{\mathrm{GCD}}\left(7,5\right)&=1\\ +\end{align*}$$* + +*Again the greatest common divisor is the smallest value that divides +all of the inputs $a,b,c$ and $d$. The smallest such number here is $1$ +so $\mathop{\mathrm{GCD}}\left(14,35,7,5\right)=1$.* +::: + +In these examples, we made use of the fact that the greatest common +divisor of two numbers is the smallest number that divides both of the +input numbers. We then looked at all of the possible combinations of the +inputs and took the smallest value that occurred. This is to be +consistent with two variable version of the $\mathop{\mathrm{GCD}}$ that +we have already developed. This was shown explicitly in the first +example with + +$$\begin{equation*} + \mathop{\mathrm{GCD}}\left(2,4,6\right)=\mathop{\mathrm{GCD}}\left(\mathop{\mathrm{GCD}}\left(2,4\right),6\right)=\mathop{\mathrm{GCD}}\left(2,\mathop{\mathrm{GCD}}\left(4,6\right)\right) +\end{equation*}$$ + +Hence an immediate property that we can deduce is that the +$\mathop{\mathrm{GCD}}$ is associative, in the sense that computing the +$\mathop{\mathrm{GCD}}$ of three numbers is equivalent to computing the +$\mathop{\mathrm{GCD}}$ of two of the inputs with the remaining input. + +::: proposition +**Proposition 131**. *$\mathop{\mathrm{GCD}}$ is associative* + +*Let $a,b,c\in\mathbb{Z}$. We have that* + +*$$\begin{equation*} + \mathop{\mathrm{GCD}}\left(a,\mathop{\mathrm{GCD}}\left(b,c\right)\right)=\mathop{\mathrm{GCD}}\left(\mathop{\mathrm{GCD}}\left(a,b\right),c\right) +\end{equation*}$$* + +*Proof:* + +*Let +$x=\mathop{\mathrm{GCD}}\left(a,\mathop{\mathrm{GCD}}\left(b,c\right)\right)$ +and +$y=\mathop{\mathrm{GCD}}\left(\mathop{\mathrm{GCD}}\left(a,b\right),c\right)$, +We need to show that $x\mid y$ and $y\mid x$ then we can conclude that +$x=y$.* + +*As +$x=\mathop{\mathrm{GCD}}\left(a,\mathop{\mathrm{GCD}}\left(b,c\right)\right)$ +then by definition of the greatest common divisor, we have that +$x\mid a$ and $x\mid\mathop{\mathrm{GCD}}\left(b,c\right)$. Moreover as +$x\mid\mathop{\mathrm{GCD}}\left(b,c\right)$ then again by definition of +the greatest common divisor we have that $x\mid b$ and $x\mid c$.* + +*As $x\mid a$ and $x\mid b$ then +$x\mid\mathop{\mathrm{GCD}}\left(a,b\right)$ and likewise $x\mid c$ so +$x\mid\mathop{\mathrm{GCD}}\left(\mathop{\mathrm{GCD}}\left(a,b\right),c\right)$ +by definition and so $x\mid y$. The proof that $y\mid x$ is similar.* + +*As $x\mid y$ and $y\mid x$ and $x>0$ and $y>0$ we conclude that $x=y$ +as required. $\qed$* +::: + +To extend our definition of the greatest common divisor to more than two +inputs, we will use the definition of the $\mathop{\mathrm{GCD}}$ given +by the decomposition of primes. That is to say, given +$a,b\in\mathbb{Z}$, we know that there exists a set of primes + +$$\begin{equation*} + T=\left\{t_1,t_2,\dots,t_v\right\} +\end{equation*}$$ + +So that $a$ and $b$ can be represented by a prime factorisation of +primes $t_i\in T$. That is + +$$\begin{align*} + a&=\prod_{i=1}^v t_i^{e_i}\\ + b&=\prod_{i=1}^v t_i^{f_i}\\ +\end{align*}$$ + +We then have that the greatest common divisor is given by + +$$\begin{equation*} + \mathop{\mathrm{GCD}}\left(a,b\right)=t_1^{\min\left(e_1,f_1\right)}t_2^{\min\left(e_2,f_2\right)}t_3^{\min\left(e_3,f_3\right)}\dots t_v^{\min\left(e_v,f_v\right)} +\end{equation*}$$ + +Firstly, we will extend the result of proposition +[115](#prop:NT_express_primes_in_common_basis){reference-type="ref" +reference="prop:NT_express_primes_in_common_basis"} to the case of $n$ +integers, the proof is similar to proposition +[115](#prop:NT_express_primes_in_common_basis){reference-type="ref" +reference="prop:NT_express_primes_in_common_basis"}. + +::: {#prop:NT_General_express_primes_in_common_basis .proposition} +**Proposition 132**. *Expression of set of integers as powers of same +primes* + +*Let $S=\left\{a_1,a_2,a_3,\dots,a_n\right\}$ be such that +$a_i\in\mathbb{Z}$ and $a_i>2$ for $1\leq i\leq n$. For each $a_i$ let +its prime factorisation be denoted by* + +*$$\begin{equation*} + \mathlarger{a_i=\prod_{\substack{p_{\left(i,k\right)\mid a_i} \\ p_{\left(i,k\right)}\text{ is prime}}} p_{\left(i,k\right)}^{e_{\left(i,k\right)}}} +\end{equation*}$$* + +*where $\left(i,k\right)$ is a index tuple with $i$ denoting one of the +primes and $k$ denoting the $k$-th element of $a_i$'s prime +factorisation. Then there exists a set of primes* + +*$$\begin{equation*} + T=\left\{t_1,t_2,t_3\dots,t_v\right\} +\end{equation*}$$* + +*with $t_12$, then we have that* + +*$$\begin{equation*} + \min\left(\min\left(a_1,a_2,a_3,\dots,a_{k-1}\right),a_k\right) +\end{equation*}$$* + +*is well-defined. We show that* + +*$$\begin{equation*} + \min\left(\min\left(a_1,a_2,a_3,\dots,a_{k}\right),a_{k+1}\right) +\end{equation*}$$* + +*is well-defined. Evaluating the inner +$\min\left(a_1,a_2,a_3,\dots,a_{k}\right)$ we have by definition that* + +*$$\begin{equation*} + \min\left(a_1,a_2,a_3,\dots,a_{k}\right)=\min\left(\min\left(a_1,a_2,a_3,\dots,a_{k-1}\right),a_k\right) +\end{equation*}$$* + +*Which by hypothesis is well-defined. Hence +$\min\left(a_1,a_2,a_3,\dots,a_{k}\right)=m$ for some $m\in\mathbb{Z}$. +Hence we have that* + +*$$\begin{equation*} + \min\left(\min\left(a_1,a_2,a_3,\dots,a_{k}\right),a_{k+1}\right)=\min\left(m,a_{k+1}\right) +\end{equation*}$$* + +*Which is well-defined. Hence by induction, we have that the general +minimum function on the integers is well-defined. $\qed$* +::: + +We also have the following proposition. + +::: {#prop:NT_general_min_function_on_integers_is_associative .proposition} +**Proposition 134**. *The general minimum function is associative* + +*Let $S=\left(a_1,a_2,a_3,\dots,a_n\right)\in\mathbb{Z}^n$ be a +$n$-tuple of integers. We have that* + +*$$\begin{equation*} + \min\left(\min\left(a_1,a_2,a_3,\dots,a_{n-1}\right),a_n\right)=\min\left(a_1,\min\left(a_2,a_3,\dots,a_{n-1},a_n\right)\right) +\end{equation*}$$* + +*Proof:* + +*We argue by induction on $n$. The case $n=1$ has nothing to prove. +Likewise for $n=2$, so we shall show it holds for $n=3$. That is* + +*$$\begin{equation*} + \min\left(\min\left(a_1,a_2\right),a_3\right)=\min\left(a_1,\min\left(a_2,a_3\right)\right) +\end{equation*}$$* + +*There are $6$ cases to consider.* + +1. *$a_1\leq a_2\leq a_3$* + +2. *$a_1\leq a_3\leq a_2$* + +3. *$a_2\leq a_1\leq a_3$* + +4. *$a_2\leq a_3\leq a_1$* + +5. *$a_3\leq a_1\leq a_2$* + +6. *$a_3\leq a_2\leq a_1$* + + + +1. *$a_1\leq a_2\leq a_3$:* + + *We have that* + + *$$\begin{align*} + \min\left(\min\left(a_1,a_2\right),a_3\right)&=\min\left(a_1,a_3\right)=a_1\\ + \min\left(a_1,\min\left(a_2,a_3\right)\right)&=\min\left(a_1,a_2\right)=a_1\\ + \end{align*}$$* + +2. *$a_1\leq a_3\leq a_2$:* + + *$$\begin{align*} + \min\left(\min\left(a_1,a_2\right),a_3\right)&=\min\left(a_1,a_3\right)=a_1\\ + \min\left(a_1,\min\left(a_2,a_3\right)\right)&=\min\left(a_1,a_3\right)=a_1\\ + \end{align*}$$* + +3. *$a_2\leq a_1\leq a_3$:* + + *$$\begin{align*} + \min\left(\min\left(a_1,a_2\right),a_3\right)&=\min\left(a_2,a_3\right)=a_2\\ + \min\left(a_1,\min\left(a_2,a_3\right)\right)&=\min\left(a_1,a_2\right)=a_2\\ + \end{align*}$$* + +4. *$a_2\leq a_3\leq a_1$:* + + *$$\begin{align*} + \min\left(\min\left(a_1,a_2\right),a_3\right)&=\min\left(a_2,a_3\right)=a_2\\ + \min\left(a_1,\min\left(a_2,a_3\right)\right)&=\min\left(a_1,a_2\right)=a_2\\ + \end{align*}$$* + +5. *$a_3\leq a_1\leq a_2$:* + + *$$\begin{align*} + \min\left(\min\left(a_1,a_2\right),a_3\right)&=\min\left(a_1,a_3\right)=a_3\\ + \min\left(a_1,\min\left(a_2,a_3\right)\right)&=\min\left(a_1,a_3\right)=a_3\\ + \end{align*}$$* + +6. *$a_3\leq a_2\leq a_1$:* + + *$$\begin{align*} + \min\left(\min\left(a_1,a_2\right),a_3\right)&=\min\left(a_2,a_3\right)=a_3\\ + \min\left(a_1,\min\left(a_2,a_3\right)\right)&=\min\left(a_2,a_3\right)=a_3\\ + \end{align*}$$* + +*Hence the base case is shown. Now suppose that the proposition holds +for some $k>3$, that is* + +*$$\begin{equation*} + \min\left(\min\left(a_1,a_2,a_3,\dots,a_{k-1}\right),k_n\right)=\min\left(a_1,\min\left(a_2,a_3,\dots,a_{k-1},a_k\right)\right) +\end{equation*}$$* + +*we show that it holds for $k+1$, i.e.* + +*$$\begin{equation*} + \min\left(\min\left(a_1,a_2,a_3,\dots,a_{k}\right),a_{k+1}\right)=\min\left(a_1,\min\left(a_2,a_3,\dots,a_{k},a_{k+1}\right)\right) +\end{equation*}$$* + +*We have by evaluating the inner minimum of the left-hand side we get* + +*$$\begin{equation*} + \min\left(a_1,a_2,a_3,\dots,a_{k}\right)=\min\left(\min\left(a_1,a_2,a_3,\dots,a_{k-1}\right),a_{k}\right) +\end{equation*}$$* + +*And so by the induction hypothesis, we have that* + +*$$\begin{align*} + \min\left(\min\left(a_1,a_2,a_3,\dots,a_{k}\right),a_{k+1}\right)&=\min\left(\min\left(\min\left(a_1,a_2,a_3,\dots,a_{k-1}\right),a_{k}\right),a_{k+1}\right)\\ + &=\min\left(\min\left(a_1,\min\left(a_2,a_3,\dots,a_{k-1},a_{k}\right)\right),a_{k+1}\right),\ \text{Induction hypothesis}\\ +\end{align*}$$* + +*As $\min\left(a_2,a_3,\dots,a_{k-1},a_{k}\right)$ is well-defined by +proposition +[133](#prop:NT_general_min_on_integers_is_well_defined){reference-type="ref" +reference="prop:NT_general_min_on_integers_is_well_defined"} then +$\min\left(a_2,a_3,\dots,a_{k-1},a_{k}\right)=M$ say where +$M\in\mathbb{Z}$. Therefore, on substituting +$\min\left(a_2,a_3,\dots,a_{k-1},a_{k}\right)$ for $M$ for ease of +reading we have* + +*$$\begin{align*} + \min\left(\min\left(a_1,\min\left(a_2,a_3,\dots,a_{k-1},a_{k}\right)\right),a_{k+1}\right)&=\min\left(\min\left(a_1,M\right),a_{k+1}\right)\\ + &=\min\left(a_1,\min\left(M,a_{k+1}\right)\right)\\ + &=\min\left(a_1,\min\left(\min\left(a_2,a_3,\dots, a_{k-1},a_{k}\right),a_{k+1}\right)\right)\\ + &=\min\left(a_1,\min\left(a_2,a_3,\dots,a_{k},a_{k+1}\right)\right) +\end{align*}$$* + +*The result now follows by induction. $\qed$* +::: + +Proposition +[134](#prop:NT_general_min_function_on_integers_is_associative){reference-type="ref" +reference="prop:NT_general_min_function_on_integers_is_associative"} is +a useful proposition, it allows us to discard the cumbersome notation of +the definition of the general minimum function on the Integers. That is +to say, we can now simply, and more easily write + +$$\begin{equation*} + \min\left(a_1,a_2,a_3,\dots,a_n\right) +\end{equation*}$$ + +For convenience, we also define the minimum function for a subset of $n$ +integers. + +::: definition +**Definition 174**. *General minimum function for a subset of integers* + +*Let $A=\left\{a_1,a_2,a_3,\dots,a_n\right\}\subset\mathbb{Z}$ be a +subset of $n$ integers. Let +$S=\left(a_1,a_2,a_3,\dots,a_n\right)\in A^n$. We define the minimum of +the set of integers $A$ by* + +*$$\begin{equation*} + \min\left(A\right)=\min\left(S\right)=\min\left(a_1,a_2,a_3,\dots,a_n\right) +\end{equation*}$$* + +*That is, we simply take the element of $A^n$ which corresponds to the +set.* +::: + +::: example +**Example 140**. *Let $A=\left\{2,3\right\}$. We have that* + +*$$\begin{equation*} + A^2=\left\{\left(2,2\right), \left(2,3\right), \left(3,2\right),\left(3,3\right)\right\} +\end{equation*}$$* + +*We have that $S=\left(2,3\right)\in A^2$ and* + +*$$\begin{equation*} + \min\left(A\right)=\min\left(S\right)=\min\left(2,3\right)=2 +\end{equation*}$$* +::: + +We have all the ingredients required to extend the +$\mathop{\mathrm{GCD}}$ function. We use a method similar to how we +extended the minimum function. + +::: definition +**Definition 175**. *Generalised greatest common divisor* + +*Let $S=\left(a_1,a_2,a_3,\dots,a_n\right)\in\mathbb{Z}^n$ be a +$n$-tuple of integers. We define the greatest common divisor function on +$S$ by* + +*$$\begin{align*} + \mathop{\mathrm{GCD}}:\mathbb{Z}^n&\rightarrow\mathbb{Z}\\ + S&\mapsto\mathop{\mathrm{GCD}}\left(S\right)=\begin{cases} + a_1,\ &\text{If } n=1\\ + \mathop{\mathrm{GCD}}\left(a_1,a_2\right),\ &\text{If } n=2\\ + \mathop{\mathrm{GCD}}\left(\mathop{\mathrm{GCD}}\left(a_1,a_2,a_3,\dots,a_{n-1}\right),a_n\right),\ &\text{If } n\geq 3\\ + \end{cases} +\end{align*}$$* +::: + +We show that this is well-defined. + +::: {#prop:NT_general_gcd_on_integers_is_well_defined .proposition} +**Proposition 135**. *Generalised greatest common divisor function for +the integers is well-defined* + +*Let $S=\left(a_1,a_2,a_3,\dots,a_n\right)\in\mathbb{Z}^n$ be a +$n$-tuple of integers. We have that $\gcd\left(S\right)$ is +well-defined.* + +*Proof:* + +*The argument is by induction on $n$. The base case is $n=2$ which is +well-defined by theorem [32](#thm:NT_gcd_exists){reference-type="ref" +reference="thm:NT_gcd_exists"}. Now suppose the result is true for some +$k>2$, that is* + +*$$\begin{equation*} + \mathop{\mathrm{GCD}}\left(\mathop{\mathrm{GCD}}\left(a_1,a_2,a_3,\dots,a_{k-1}\right),a_k\right) +\end{equation*}$$* + +*is well-defined. We show that* + +*$$\begin{equation*} + \mathop{\mathrm{GCD}}\left(\mathop{\mathrm{GCD}}\left(a_1,a_2,a_3,\dots,a_{k}\right),a_{k+1}\right) +\end{equation*}$$* + +*is well-defined. Evaluating the inner +$\mathop{\mathrm{GCD}}\left(a_1,a_2,a_3,\dots,a_{k}\right)$ we have by +definition that* + +*$$\begin{equation*} + \mathop{\mathrm{GCD}}\left(a_1,a_2,a_3,\dots,a_{k}\right)=\mathop{\mathrm{GCD}}\left(\mathop{\mathrm{GCD}}\left(a_1,a_2,a_3,\dots,a_{k-1}\right),a_k\right) +\end{equation*}$$* + +*Which by hypothesis is well-defined. Hence +$\mathop{\mathrm{GCD}}\left(a_1,a_2,a_3,\dots,a_{k}\right)=d$ for some +$d\in\mathbb{Z}$. Hence we have that* + +*$$\begin{equation*} + \mathop{\mathrm{GCD}}\left(\mathop{\mathrm{GCD}}\left(a_1,a_2,a_3,\dots,a_{k}\right),a_{k+1}\right)=\mathop{\mathrm{GCD}}\left(d,a_{k+1}\right) +\end{equation*}$$* + +*Which is well-defined. The result now follows by induction. $\qed$* +::: + +As with the minimum function, to avoid cumbersome notation we can show +that the generalised greatest common divisor is associative. + +::: {#prop:NT_general_gcd_on_integers_is_associative .proposition} +**Proposition 136**. *Generalised $\mathop{\mathrm{GCD}}$ is +associative* + +*Let $S=\left(a_1,a_2,a_3,\dots,a_n\right)\in\mathbb{Z}^n$ be a +$n$-tuple of integers. We have that* + +*$$\begin{equation*} + \mathop{\mathrm{GCD}}\left(\mathop{\mathrm{GCD}}\left(a_1,a_2,a_3,\dots,a_{n-1}\right),a_n\right)=\mathop{\mathrm{GCD}}\left(a_1,\mathop{\mathrm{GCD}}\left(a_2,a_3,\dots,a_{n-1},a_n\right)\right) +\end{equation*}$$* + +*Proof:* + +*We argue by induction on $n$. The cases of $n=1$ and $n=2$ are trivial, +so we show it holds for $n=3$.* + +*Let +$x=\mathop{\mathrm{GCD}}\left(a_1,\mathop{\mathrm{GCD}}\left(a_2,a_3\right)\right)$ +and +$y=\mathop{\mathrm{GCD}}\left(\mathop{\mathrm{GCD}}\left(a_1,a_2\right),a_3\right)$, +We need to show that $x\mid y$ and $y\mid x$ then we can conclude that +$x=y$.* + +*As +$x=\mathop{\mathrm{GCD}}\left(a_1,\mathop{\mathrm{GCD}}\left(a_2,a_3\right)\right)$ +then by definition of the greatest common divisor, we have that +$x\mid a_1$ and $x\mid\mathop{\mathrm{GCD}}\left(a_2,a_3\right)$. +Moreover, as $x\mid\mathop{\mathrm{GCD}}\left(a_2,a_3\right)$ then again +by definition of the greatest common divisor we have that $x\mid a_2$ +and $x\mid a_3$.* + +*As $x\mid a_1$ and $x\mid a_2$ then +$x\mid\mathop{\mathrm{GCD}}\left(a_1,a_2\right)$ and likewise +$x\mid a_3$ so +$x\mid\mathop{\mathrm{GCD}}\left(\mathop{\mathrm{GCD}}\left(a_1,a_2\right),a_3\right)$ +by definition and so $x\mid y$. The proof that $y\mid x$ is similar.* + +*As $x\mid y$ and $y\mid x$ and $x>0$ and $y>0$ we conclude that $x=y$ +as required.* + +*Now suppose the result is true for some $k>2$. That is* + +*$$\begin{equation*} + \mathop{\mathrm{GCD}}\left(\mathop{\mathrm{GCD}}\left(a_1,a_2,a_3,\dots,a_{k-1}\right),a_k\right)=\mathop{\mathrm{GCD}}\left(a_1,\mathop{\mathrm{GCD}}\left(a_2,a_3,\dots,a_{k-1},a_k\right)\right) +\end{equation*}$$* + +*we show that* + +*$$\begin{equation*} + \mathop{\mathrm{GCD}}\left(\mathop{\mathrm{GCD}}\left(a_1,a_2,a_3,\dots,a_{k}\right),a_{k+1}\right)=\mathop{\mathrm{GCD}}\left(a_1,\mathop{\mathrm{GCD}}\left(a_2,a_3,\dots,a_{k},a_{k+1}\right)\right) +\end{equation*}$$* + +*Evaluation of the inner $\mathop{\mathrm{GCD}}$ of the left-hand side +yields* + +*$$\begin{equation*} + \mathop{\mathrm{GCD}}\left(a_1,a_2,a_3,\dots,a_{k}\right)=\mathop{\mathrm{GCD}}\left(\mathop{\mathrm{GCD}}\left(a_1,a_2,a_3,\dots,a_{k-1}\right)a_{k}\right) +\end{equation*}$$* + +*So by the induction hypothesis, we have that* + +*$$\begin{align*} + \mathop{\mathrm{GCD}}\left(\mathop{\mathrm{GCD}}\left(a_1,a_2,a_3,\dots,a_{k}\right),a_{k+1}\right)&=\mathop{\mathrm{GCD}}\left(\mathop{\mathrm{GCD}}\left(\mathop{\mathrm{GCD}}\left(a_1,a_2,a_3,\dots,a_{k-1}\right)a_{k}\right),a_{k+1}\right)\\ + &=\mathop{\mathrm{GCD}}\left(\mathop{\mathrm{GCD}}\left(a_1,\mathop{\mathrm{GCD}}\left(a_2,a_3,\dots,a_{k-1},a_{k}\right)\right),a_{k+1}\right),\ \text{By hypothesis}\\ +\end{align*}$$* + +*As $\mathop{\mathrm{GCD}}\left(a_2,a_3,\dots,a_{k-1},a_{k}\right)$ is +well-defined by proposition +[135](#prop:NT_general_gcd_on_integers_is_well_defined){reference-type="ref" +reference="prop:NT_general_gcd_on_integers_is_well_defined"}, we have +$\mathop{\mathrm{GCD}}\left(a_2,a_3,\dots,a_{k-1},a_{k}\right)=d$ with +$d\in\mathbb{Z}$. Hence we have* + +*$$\begin{align*} + \mathop{\mathrm{GCD}}\left(\mathop{\mathrm{GCD}}\left(a_1,\mathop{\mathrm{GCD}}\left(a_2,a_3,\dots,a_{k-1},a_{k}\right)\right),a_{k+1}\right)&=\mathop{\mathrm{GCD}}\left(\mathop{\mathrm{GCD}}\left(a_1,d\right),a_{k+1}\right)\\ + &=\mathop{\mathrm{GCD}}\left(a_1,\mathop{\mathrm{GCD}}\left(d,a_{k+1}\right)\right)\\ + &=\mathop{\mathrm{GCD}}\left(a_1,\mathop{\mathrm{GCD}}\left(\mathop{\mathrm{GCD}}\left(a_2,a_3,\dots,a_{k-1},a_{k}\right),a_{k+1}\right)\right)\\ +\end{align*}$$* + +*As required. $\qed$* +::: + +As with the minimum function, we can now simply write + +$$\begin{equation*} + \mathop{\mathrm{GCD}}\left(a_1,a_2,a_3,\dots,a_{n-1},a_n\right) +\end{equation*}$$ Likewise for convenience, we define the +$\mathop{\mathrm{GCD}}$ function for a subset of $n$ integers. + +::: definition +**Definition 176**. *General greatest common divisor function for a +subset of integers* + +*Let $A=\left\{a_1,a_2,a_3,\dots,a_n\right\}\subset\mathbb{Z}$ be a +subset of $n$ integers. Let +$S=\left(a_1,a_2,a_3,\dots,a_n\right)\in A^n$. We define the +$\mathop{\mathrm{GCD}}$ of the set of integers $A$ by* + +*$$\begin{equation*} + \mathop{\mathrm{GCD}}\left(A\right)=\mathop{\mathrm{GCD}}\left(S\right)=\mathop{\mathrm{GCD}}\left(a_1,a_2,a_3,\dots,a_n\right) +\end{equation*}$$* + +*That is, we simply take the element of $A^n$ which corresponds to the +set.* +::: + +::: example +**Example 141**. *Let $A=\left\{2,3\right\}$. We have that* + +*$$\begin{equation*} + A^2=\left\{\left(2,2\right), \left(2,3\right), \left(3,2\right),\left(3,3\right)\right\} +\end{equation*}$$* + +*We have that $S=\left(2,3\right)\in A^2$ and* + +*$$\begin{equation*} + \mathop{\mathrm{GCD}}\left(A\right)=\mathop{\mathrm{GCD}}\left(S\right)=\mathop{\mathrm{GCD}}\left(2,3\right)=1 +\end{equation*}$$* +::: + +We can now finally generalise the computation of the greatest common +divisor from the prime factorisation of the inputs. + +::: {#prop:NT_general_gcd_can_be_computed_by_primes .proposition} +**Proposition 137**. *Generalised version of the greatest common divisor +from prime factorisation* + +*Let $S=\left\{a_1,a_2,a_3,\dots,a_n\right\}\subset\mathbb{Z}$ be a set +of integers so that at least one $a_i\neq 0$ for $1\leq i\leq n$. By +proposition +[132](#prop:NT_General_express_primes_in_common_basis){reference-type="ref" +reference="prop:NT_General_express_primes_in_common_basis"}, we know +that there exists a set of primes* + +*$$\begin{equation*} + T=\left\{t_1,t_2,t_3,\dots,t_v\right\} +\end{equation*}$$* + +*so that for each $a_i$ we have prime factorisations given by* + +*$$\begin{equation*} + \mathlarger{a_i=\prod_{j=1}^v t_{j}^{f_{\left(i,j\right)}}} +\end{equation*}$$ For $1\leq i\leq n$. Define the family of sets for +each $1\leq j\leq v$* + +*$$\begin{equation*} + P_j=\left\{f_{\left(i,j\right)} : 1\leq i\leq n\right\} +\end{equation*}$$* + +*We have that the greatest common divisor +$\mathop{\mathrm{GCD}}\left(a_1,a_2,a_3,dots,a_n\right)$ is given by* + +*$$\begin{equation*} + \mathop{\mathrm{GCD}}\left(a_1,a_2,a_3,\dots,a_n\right)=t_1^{\min\left(P_1\right)}t_2^{\min\left(P_2\right)}t_3^{\min\left(P_3\right)}\dots t_v^{\min\left(P_v\right)} +\end{equation*}$$* + +*Proof:* + +*The proof is similar to that of proposition +[116](#prop:NT_gcd_can_be_computed_by_primes){reference-type="ref" +reference="prop:NT_gcd_can_be_computed_by_primes"}. Let +$S=\left\{a_1,a_2,a_3,\dots,a_n\right\}\subset\mathbb{Z}$ be as given so +by proposition +[132](#prop:NT_General_express_primes_in_common_basis){reference-type="ref" +reference="prop:NT_General_express_primes_in_common_basis"} we have a +set of primes* + +*$$\begin{equation*} + T=\left\{t_1,t_2,t_3,\dots,t_v\right\} +\end{equation*}$$* + +*so that for each $a_i$ we have prime factorisations given by* + +*$$\begin{equation*} + \mathlarger{a_i=\prod_{j=1}^v t_{j}^{f_{\left(i,j\right)}}} +\end{equation*}$$* + +*Now, let $d=\mathop{\mathrm{GCD}}\left(a_1,a_2,a_3,\dots,a_n\right)$ +and let +$D = t_1^{\min\left(P_1\right)}t_2^{\min\left(P_2\right)}t_3^{\min\left(P_3\right)}\dots t_v^{\min\left(P_v\right)}$, +we show that $d\leq D$ and $D\leq d$. Define +$\sigma_j=\min\left(\left\{f_{\left(i,j\right)}: 1\leq i\leq n\right\}\right)$ +for $1\leq j\leq v$.* + +1. *$D\leq d$:* + + *By the definition of the minimum, we have that + $\sigma_j\leq f_{\left(i,j\right)}$ for each $1\leq i\leq n$. Hence, + for each $i$ and $j$ there exists + $k_{\left(i,j\right)}\in\mathbb{Z}$ so that* + + *$$\begin{equation*} + f_{\left(i,j\right)} = \sigma_j + k_{\left(i,j\right)} + \end{equation*}$$* + + *So that $a_i$ can be expressed as* + + *$$\begin{align*} + a_i&=\prod_{j=1}^v t_j^{f_{\left(i,j\right)}}\\ + &=\prod_{j=1}^v t_j^{\sigma_j+k_{\left(i,j\right)}}\\ + &=\prod_{j=1}^v t_j^{\sigma_j} t_j^{k_{\left(i,j\right)}}\\ + &=\prod_{j=1}^v t_j^{\sigma_j} \prod_{j=1}^vt_j^{k_{\left(i,j\right)}}\\ + &= D * \prod_{j=1}^vt_j^{k_{\left(i,j\right)}} + \end{align*}$$* + + *As $a_i$ was arbitrary this argument holds for each + $1\leq i\leq n$. Hence, we have that $D\mid a_i$ for each $i$, so + $D$ is a common divisor of each $a_i$. We conclude that $D\leq d$.* + +2. *$d\leq D$:* + + *Suppose that $d\mid D$ then $\exists k\in\mathbb{Z}$ so that* + + *$$\begin{equation*} + d=DK + \end{equation*}$$* + + *Now, $k$ has a factorisation into primes by the fundamental theorem + of arithmetic. Moreover, $k$ could have primes in common with $D$, + so we can take those primes that are in common with $D$ and $k$ and + place them into the factorisation of $D$. That is* + + *$$\begin{align*} + d&=Dk\\ + d&=t_1^{\sigma_1}t_2^{\sigma_1}t_3^{\sigma_3}\dots t_v^{\sigma_v}k\\ + d&=t_1^{\lambda_1}t_2^{\lambda_1}t_3^{\lambda_3}\dots t_v^{\lambda_v}k'\\ + \end{align*}$$* + + *Where $\lambda_j$ are the new values for each prime after + extracting the primes in common with $D$ and $k$ into $D$. $k'$ are + the primes that are not in common. We need to show that* + + 1. *$k'=1$* + + 2. *$\lambda_j\leq \sigma_j$ for all $1\leq j\leq v$* + + + + 1. *$k'=1$:* + + *Suppose for a contradiction that $k'\neq 1$. As $d>0$ and $D>0$ + then $k>0$ and so $k'>0$. Now as $k'\neq 1$ we have $k'>1$ and + so by the fundamental theorem of arithmetic we have that $k'$ + has a factorisation into primes, say* + + *$$\begin{equation*} + k'=q_1^{r_1}q_2^{r_2}q_3^{r_3}\dots q_c^{r_c} + \end{equation*}$$* + + *Now, no $q_l=t_j$ as $k'$ has no primes in common with + $t_1^{\lambda_1}t_2^{\lambda_1}t_3^{\lambda_3}\dots t_v^{\lambda_v}$. + Pick one of the primes in $k'$, say $q=q_l$ then $q\mid d$. Now + as $d=\gcd\left(a_1,a_2,a_3,\dots,a_n\right)$ then we have + $q\mid a_i$ for at least one $a_i$. This is a contradiction as + then $q$ is one of the primes $t_j$. We conclude that $k'=1$* + + 2. *$\lambda_j\leq \sigma_j$ for all $1\leq j\leq v$:* + + *Suppose for contraction that $\lambda_j>\sigma_j$ for all + $1\leq j\leq v$. Without loss of generality, take $j=1$, for if + not re-label the primes.* + + *By definition of $\sigma_1$, we have that + $\sigma_1=\min\left(\left\{f_{\left(i,1\right)}: 1\leq i\leq n\right\}\right)$, + without loss of generality take $i=1$ as the case for the other + values of $i$ are similar. We have that + $\sigma_1=f_{\left(1,1\right)}$ and so + $\lambda_1>f_{\left(1,1\right)}$. As $d$ is the greatest common + divisor of $a_1$ then there is an $s\in\mathbb{Z}$ so that + $ds=a$ where $s>0$ as both $a$ and $d$ are.* + + *Comparing the prime factorisations, we get that* + + *$$\begin{equation*} + s*t_1^{\lambda_1}t_2^{\lambda_2}t_3^{\lambda_3}\dots t_v^{\lambda_v}=t_1^{f_{\left(1,1\right)}}t_2^{f_{\left(1,2\right)}}t_3^{f_{\left(1,3\right)}}\dots t_v^{f_{\left(1,v\right)}} + \end{equation*}$$* + + *Dividing by $\displaystyle t_1^{f_{\left(1,1\right)}}$ we get + that* + + *$$\begin{equation*} + s*t_1^{\lambda_1-f_{\left(1,1\right)}}t_2^{\lambda_2}t_3^{\lambda_3}\dots t_v^{\lambda_v}=t_1^{f_{\left(1,1\right)}-f_{\left(1,1\right)}}t_2^{f_{\left(1,2\right)}}t_3^{f_{\left(1,3\right)}}\dots t_v^{f_{\left(1,v\right)}} + \end{equation*}$$* + + *Where clearly + $\displaystyle t_1^{f_{\left(1,1\right)}-f_{\left(1,1\right)}}=1$. + So this can be re-written as* + + *$$\begin{equation*} + s*t_1^{\lambda_1-f_{\left(1,1\right)}}t_2^{\lambda_2}t_3^{\lambda_3}\dots t_v^{\lambda_v}=t_2^{f_{\left(1,2\right)}}t_3^{f_{\left(1,3\right)}}\dots t_v^{f_{\left(1,v\right)}} + \end{equation*}$$* + + *As $\lambda_1>f_{\left(1,1\right)}$ then + $\lambda_1-f_{\left(1,1\right)}>0$ and so $t_1$ divides the + left-hand side of the equation. By the fundamental theorem of + arithmetic, $t_1$ divides the left-hand side it must also divide + the right-hand side and therefore be in the factorisation. It is + not in the factorisation on the right-hand side which is a + contradiction. It follows $\lambda_j\leq\sigma_j$ for all + $1\leq j\leq v$* + + *Therefore we conclude that $d\leq D$.* + + *As $d\leq D$ and $D\leq d$ we have that $d=D$ and the result is + shown. $\qed$* +::: + +These last few results were somewhat technical. To show that our new +generalised $\mathop{\mathrm{GCD}}$ works we give an example. + +::: example +**Example 142**. *We compute +$\mathop{\mathrm{GCD}}\left(54,78,35,144,50\right)$. By inspection of +each of the numbers we have that* + +*$$\begin{align*} + 54&=2*3^3\\ + 78&=2*3*13\\ + 35&=5*7\\ + 144&=2^4*3^2\\ + 50&=2*5^2 +\end{align*}$$* + +*Hence, the set of primes $T$ is given by* + +*$$\begin{equation*} + T=\left\{2,3,5,7,13\right\} +\end{equation*}$$* + +*Now, by the proposition, we know that* + +*$$\begin{equation*} + \mathop{\mathrm{GCD}}\left(54,78,35,144,50\right)=t_1^{\min\left(P_1\right)}t_2^{\min\left(P_2\right)}t_3^{\min\left(P_3\right)}t_4^{\min\left(P_4\right)}t_5^{\min\left(P_5\right)} +\end{equation*}$$* + +*Where $P_j$ will be the powers of the prime $t_j$ that appear in the +factorisation of each of the inputs. Taking $t_1=2, t_2=3, t_3=5, t_4=7$ +and $t_5=13$ we have* + +*$$\begin{align*} + P_1&=\left\{1,1,0,4,1\right\}=\left\{0,1,4\right\}\\ + P_2&=\left\{3,1,0,2,0\right\}=\left\{0,1,2,3\right\}\\ + P_3&=\left\{0,0,1,0,2\right\}=\left\{0,1,2\right\}\\ + P_4&=\left\{0,0,1,0,0\right\}=\left\{0,1\right\}\\ + P_5&=\left\{0,1,0,0,0\right\}=\left\{0,1\right\}\\ +\end{align*}$$ From which it is clear that the minimum of every $P_j$ is +$0$. So that* + +*$$\begin{equation*} + \mathop{\mathrm{GCD}}\left(54,78,35,144,50\right)=1 +\end{equation*}$$* +::: + +With a generalised $\mathop{\mathrm{GCD}}$ function, we can extend +Bézout's Identity. + +::: {#thm:NT_general_bezout_idenity .theorem} +**Theorem 44**. *Generalised Bézout's Identity* + +*Let $S=\left\{a_1,a_2,a_3,\dots,a_n\right\}\subset\mathbb{Z}$ be a set +of integers so that at least one $a_i\neq 0$ for $1\leq i\leq n$. +Consider $d=\mathop{\mathrm{GCD}}\left(a_1,a_2,a_3,\dots ,a_n\right)$. +Then, for $i\leq 1\leq n$ we have $\exists x_i\in\mathbb{Z}$ so that* + +*$$\begin{equation*} + d=a_1x_1+a_2x_2+a_2x_2+\dots+a_nx_n=\sum_{i=1}^n a_ix_n +\end{equation*}$$* + +*Proof:* + +*Let $S$ be as given by the hypothesis and let +$d=\mathop{\mathrm{GCD}}\left(a_1,a_2,a_3,\dots ,a_n\right)$. By +definition, we have that as $d\mid a_i$ for each $1\leq i\leq n$ then by +proposition +[103](#prop:NT_Divisor_dividing_all_in_set_divides_linear_combination){reference-type="ref" +reference="prop:NT_Divisor_dividing_all_in_set_divides_linear_combination"} +we have that* + +*$$\begin{equation*} + d\mid\sum_{i=1}^n a_ix_n +\end{equation*}$$* + +*for any $x_i\in\mathbb{Z}$. Define the set $A$ by* + +*$$\begin{equation*} + G=\left\{\sum_{i=1}^n a_ix_n : x_i\in\mathbb{Z}\right\} +\end{equation*}$$* + +*Clearly, there are both positive and negative elements in $G$, +additionally $0\in G$ by taking each $x_i=0$. Define $\Tilde{G}$ by* + +*$$\begin{equation*} + \Tilde{G}=\left\{g\in G: g>0\right\} +\end{equation*}$$* + +*It follows that $\Tilde{G}\subset\mathbb{Z}$ and so by the +well-ordering principle it has a smallest element $\Tilde{g}$ of the +form* + +*$$\begin{equation*} + \Tilde{g}=\sum_{i=1}^n a_ix_n +\end{equation*}$$* + +*We must show that $\Tilde{g}\mid a_i$ for each $i$. Suppose for +contradiction and without loss of generality that $\Tilde{g}\nmid a_1$. +By the division algorithm, we have that* + +*$$\begin{equation*} + a_1=q\Tilde{g}+r +\end{equation*}$$* + +*with $0\deg\left(Q\right)$ where $\deg\left(P\right)=n$ and +$\deg\left(P\right)=m$. Then as tuples we have that* + +*$$\begin{align*} + P=\left(p_0,p_1,p_2,\dots,p_{n-1},p_n\right)\\ + Q=\left(q_0,q_1,q_2,\dots,q_{m-1},q_m\right)\\ +\end{align*}$$* + +*As $\deg\left(Q\right)<\deg\left(P\right)$ we use the tuple extension +mapping $E_m^n$ on $Q$ and we have that +$\deg\left(E_m^n\left(Q\right)\right)\leq n$. Hence* + +*$$\begin{equation*} + \deg\left(P\oplus_S Q\right)\leq \deg\left(P\oplus_S E\left(Q\right)\right)\leq n = \max\left(\deg\left(P\right),\deg\left(Q\right)\right) +\end{equation*}$$* + +*$\qed$* +::: + +We are getting an idea for our problem with the polynomial given by + +$$\begin{equation*} + P=0+0*X+0*X^2+0*X^3+\dots+0*X^n +\end{equation*}$$ + +If we want lemma +[12](#lem:NT_Polynomial_degree_addition){reference-type="ref" +reference="lem:NT_Polynomial_degree_addition"} to be consistent, we +should define the degree of $P$ to be such that it is no larger than the +degree of any other polynomial. In particular, for $c\in S$ we have that +$Q=c$ with $Q\in S\left[X\right]$ has degree $0$, we must have that +$\deg\left(P\right)< \deg\left(Q\right)=0$. This still doesn't fully +answer the question, which negative integer should we take for the +degree of $P$? Maybe, once we have a definition for the multiplication +of polynomials, it will provide further insight. + +Now, given a potential candidate for defining the addition of two +polynomials, we can also consider a potential candidate for defining the +subtraction of two polynomials. As before, we take inspiration from +$\mathbb{Z}$. + +As we have shown that the addition of integers is closed and +well-defined, additionally, for every $x\in\mathbb{Z}$ we have that +$\exists y$ so that $x+y=0$. In particular, we take $y=-x$ so that the +expression becomes $x-x=0$. A sensible definition for polynomial +subtraction should also respect these properties; subtracting two +polynomials should give another polynomial. This raises a question; +suppose $P\in S\left[X\right]$, what is $P-P$? + +We know that in $\mathbb{N}$, $\mathbb{Z}$ and $\mathbb{Q}$, that for an +element $x$ that $x-x$ should be $0$, but what does it mean for $0$ to +be an element of $S$ and by extension $S\left[X\right]$? In particular +is it the same $0$ as for $\mathbb{N}$, $\mathbb{Z}$ and $\mathbb{Q}$? + +On the other hand, we know that for any $x$ in $\mathbb{N}$, +$\mathbb{Z}$ and $\mathbb{Q}$ that $x+0=x=0+x$, a similar sort of +element of $S$ would be useful and clearly plays an important role for +defining a similar element for $S\left[X\right]$. This idea is general +enough, assuming we have a well-behaved $+_S$, that we can apply it to a +set $S$. + +::: definition +**Definition 186**. *Additive Identity of a set $S$* + +*Let $S$ be a set so that there is an operation $+_S:S^2\rightarrow S$ +such that $+_S$ is closed and well-defined. Let $e\in S$. If we have +that $\forall s \in S$ that $s+_S e=s$, then we say that $e$ is a right +additive identity element of $S$.* + +*Similarly, if $\forall s \in S$ we have that $e+_Ss=s$, then we say +that $e$ is a left additive identity element of $S$.* + +*If we have that $\forall s\in S$ that $e+_S s=s=s+_S e$, we simply call +$e$ an additive identity element.* + +*If we need to be clear which set the additive inverse belongs to, we +will write $e_S$* +::: + +It is an immediate consequence of $+_S$ that the identity element is +unique. + +::: proposition +**Proposition 140**. *The additive identity element of a set $S$ is +unique* + +*Let $S$ be a set so that there is an operation $+_S:S^2\rightarrow S$ +such that $+_S$ is closed and well-defined. Let $e,f\in S$ be additive +identity elements of $S$.* + +*We have that $e=f$.* + +*Proof:* + +*Let $S$ and $+_S:S^2\rightarrow S$ be as given, and let $e,f\in S$ be +additive identity elements of $S$.* + +*By definition, we have that* + +*$$\begin{equation*} + e=e+_s f=f +\end{equation*}$$* + +*As $+_S$ is well-defined and closed, we have that $e=f$ as required. +$\qed$* +::: + +From this, we can immediately identify that $0$ in $\mathbb{N}$, +$\mathbb{Z}$ and $\mathbb{Q}$ is unique. + +We have resolved one part of this problem, that in $\mathbb{N}$, +$\mathbb{Z}$ and $\mathbb{Q}$, for an element $x$ that $x-x=0$. We have +answered what it means for \"$0$\" to be in $S$, but what does it mean +for $-x\in S$ given $x\in S$?. Noting that $x-x=x+_S\left(-x\right)$, +this is precisely what it means for $x$ to be invertible in $S$ at least +with respect to $+_S$. As with the additive identity of $S$, this idea +is also general enough to apply to a more general set $S$. + +::: definition +**Definition 187**. *Additive Inverse of a set $S$* + +*Let $S$ be a set so that there is an operation $+_S:S^2\rightarrow S$ +such that $+_S$ is closed and well-defined. Let $s\in S$.* + +*If we have that $\exists x\in S$ such that $s+_S x=e$, then we say that +$x$ is a right additive inverse element of $s$ in $S$.* + +*Similarly, if $\exists x\in S$ such that $x+_S s=e$, then we say that +$x$ is a left additive inverse element of $s$ in $S$.* + +*If we have that $\exists x\in S$ that $x+_S s=s=s+_S x$, we simply call +$x$ an additive inverse element of $s$ in $S$.* +::: + +As with the additive identity element, we have an immediate consequence +that the inverse of an element $s\in S$ is unique. + +::: proposition +**Proposition 141**. *The additive inverse element of an element of $S$ +is unique* + +*Let $S$ be a set so that there is an operation $+_S:S^2\rightarrow S$ +such that $+_S$ is closed and well-defined. Let $s\in S$ be an arbitrary +element of $S$.* + +*We have that the additive inverse of $s$ is unique.* + +*Proof:* + +*Let $S$ and $+_S:S^2\rightarrow S$ be as given, and let $s\in S$ be an +arbitrary element of $S$ and suppose that $s$ has two inverses $x$ and +$y$.* + +*By definition, we have that* + +*$$\begin{align*} + x&=x+_S e\\ + &=x+_S\left(s+_S y\right)\\ + &= +\end{align*}$$* + +*As $+_S$ is well-defined and closed, we have that $e=f$ as required. +$\qed$* +::: + +It would also be useful to undo the addition of polynomials via +polynomial subtraction. The only requirement is that we need $+_S$ to be +invertible In particular, as we are using a well-defined and closed +operation on $S$, that is $+_S$, we have gained a definition of +subtraction for free! Using $-_S$ to denote subtraction in $S$, we have + +$$\begin{align*} + \ominus_S:s^n\times s^n&\mathlarger{\mathlarger{\rightarrow}}s^n\\ + \left(P, E\left(Q\right)\right)&\mapsto\ominus_S\left(P,E\left(Q\right)\right)=\left(p_0-_S q_0,p_1-_S q_1, p_2-_S q_2,\dots, p_{n-1}-_S0, p_n-_S0\right) +\end{align*}$$ + +Given a notion of subtraction, we can also define what it means for two +polynomials to be equal. Firstly, recall what it means for + +::: definition +**Definition 188**. *Equality of Polynomials* + +*Let $P,Q\in S\left[X\right]$ where $\deg\left(P\right)=n$ and +$\deg\left(Q\right)=m$ where without loss of generality $m\leq n$.* + +*We say that $P$ and $Q$ are equal as polynomials, written $P=Q$, if and +only if* + +*$$\begin{equation*} + P\ominus_S Q = 0 = \left(\underbrace{0,0,0,\dots, 0,0}_{n+1 \text{ times}}\right) +\end{equation*}$$* + +*That is, if the difference between the two is the zero polynomial.* +::: + +We can therefore define the following relation. + +::: definition +**Definition 189**. +::: + +It is immediate that a polynomial therefore has a unique representation +as an $n+1$-tuple. + +##### Defining multiplication between two polynomials + +We can use the same idea of the $n+1$-tuples to define multiplication of +polynomials. Recall that we observed that we can express the +intermediate $X$, and powers of it, as follows + +$$\begin{align*} + P\left(X\right)=1=X^0 &\iff a=\left(1\right)\\ + P\left(X\right)=X &\iff a=\left(0,1\right)\\ + P\left(X\right)=X^2 &\iff a=\left(0,0,1\right)\\ + P\left(X\right)=X^3 &\iff a=\left(0,0,0,1\right)\\ + &\dots +\end{align*}$$ + +Intuitively, we want $X^2=X*X$, $X^3=X^2*X$ and so on. That is + +$$\begin{align*} + X*X=\left(0,1\right)*\left(0,1\right)&=\left(0,0,1\right)=X^2\\ + X^2*X=\left(0,0,1\right)*\left(0,1\right)&=\left(0,0,0,1\right)=X^3\\ + &\dots +\end{align*}$$ + +What about more complex expressions? Say $X*\left(X+X^2\right)$. The +answer to this would depend on if multiplication is distributive over +addition with respect to the indeterminate, and additionally on +multiplication is commutative!. For now, let us assume that this is the +case, + +It seems therefore that multiplication by $X$ has the effect of +"shifting" to the right + +[^1]: *We are clearly not talking about sunsets* + +[^2]: If we are being logical and don't want to get soaked before we get + to our destination. + +[^3]: By exist we mean in the abstract sense. + +[^4]: *Without loss of generality means we have made a choice in the + proof which allows us to consider a single case as the other cases + have the same argument just with the notation changed to reflect the + different choice.* + +[^5]: *Unless you are either not a human or somehow reading this in some + unknown form of existence* + +[^6]: *We will first need to prove that in order to speak of the inverse + of a mapping that we will need the left and right inverses to be + equal* + +[^7]: Hence the similar names. + +[^8]: Hopefully not all at once! + +[^9]: We can think of this as some sort of singularity + +[^10]: *Phew!* + +[^11]: Until someone manages to find a way to get past the elegant + mathematics of the encryption scheme! + +[^12]: If there is only one theorem you learn when studying Number + Theory, it has to be this one! + +[^13]: I prefer this way of thinking. + +[^14]: RSA stands for Rivest--Shamir--Adleman + +[^15]: *Named after the 3rd-century mathematician Diophantus of + Alexandria* + +[^16]: When we have fully defined polynomial addition, we will go with + the usual convention of just using $+$ to denote addition