diff --git a/Vol1/vol1.md b/Vol1/vol1.md deleted file mode 100644 index f6c5d26..0000000 --- a/Vol1/vol1.md +++ /dev/null @@ -1,23080 +0,0 @@ -# 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