Number Systems

Naturals and Integers

Natural Numbers

We define the set of natural numbers \require{mhchem}\newcommand{\R}{\mathbb{R}}\newcommand{\N}{\mathbb{N}}\newcommand{\Z}{\mathbb{Z}}\newcommand{\Q}{\mathbb{Q}} \N by the following Peano axioms with successor function $S:\mathbb{N}\to \mathbb{N}$.

• $0$ is a natural number.
• For every natural number $n$, $S(n)$ is a natural number. That is, the natural numbers are closed under $S$.
• For all natural numbers $m$ and $n$, if $S(m)=S(n)$, then $m=n$. That is, $S$ is injective.
• There is no natural number whose successor is $0$.
• If $M$ is a set such that $0\in M$, and for every $n\in \mathbb{N}$, $n\in M⟹S(n)\in M$, then $M$ contains every natural number. That is the successor function generates all the natural numbers different from $0$.

Def Addition on Naturals Addition is a function $+:\mathbb{N}\times \mathbb{N}\to \mathbb{N}$, defined recursively by $a+0=a,a+S(b)=S(a+b)$ Def Multiplication on Naturals Multiplication is a function $\times :\mathbb{N}\times \mathbb{N}\to \mathbb{N}$. Given addition, it is defined recursively as: $a\times 0=0,a\times S(b)=a+(a\times b)$

Prop $S(0)$ is the multiplicative identity.

Def Total Order on Naturals The standard total order relation $\le$ on natural numbers can be defined as follows,$a\le b\text{ if }\exists c\in \mathbb{N}\text{ s.t. }a+c=b$

Integers

We define the integers $\mathbb{Z}$ as a field of equivalence classes:$\mathbb{Z}=(\mathbb{N}\times \mathbb{N})/\sim$where $\sim$ is an equivalence relation defined on $\mathbb{N}\times \mathbb{N}$ as $(a,b)\sim (c,d)⟺a+d=b+c$We write $[(a,b){]}_{\mathbb{Z}}$ as the corresponding equivalence class. And define $[(a,b){]}_{\mathbb{Z}}+[(c,d){]}_{\mathbb{Z}}=[(a+c,b+d){]}_{\mathbb{Z}},[(a,b){]}_{\mathbb{Z}}\cdot [(c,d){]}_{\mathbb{Z}}=[(ac+bd,ad+bc){]}_{\mathbb{Z}}$

Rationals

Definition

Def Rational Numbers We define the rational numbers $\text{Q}$ as a field of equivalence classes:$\text{Q}=(\mathbb{Z}\times \mathbb{Z})/\sim$where $\sim$ is an equivalence relation defined on $\mathbb{Z}\times \mathbb{Z}$ as $(a,b)\sim (c,d)⟺ad=bc$We write $[(a,b){]}_{\text{Q}}$ as the corresponding equivalence class. And thus define $\begin{array}{r}[][(a,b){]}_{\text{Q}}+[(c,d){]}_{\text{Q}}=[(ad+bc,bd){]}_{\text{Q}}\\ [(a,b){]}_{\text{Q}}\cdot [(c,d){]}_{\text{Q}}=[(ac,bd){]}_{\text{Q}}\\ 0_{\text{Q}}=[(0,1){]}_{\text{Q}}\\ 1_{\text{Q}}=[(1,1){]}_{\text{Q}}\\ [(a,b){]}_{\text{Q}}\le [(c,d){]}_{\text{Q}}\text{ if }(bc-ad)bd\ge 0\end{array}$

Reals

$\text{Q}$-Cauchy Equivalence

For two sequences $(a_n)$ and $(b_n)$ in $\text{Q}$, we say that $(a_n)$ and $(b_n)$ are $\text{Q}$-Cauchy equivalent if$\forall \epsilon >0\in \text{Q}.\exists N\in \mathbb{N}.n\ge N⟹|a_n-b_n|<\epsilon$

Definition

$\text{Q}$-Cauchy equivalence is an equivalence relation.

Proof Clearly it is reflexive as $|a_n-a_n|=0<\epsilon$. It is symmetric since$|a_n-b_n|<\epsilon ⟺|b_n-a_n|<\epsilon$Suppose $(a_n)\sim (b_n)$ and $(b_n)\sim (c_n)$. For all $\epsilon >0$, exists $N_1,N_2\in \mathbb{N}$ such that$\begin{array}{r}n\ge N_1⟹|a_n-b_n|<\frac{\epsilon }{2}\\ n\ge N_2⟹|b_n-c_n|<\frac{\epsilon }{2}\end{array}$Let $N=\max \{N_1,N_2\}$. Suppose $n\ge N$, we have:$|a_n-c_n|\le |a_n-b_n|+|b_n-c_n|<\frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon$Hence $(a_n)\sim (c_n)$, proved transitivity. $\square$

Real Numbers

The real numbers $\mathbb{R}$ is defined as a field of equivalence classes of $\text{Q}$-Cauchy sequences. We write $[(a_n){]}_{\mathbb{R}}$ as the corresponding equivalence class. And thus define: $[(a_n{)}_{\mathbb{R}}]\ge [(b_n{)}_{\mathbb{R}}]\text{ if }\exists N\in \mathbb{N}\text{ s.t. }{a}_n\ge b_n\text{ for all }n>N.$

Proposition

Every Cauchy sequence of real numbers converges to a real number.

Interval

A set $I\subset \mathbb{R}$ is an (closed) interval if $a,b\in I$ and $a<x<b$ imply $x\in I$.

Archimedean Property

Archimedean Property

In any ordered field $F$, define $f:F\to \text{Q}$ such that $0_F\mapsto 0,1_F\mapsto 1$respect $+,\times$ and order in $\text{Q}$. We say that it has Archimedean property if one of the following equivalent properties hold:

• $\mathbb{N}$ is not bounded above by any element of $f(F)$.
• For all $y\in F$ with $y{>}_F{0}_F$, there is $n\in \mathbb{N}$ such that $\frac{1}{n}{<}_{\text{Q}}f(y)$.
• For all $x,y\in F$ with $y{>}_F{0}_F$, then there is $n\in \mathbb{N}$ such that $nf(y){>}_{\text{Q}}f(x)$.

Proposition

Prop Archimedean property holds for $\mathbb{R}$. Proof

Complex Numbers

Although $\mathbb{R}$ is good enough, it is not quite sufficient, for example, it is not algebraically closed. So complex numbers come into play. There are mainly two ways to construct complex numbers. The first one is to define them as pairs of real numbers (See here), and the second one is to define them as equivalence classes of polynomials (See here). The first one is more intuitive (geometric), while the second one is more algebraic.

Following convention, we take the first as our definition:

The field of complex numbers, denoted by $\mathbb{C}$, is the usual Euclidean space \newcommand{\R}{\mathbb{R}}\R^{2} endowed with the additional operation of multiplication of vectors defined as follows: $(x_1,y_1)\cdot (x_2,y_2)=(x_1{x}_2-y_1{y}_2,x_1{y}_2+x_2{y}_1)$And we write $i$ for $(0,1)$, $1$ for $(1,0)$. Therefore $x+yi$ denotes $(x,y)$.

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