Construction of the Real Numbers

Definition

Axiomatic Definition

Let $\mathbb{R}$ denote the set of all real numbers, then

• The set $\mathbb{R}$ is a Field
• The set $\mathbb{R}$ satisfies Order Axiom
• The set $\mathbb{R}$ satisfies Completeness Axioms

Construction from Cauchy Sequences

Let $R$ be the set of Cauchy Sequence of rational numbers

$(a_n)\sim (b_n)\iff \underset{n\to \infty }{\lim }(a_n-b_n)=0$ Define an Equivalence Relation $\sim$ on the $R$ The two cauchy sequences $(a_n),(b_n)$ are equivalent if the sequence $(a_n-b_n)$ converges to $0$.

The set of real numbers, denoted by $\mathbb{R}$, is defined as the set of equivalence class under the relation $\sim$

$(\underset{n\to \infty }{\lim }{a}_n=\alpha )\Rightarrow ([(a_n)]:=\alpha )$ and denote the value represented by the Equivalence Class as a limit of the Cauchy Sequence

The operations of addition and multiplication on $\mathbb{R}$ is defined as follows

• $[(a_n)]+[(b_n)]=[(a_n)+(b_n)]$
• $[(a_n)]\cdot [(b_n)]=[(a_n)\cdot (b_n)]$

Construction by Dedekind Cuts

Let $C$ be the subset of the set of Rational Number $\mathbb{Q}$ If $C$ fulfills the following conditions, then $C$ is a real number

• $C\ne \varnothing$
• $\forall s,t\in C,t<s\Rightarrow (s\in C\Rightarrow t\in C)$
• $\nexists u\in C,\forall s\in C,s\le u$
• $\forall c\in C,\exists x\in \mathbb{Q},c<x$ or $C\ne \mathbb{Q}$

By the conditions, we have properties $\forall s\notin C,\nexists t\in C,t>s$ $\forall r\in C,\forall s\notin C,r<s$

We define a Total Order Relation on the real numbers as follows, $x\le y\iff x\subset y$

Operations

Let $A,B\in \mathbb{R}$

• $A+B:=\{a+b|a\in A\wedge b\in B\}$
• $A-B:=\{a-b|a\in A\wedge b\in (\mathbb{Q}\setminus B)\}$
• $-B:=\{a-b|a<0\wedge b\in (\mathbb{Q}\setminus B)\}$
• If $A,B\ge 0$, $A\times B:=\{a\times b|a\in A\wedge b\in B\}$
• If $A\ge 0,B\le 0$, then $A\times B:=-(A\times (-B))$
• If $A\le 0,B\ge 0$, then $A\times B:=-((-A)\times B)$
• If $A,B\le 0$, then $A\times B:=(-A)\times (-B)$
• $\frac{1}{B}:=\{\frac{a}{b}|a<1\wedge b\in (\mathbb{Q}\setminus B)\}$
• $A÷B=\frac{A}{B}:=A\times \frac{1}{B}$