rational numbers are real numbers

Let us first show that the natural numbers $0,1,2,\mathrm{\dots }$ arecontained in the real numbers as constructed above.Heuristically, this should be clear. We start with $0$.By adding $1$ repeatedly we obtain the natural numbers

$$0,0+1,(0+1)+1,((0+1)+1)+1,\mathrm{\dots },$$

To make this precise, let $\mathbb{N}$ be the natural numbers.(We assume that these exist. For example, all the usual constructionsof $ℝ$ rely on the existence of the natural numbers.)Then we can define a map $f:\mathbb{N}\to ℝ$ as

1. $f(0)=0$, or more precisely, $f(0_{\mathbb{N}})=0_{ℝ}$,

2. $f(a+1)=f(a)+1$ for $a\in \mathbb{N}$.

By induction on $a$ one can prove that

$f(a+b)$$=$$f(a)+f(b),$

$f(ab)$$=$$f(a)f(b),a,b\in \mathbb{N}$

and

$f(a)$$\ge$$0,a\in \mathbb{N}\text{with equality only when}a=0.$

The last claim follows since $f(a)>0$ for $a=1,2,\mathrm{\dots }$ (by induction),and $f(0)=0$.It follows that $f$ is an injection: If $a\le b$, then $f(a)=f(b)$ impliesthat $f(a)=f(a)+f(b-a)$, so $a=b$.

To conclude, let us show that$f(\mathbb{N})\subset ℝ$ satisfies the Peano axioms with zero element $f(0)$ andsucessor operator

$S:f(\mathbb{N})$$\to$$f(\mathbb{N})$

$k$$\mapsto$$f(f^{-1}(k)+1)$

First, as $f$ is a bijection, $x=y$ if and only if $S(x)=S(y)$is clear.Second, if $S(k)=0$ for some $k=f(a)\in f(\mathbb{N})$, then $a+1=0$; a contradiction.Lastly, the axiom of induction follows since $\mathbb{N}$ satisfies this axiom.We have shown that $f(\mathbb{N})$ are a subset of the real numbers thatbehave as the natural numbers.

From the natural numbers, the integers and rationals canbe defined as

$\mathbb{Z}$$=$$\mathbb{N}\cup \{-z\in ℝ:z\in \mathbb{N}\},$

$\mathbb{Q}$$=$$\{{\displaystyle \frac{a}{b}}:a\in \mathbb{Z},b\in \mathbb{N}\setminus \{0\}\}.$

Mathematically, $\mathbb{Z}$ and $\mathbb{Q}$ are subrings of $ℝ$ that arering isomorphic to the integers and rationals, respectively.

Other constructions

The above construction follows [1]. However, there are alsoother constructions. For example, in [2], natural numbers in $ℝ$are defined as follows. First, a set $L\subseteq ℝ$ is inductive if

1. $0\in L$,

2. if $a\in L$, then $a+1\in L$.

Then the natural numbers are defined as real numbers that are contained in allinductive sets.A third approach is to explicitly exhibit the natural numbers whenconstructing the real numbers. For example, in [3],it is shown that the rational numbers form a subfield of $ℝ$using explicit Dedekind cuts.