rational numbers are real numbers
Let us first show that the natural numbers^{} $0,1,2,\mathrm{\dots}$ are contained 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 constructions of $\mathbb{R}$ rely on the existence of the natural numbers.) Then we can define a map $f:\mathbb{N}\to \mathbb{R}$ as

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

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 $f(a)=f(b)$ implies that $f(a)=f(a)+f(ba)$, so $a=b$.
To conclude, let us show that $f(\mathbb{N})\subset \mathbb{R}$ satisfies the Peano axioms^{} with zero element $f(0)$ and sucessor 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 that behave as the natural numbers.
From the natural numbers, the integers and rationals can be defined as
$\mathbb{Z}$  $=$  $\mathbb{N}\cup \{z\in \mathbb{R}:z\in \mathbb{N}\},$  
$\mathbb{Q}$  $=$  $\{{\displaystyle \frac{a}{b}}:a\in \mathbb{Z},b\in \mathbb{N}\setminus \{0\}\}.$ 
Mathematically, $\mathbb{Z}$ and $\mathbb{Q}$ are subrings of $\mathbb{R}$ that are ring isomorphic^{} to the integers and rationals, respectively.
Other constructions
The above construction follows [1]. However, there are also other constructions. For example, in [2], natural numbers in $\mathbb{R}$ are defined as follows. First, a set $L\subseteq \mathbb{R}$ 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 all inductive sets^{}. A third approach is to explicitly exhibit the natural numbers when constructing the real numbers. For example, in [3], it is shown that the rational numbers form a subfield^{} of $\mathbb{R}$ using explicit Dedekind cuts^{}.
References
 1 H.L. Royden, Real analysis, Prentice Hall, 1988.
 2 M. Spivak, Calculus, Publish or Perish.
 3 W. Rudin, Principles of mathematical analysis, McGrawHill, 1976.
Title  rational numbers are real numbers 

Canonical name  RationalNumbersAreRealNumbers 
Date of creation  20130322 15:45:49 
Last modified on  20130322 15:45:49 
Owner  matte (1858) 
Last modified by  matte (1858) 
Numerical id  6 
Author  matte (1858) 
Entry type  Result 
Classification  msc 54C30 
Classification  msc 2600 
Classification  msc 12D99 
Related topic  GroundField 