rational numbers are real numbers

Let us first show that the natural numbers $0,1,2,\ldots$ 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

$0,\quad 0+1,\quad(0+1)+1,\quad((0+1)+1)+1,\ldots,$

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

1.

$f(0)=0$ , or more precisely, $f(0_{\mathbbmss{N}})=0_{\mathbbmss{R}}$ ,
2.

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

By induction on $a$ one can prove that

	$\displaystyle f(a+b)$	$\displaystyle=$	$\displaystyle f(a)+f(b),$
	$\displaystyle f(ab)$	$\displaystyle=$	$\displaystyle f(a)f(b),\quad a,b\in\mathbbmss{N}$

and

$\displaystyle f(a)$

$\displaystyle\geq$

$\displaystyle 0,\ a\in\mathbbmss{N}\ \mbox{with equality only when}\ a=0.$

The last claim follows since $f(a)>0$ for $a=1,2,\ldots$ (by induction), and $f(0)=0$ . It follows that $f$ is an injection: If $a\leq b$ , then $f(a)=f(b)$ implies that $f(a)=f(a)+f(b-a)$ , so $a=b$ .

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

	$\displaystyle S\colon f(\mathbbmss{N})$	$\displaystyle\to$	$\displaystyle f(\mathbbmss{N})$
	$\displaystyle k$	$\displaystyle\mapsto$	$\displaystyle f(f^{-1}(k)+1)$

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 $k=f(a)\in f(\mathbbmss{N})$ , then $a+1=0$ ; a contradiction. Lastly, the axiom of induction follows since $\mathbbmss{N}$ satisfies this axiom. We have shown that $f(\mathbbmss{N})$ 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

	$\displaystyle\mathbbmss{Z}$	$\displaystyle=$	$\displaystyle\mathbbmss{N}\cup\{-z\in\mathbbmss{R}:z\in\mathbbmss{N}\},$
	$\displaystyle\mathbbmss{Q}$	$\displaystyle=$	$\displaystyle\left\{\frac{a}{b}:a\in\mathbbmss{Z},b\in\mathbbmss{N}\setminus\{% 0\}\right\}.$

Mathematically, $\mathbbmss{Z}$ and $\mathbbmss{Q}$ are subrings of $\mathbbmss{R}$ 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 $\mathbbmss{R}$ are defined as follows. First, a set $L\subseteq\mathbbmss{R}$ 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 $\mathbbmss{R}$ 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, McGraw-Hill, 1976.

Title	rational numbers are real numbers
Canonical name	RationalNumbersAreRealNumbers
Date of creation	2013-03-22 15:45:49
Last modified on	2013-03-22 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 26-00
Classification	msc 12D99
Related topic	GroundField