## You are here

Homerational numbers are real numbers

## Primary tabs

# 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.

## Mathematics Subject Classification

54C30*no label found*26-00

*no label found*12D99

*no label found*

- Forums
- Planetary Bugs
- HS/Secondary
- University/Tertiary
- Graduate/Advanced
- Industry/Practice
- Research Topics
- LaTeX help
- Math Comptetitions
- Math History
- Math Humor
- PlanetMath Comments
- PlanetMath System Updates and News
- PlanetMath help
- PlanetMath.ORG
- Strategic Communications Development
- The Math Pub
- Testing messages (ignore)

- Other useful stuff
- Corrections