## You are here

HomeArchimedean property

## Primary tabs

# Archimedean property

Let $x$ be any real number. Then there exists a natural number $n$ such that $n>x$.

This theorem is known as the *Archimedean property* of real numbers. It is also sometimes called the axiom of Archimedes, although this name is doubly deceptive: it is neither an axiom (it is rather a consequence of the least upper bound property) nor attributed to Archimedes (in fact, Archimedes credits it to Eudoxus).

###### Proof.

Let $x$ be a real number, and let $S=\{a\in\mathbb{N}:a\leq x\}$. If $S$ is empty, let $n=1$; note that $x<n$ (otherwise $1\in S$).

Assume $S$ is nonempty. Since $S$ has an upper bound, $S$ must have a least upper bound; call it $b$. Now consider $b-1$. Since $b$ is the least upper bound, $b-1$ cannot be an upper bound of $S$; therefore, there exists some $y\in S$ such that $y>b-1$. Let $n=y+1$; then $n>b$. But $y$ is a natural, so $n$ must also be a natural. Since $n>b$, we know $n\not\in S$; since $n\not\in S$, we know $n>x$. Thus we have a natural greater than $x$. ∎

###### Corollary 1.

If $x$ and $y$ are real numbers with $x>0$, there exists a natural $n$ such that $nx>y$.

###### Proof.

###### Corollary 2.

If $w$ is a real number greater than $0$, there exists a natural $n$ such that $0<1/n<w$.

###### Proof.

Using Corollary 1, choose $n\in\mathbb{N}$ satisfying $nw>1$. Then $0<1/n<w$. ∎

###### Corollary 3.

If $x$ and $y$ are real numbers with $x<y$, there exists a rational number $a$ such that $x<a<y$.

###### Proof.

First examine the case where $0\leq x$. Using Corollary 2, find a natural $n$ satisfying $0<1/n<(y-x)$. Let $S=\{m\in\mathbb{N}:m/n\geq y\}$. By Corollary 1 $S$ is non-empty, so let $m_{0}$ be the least element of $S$ and let $a=(m_{0}-1)/n$. Then $a<y$. Furthermore, since $y\leq m_{0}/n$, we have $y-1/n<a$; and $x<y-1/n<a$. Thus $a$ satisfies $x<a<y$.

Now examine the case where $x<0<y$. Take $a=0$.

Finally consider the case where $x<y\leq 0$. Using the first case, let $b$ be a rational satisfying $-y<b<-x$. Then let $a=-b$. ∎

## Mathematics Subject Classification

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

## Recent Activity

new correction: Error in proof of Proposition 2 by alex2907

Jun 24

new question: A good question by Ron Castillo

Jun 23

new question: A trascendental number. by Ron Castillo

Jun 19

new question: Banach lattice valued Bochner integrals by math ias

## Comments

## Axiom vs. Theorem

While the phrase "is neither an axiomn nor attributed to Archimedes"

is fine rhetoric, the first half of it needs to be taken with a

teaspoon of salt. Whether or not a statement is an axiom depends

on how one chooses to axiomatize one's theory. To be sure, if one

chooses the least upper bound property as an axiom, then the

Archimedean property follows as a consequence, hence is a theorem.

However, there is more than one way to axiomatize the theory of

real numbers. For instance, I could also characterize the real

number system as follows:

* The real number system is an Archimedean ordered field.

* No proper extension field of the real number system is Archimedean.

In this case, the roles are reversed --- the Archimedean property is

now an axiom and the least upper bound property is now a theorem.