# 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 (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.

Since $x$ and $y$ are reals, and $x\neq 0$, $y/x$ is a real. By the Archimedean property, we can choose an $n\in\mathbb{N}$ such that $n>y/x$. Then $nx>y$. ∎

###### Corollary 2.

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

###### Proof.

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

###### Corollary 3.

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

###### 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. Furthermore, since $y\leq m_{0}/n$, we have $y-1/n; and $x. Thus $a$ satisfies $x.

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

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

Title Archimedean property ArchimedeanProperty 2013-03-22 13:00:47 2013-03-22 13:00:47 Daume (40) Daume (40) 9 Daume (40) Theorem msc 12D99 axiom of Archimedes Archimedean principle ArchimedeanSemigroup ExistenceOfSquareRootsOfNonNegativeRealNumbers