# 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 $S=\{a\in \mathbb{N}:a\le x\}$. If $S$ is empty, let $n=1$; note that $$ (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 $n\notin S$; since $n\notin S$, we know $n>x$. Thus we have a natural greater than $x$. ∎

###### Corollary 1.

If $x$ and $y$ are real numbers with $x\mathrm{>}\mathrm{0}$, there exists a natural $n$ such that $n\mathit{}x\mathrm{>}y$.

###### Proof.

Since $x$ and $y$ are reals, and $x\ne 0$, $y/x$ 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 $\mathrm{0}$, there exists a natural $n$ such that $$.

###### Proof.

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

###### Corollary 3.

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

###### Proof.

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

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

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

Title | Archimedean property |
---|---|

Canonical name | ArchimedeanProperty |

Date of creation | 2013-03-22 13:00:47 |

Last modified on | 2013-03-22 13:00:47 |

Owner | Daume (40) |

Last modified by | Daume (40) |

Numerical id | 9 |

Author | Daume (40) |

Entry type | Theorem |

Classification | msc 12D99 |

Synonym | axiom of Archimedes |

Synonym | Archimedean principle |

Related topic | ArchimedeanSemigroup |

Related topic | ExistenceOfSquareRootsOfNonNegativeRealNumbers |