Archimedean propertyArchimedeanProperty2002-08-29 23:10:552007-01-15 05:11:46Theorem\usepackage{amssymb}
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Let $x$ be any real number. Then there exists a natural number $n$ such that $n > x$.
This theorem is known as the \emph{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).
\begin{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 $x$.
\end{proof}
\begin{corollary}
If $x$ and $y$ are real numbers with $x > 0$, there exists a natural $n$ such that $nx > y$.
\end{corollary}
\begin{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 $nx > y$.
\end{proof}
\begin{corollary}
If $w$ is a real number greater than $0$, there exists a natural $n$ such that $0 < 1/n < w$.
\end{corollary}
\begin{proof}
Using Corollary 1, choose $n \in \mathbb{N}$ satisfying $nw > 1$. Then $0 < 1/n < w$.
\end{proof}
\begin{corollary}
If $x$ and $y$ are real numbers with $x < y$, there exists a rational number $a$ such that $x < a < y$.
\end{corollary}
\begin{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 $a = -b$.
\end{proof}