Axiom of Archimedes

Theorem
Let $x$ be a real number.

Then there exists a natural number greater than $x$.


 * $\forall x \in \R: \exists n \in \N: n > x$

That is, the set of natural numbers is unbounded above.

Proof
Let $x \in \R$.

Let $S$ be the set of all natural numbers less than or equal to $x$:
 * $S = \set {a \in \N: a \le x}$

It is possible that $S = \O$.

Suppose $0 \le x$.

Then by definition, $0 \in S$.

But $S = \O$, so this is a contradiction.

From the Trichotomy Law for Real Numbers it follows that $0 > x$.

Thus we have the element $0 \in \N$ such that $0 > x$.

Now suppose $S \ne \O$.

Then $S$ is bounded above (by $x$, for example).

Thus by the Continuum Property of $\R$, $S$ has a supremum in $\R$.

Let $s = \map \sup S$.

Now consider the number $s - 1$.

Since $s$ is the supremum of $S$, $s - 1$ cannot be an upper bound of $S$ by definition.

So $\exists m \in S: m > s - 1 \implies m + 1 > s$.

But as $m \in \N$, it follows that $m + 1 \in \N$.

Because $m + 1 > s$, it follows that $m + 1 \notin S$ and so $m + 1 > x$.

Also known as
This result is also known as:
 * the Archimedean law
 * the Archimedean property (of the natural numbers)
 * the Archimedean ordering property (of the real line)
 * the axiom of Archimedes.

Also see

 * The Archimedean property, which may or may not be satisfied by an abstract algebraic structure.


 * In Equivalence of Archimedean Property and Archimedean Law it is shown that on the field of real numbers the two are equivalent.

Not to be confused with the better-known (outside the field of mathematics) Archimedes' Principle.