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.

Also known as the Archimedean law, or the Archimedean Property of Natural Numbers.

Proof
Let $x \in \R$.

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


 * It is possible that $S = \varnothing$. If this is the case, then $0 \in \N$ such that $0 > x$.

Otherwise, by the Trichotomy Law for Real Numbers, $0 \le x$ and so $0 \in S$ (and it can't be, because we've just said that $S = \varnothing$).


 * Now suppose $S \ne \varnothing$.

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

Thus by the Least Upper Bound Property of $\R$, $S$ has a supremum in $\R$.

Let $s = \sup \left({S}\right)$.

Now consider the number $s - 1$.

Since $s$ is the supremum of $S$, $s-1$ can not 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 $m + 1 > x$.

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