Axiom of Archimedes

Theorem
Let $$x$$ be a real number.

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

$$\forall x \in \mathbb{R}: \exists n \in \mathbb{N}: n > x$$

Proof
Let $$x \in \mathbb{R}$$.

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


 * It's possible that $$S = \varnothing$$. If this is the case, then $$0 \in \mathbb{N}$$ 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 $$\mathbb{R}$$, $$S$$ has a supremum in $$\mathbb{R}$$.

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

But as $$m \in \mathbb{N}$$, it follows that $$m + 1 = \in \mathbb{N}$$.

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