TITLE: Archimedean property of the rational numbers
QUESTION [10 upvotes]: Can you prove the Archimedean property of the rational numbers without constructing the reals and using the least upper bound property? It seems odd to have to take this roundabout approach, but I don't know any proof that avoids it.

REPLY [7 votes]: Let $x \in \mathbb{Q}$ 
If $x < 0$ there's nothing to do.
If $x \geqslant 0$ then we can write $x$ as $\dfrac{n}{m}$ (with no common factor other than 1), such that $n \in \mathbb{N}$ and $m \in N_{\ne 0}$
Since $m \geqslant 1$ then we have $n = mx \geqslant x$ which means $n + 1 > x$