Solution

Prove the Archimedian property which states that if $a$ and $b$ are positive integers, then there exists some positive integer $n$ such that $na \ge b$.


We argue indirectly. Assume that there does not exist any positive integer $n$ such that $na \ge b$. That is to say, assume $$na < b \quad \forall n \in \mathbb{Z}^+$$

Subtracting $na$ from both sides, we then have $$b - na > 0 \quad \forall n \in \mathbb{Z}^+$$ This gives us a large collection of non-negative values with which to work -- perhaps the well-ordering principle might be useful here. Suppose we construct the set $S$ made up of all of these values. That is to say, suppose $$S = \{b - na, \textrm{ where } n \in \mathbb{Z}^+ \}$$ $S$ is clearly a non-empty set of non-negative values, so the well-ordering principle guarantees it has a least element, say $m = b - n_m a$.

But what happens if we consider $x = b - (n_m+1) a$?

Certainly, by its construction $x \in S$ and $x \lt m$. This contradicts $m$ being the least element of $S$.

Having reached a contradiction, the opposite of our assumption must be true. There must indeed exist some positive integer $n$ such that $na \ge b$.

QED.