Theorem 1

Let a and b be integers such that $b > 0$. Then there exist integers q and r such that $a = bq + r$ and $r < b$, and the integers q and r are uniquely determined by these conditions.

The conditions on $q$ and $r$ are equivalent to $bq \leqslant a < b(q + 1)$ and $r = a - bq$ (no. 2, Proposition 2). Hence we have to find $q$ such that $bq \leqslant a < b(q + 1)$; in other words, $q$ must be the smallest integer such that $a < b(q + 1)$, which shows that $q$ and $r = a - bq$ are uniquely determined. To prove their existence, we note that there exist integers $p$ such that $a < bp$, for example $a + 1$ (since $b > 0$). Let $m$ be the least of these integers. We have $m \neq 0$, and we may therefore write $m = q + 1$ with $q \leqslant m$ (§ 4, no. 2, Proposition 2); it follows that $bq \leqslant a < b(q + 1)$.