Proposition. Let $a\in \mathbb{Z}^+$, $b\in\mathbb{Z}$. Then there exist unique numbers $q,r\in\mathbb{Z}$ such that $b=qa+r; 0≤r<a.$
Proof.
Existence:
Define $S:=\{b-qa \mid q\in \mathbb{Z}, b-qa\in \mathbb{N}\}$. We show that $S$ is non-empty as follows:
- If $b≥0$, choose $q=0$,
- If $b<0$, choose $q=b$.
Then by the least integer principle, $S$ has a minimal element. Call it $r$. Thus $b=qa+r$, where $r≥0$, since $r\in\mathbb{N}$. Also, we have $r<a$ as otherwise $b-(q+1)a=r-a\in\mathbb{N}$, which contradicts the minimality of $r$.
How is the contradiction in the r≥a case established? I don't see how they even used the hypothesis $r≥a$ Wouldn't we require b-(q+1)a ≥ 0? since otherwise, it won't even be an element of S
