Proof
(related to Lemma: Division with Quotient and Remainder)
 Let $a,b\in\mathbb Z$ be integers with $a > 0.$
 We first show that there is at least one pair of integers $q,r$ fulfilling $$b=qa + r,\quad 0\le r < a.\quad\quad ( * )$$
 Among the integers $bua$ there are negative and positive ones for sufficiently large negative or positive integer $u.$
 The set $S:=\{bua\mid u\in\mathbb Z\wedge bua\ge 0\}$ of nonnegetive integers $bua$ is not empty. Since this is a nonempty subset of natural numbers $S\subset \mathbb N$ and the ordered natural numbers $(N,\le)$ are wellordered, there exists a minimum $\min(S)$ for some $u_0\in\mathbb Z.$
 We set $q:=u_0$ and $r:=bqa.$ Then we have $bqa\ge 0$ and $ra=b(q1)a < 0.$
 This shows that the constructed $q,r$ solve $( * ).$
 We now show that there is at most one pair of integers $q,r$ fulfilling $( * ).$
 If we have found some $q,r$ fulfilling $( * ),$ then if $u < q$ then $u\le q1$ and $bua\ge b(q1)a=r+a\ge a.$
 If we have found some $q,r$ fulfilling $( * ),$ then if $u > q$ then $u\ge q+1$ and $bua\le b(q+1)a=ra < 0.$
 From this it follows that $0\le bua < a$ can only be fulfilled if $u=q.$
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References
Bibliography
 Kramer Jürg, von Pippich, AnnaMaria: "Von den natürlichen Zahlen zu den Quaternionen", SpringerSpektrum, 2013