Division Theorem/Positive Divisor/Positive Dividend/Existence/Proof 1

Theorem
For every pair of integers $a, b$ where $a \ge 0$ and $b > 0$, there exist integers $q, r$ such that $a = q b + r$ and $0 \le r < b$.

That is:


 * $\forall a, b \in \Z, a \ge 0, b > 0: \exists q, r \in \Z: a = q b + r, 0 \le r < b$

Proof
Let $a, b \in \Z$ such that $a \ge 0$ and $b > 0$ be given.

Let $S$ be defined as the set of all positive integers of the form $a - z b$ where $z$ is an integer:


 * $S = \left\{{x \in \Z_{\ge 0}: \exists z \in \Z: x = a - z b}\right\}$

By setting $z = 0$ we have that $a \in S$.

Thus $S \ne \varnothing$.

We have that $S$ is bounded below by $0$.

From Set of Integers Bounded Below by Integer has Smallest Element it follows that $S$ has a smallest element $r$.

Thus:
 * $\exists q \in \Z: a - q b = r$

and so:
 * $a = q b + r$

So we have proved the existence of $q$ and $r$ such that $a = q b + r$.

It remains to be shown that $0 \le r < b$.

We have that $r \in S$ which is bounded below by $0$.

Therefore $0 \le r$.

Aiming for a contradiction, suppose $b \le r$.

So:

But then:

But $r - b < r$ contradicts the choice of $r$ as the least element of $S$.

Hence $r < b$ as required.

Thus the existence of $q$ and $r$ satisfying $a = q b + r, 0 \le r < b$ has been demonstrated.