Division Theorem/Positive Divisor/Uniqueness/Proof 3

Theorem
For every pair of integers $a, b$ where $b > 0$, the integers $q, r$ such that $a = q b + r$ and $0 \le r < b$ are unique:


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

Proof
It is given by Division Theorem: Positive Divisor: Existence that such $q$ and $r$ exist.

Suppose that:
 * $a = b q_1 + r_1 = b q_2 + r_2$

where both $0 \le r_1 < b$ and $0 \le r_2 < b$.

Suppose WLOG that $r_1 \ge r_2$.

Then:
 * $r_1 - r_2 = b \left({q_2 - q_1}\right)$

That is:
 * $b \mathop \backslash \left({r_2 - r_1}\right)$

where $\backslash$ denotes divisibility.

But:
 * $r_1 - r_2 < b$

while from Integer Absolute Value Greater than Divisors: Corollary:
 * $r_1 - r_2 \ge b$

unless from Integer Divides Zero $r_1 - r_2 = 0$.

So $r_1 = r_2$ and it follows directly that $q_1 = q_2$.