Division Theorem/Positive Divisor/Uniqueness/Proof 3

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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:

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


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 $0 \le r_2 < b$.

Without loss of generality, suppose $r_1 \ge r_2$.


$r_1 - r_2 = b \paren {q_2 - q_1}$

That is:

$b \divides \paren {r_2 - r_1}$

where $\divides$ denotes divisibility.


$r_1 - r_2 < b$

while from Absolute Value of Integer is not less 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 $q_1 = q_2$.