# Division Theorem/Positive Divisor/Positive Dividend/Uniqueness/Proof 1

## Theorem

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

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

## Proof

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

Suppose $q_1, r_1$ and $q_2, r_2$ are two pairs of $q, r$ that satisfy $a = q b + r, 0 \le r < b$.

That is:

 $\displaystyle a$ $=$ $\displaystyle q_1 b + r_1, 0 \le r_1 < b$ $\displaystyle a$ $=$ $\displaystyle q_2 b + r_2, 0 \le r_2 < b$

This gives:

$0 = b \paren {q_1 - q_2} + \paren {r_1 - r_2}$

Aiming for a contradiction, suppose that $q_1 \ne q_2$.

Without loss of generality, suppose that $q_1 > q_2$.

Then:

 $\displaystyle q_1 - q_2$ $\ge$ $\displaystyle 1$ $\displaystyle \leadsto \ \$ $\displaystyle r_2 - r_1$ $=$ $\displaystyle b \left({q_1 - q_2}\right)$ $\displaystyle$ $\ge$ $\displaystyle b \times 1$ as $b > 0$ $\displaystyle$ $=$ $\displaystyle b$ $\displaystyle \leadsto \ \$ $\displaystyle r_2$ $\ge$ $\displaystyle r_1 + b$ $\displaystyle$ $\ge$ $\displaystyle b$

This contradicts the assumption that $r_2 < b$.

A similar contradiction follows from the assumption that $q_1 < q_2$.

Therefore $q_1 = q_2$ and so $r_1 = r_2$.

Thus it follows that $q$ and $r$ are unique.

$\blacksquare$