# Division Theorem/Positive Divisor/Uniqueness

## 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:

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

In the above equation:

$a$ is the dividend
$b$ is the divisor
$q$ is the quotient
$r$ is the principal remainder, or, more usually, just the remainder.

## Proof 1

It is given by Division Theorem: Positive Divisor: 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.

$\blacksquare$

## Proof 2

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

Let $a = q b + r$ where $q, r \in \Z$ and $0 \le r < b$.

Thus:

$\dfrac a b = q + \dfrac r b$

and:

$0 \le \dfrac r b \le \dfrac {b - 1} b < 1$

So:

$q = \floor {\dfrac a b}$

and so:

$r = a - b \floor {\dfrac a b}$

Thus, given $a$ and $b$, the numbers $q$ and $r$ are unique determined.

$\blacksquare$

## Proof 3

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 $r_1 \ge r_2$.

Then:

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

That is:

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

where $\divides$ denotes divisibility.

But:

$r_1 - r_2 < b$
$r_1 - r_2 \ge b$

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

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

$\blacksquare$