# 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:

- $r_1 - r_2 < b$

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 $q_1 = q_2$.

$\blacksquare$

## Sources

- 1971: George E. Andrews:
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