# Division Theorem/Positive Divisor/Positive Dividend/Uniqueness

## 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 1

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:

\(\ds a\) | \(=\) | \(\ds q_1 b + r_1, 0 \le r_1 < b\) | ||||||||||||

\(\ds a\) | \(=\) | \(\ds q_2 b + r_2, 0 \le r_2 < b\) |

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:

\(\ds q_1 - q_2\) | \(\ge\) | \(\ds 1\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds r_2 - r_1\) | \(=\) | \(\ds b \paren {q_1 - q_2}\) | |||||||||||

\(\ds \) | \(\ge\) | \(\ds b \times 1\) | as $b > 0$ | |||||||||||

\(\ds \) | \(=\) | \(\ds b\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds r_2\) | \(\ge\) | \(\ds r_1 + b\) | |||||||||||

\(\ds \) | \(\ge\) | \(\ds b\) |

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: Positive Dividend: Existence that such $q$ and $r$ exist.

Let $a = 0$.

It is immediately apparent that $q = r = 0$ is the only possible solution with $0 \le r < b$.

$\Box$

Let $a > 0$ and $b = 1$.

Then from the condition $0 \le r < b$ it follows that $r = 0$.

Hence $r = 0, q = a$ is the only possible solution.

$\Box$

Let $a > 0$ and $b > 1$.

By the Basis Representation Theorem, $a$ has a unique representation to the base $b$:

\(\ds a\) | \(=\) | \(\ds \sum_{k \mathop = 0}^s r_k b^k\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds b \sum_{k \mathop = 0}^{s - 1} r_k b^{k - 1} + r_0\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds b q + r\) | where $0 \le r = r_0 < b$ |

Suppose a second pair $q', r'$ were to exist.

Then there would be a representation for $q'$ to the base $b$:

- $\displaystyle q' = \sum_{k \mathop = 0}^t u_k b^k$

so that:

\(\ds a\) | \(=\) | \(\ds b q' + r'\) | $0 \le r < b$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^t u_k b^{k + 1} + r'\) |

But there already exists a representation of $a$ to the base $b$:

- $\displaystyle a = \sum_{k \mathop = 0}^s r_k b^k$

By the Basis Representation Theorem, such a representation is unique.

So:

\(\ds t\) | \(=\) | \(\ds s - 1\) | ||||||||||||

\(\ds u_k\) | \(=\) | \(\ds r_{k + 1}\) | ||||||||||||

\(\ds r'\) | \(=\) | \(\ds a_0\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds r\) |

and hence $q' = q$.

$\blacksquare$