# Division Theorem/Positive Divisor/Positive Dividend/Uniqueness/Proof 2

## 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.

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 $q' = q$.

$\blacksquare$