# Division Theorem/Positive Divisor/Positive Dividend/Existence/Proof 3

## Theorem

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

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

## Proof

Let $a, b \in \Z$ such that $a \ge 0$ and $b > 0$ be given.

When $a = 0$, the integers $q = r = 0$ satisfy the conditions of the theorem.

Let $a > 0$.

For each $k \in \closedint 0 {a + 1}$, let $r_k = b k$.

(Note that here, as elsewhere in this proof, $\closedint 0 {a + 1}$ denotes an integer interval.)

Then $\sequence {r_k}_{0 \mathop \le k \mathop \le a + 1}$ is a strictly increasing sequence of positive integers.

It is also the case that:

$a + 1 \in \closedint 1 {\map b {a + 1} }$

So from Strictly Increasing Sequence induces Partition there exists $q \in \closedint 0 a$ such that:

$a + 1 \in \closedint {r_q + 1} {r_{q + 1} }$

Let $r = a - b q$.

Then $a = b q + r$.

We have:

 $\ds b q + 1$ $=$ $\ds r_q + 1$ $\ds$ $\le$ $\ds a + 1$ $\ds$ $\le$ $\ds r_{q + 1}$ $\ds$ $=$ $\ds b q + b$

and so:

 $\ds 0$ $\le$ $\ds r$ $\ds$ $=$ $\ds \paren {a + 1} - \paren {b q + 1}$ $\ds$ $\le$ $\ds \paren {b q + b} - \paren {b q + 1}$ $\ds$ $=$ $\ds b - 1$ $\ds$ $<$ $\ds b$

Hence the result.

$\blacksquare$