# Division Theorem/Positive Divisor/Positive Dividend/Existence

## 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 1

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

Let $S$ be defined as the set of all positive integers of the form $a - z b$ where $z$ is an integer:

$S = \set {x \in \Z_{\ge 0}: \exists z \in \Z: x = a - z b}$

By setting $z = 0$ we have that $a \in S$.

Thus $S \ne \O$.

We have that $S$ is bounded below by $0$.

From Set of Integers Bounded Below by Integer has Smallest Element it follows that $S$ has a smallest element $r$.

Thus:

$\exists q \in \Z: a - q b = r$

and so:

$a = q b + r$

So we have proved the existence of $q$ and $r$ such that $a = q b + r$.

It remains to be shown that $0 \le r < b$.

We have that $r \in S$ which is bounded below by $0$.

Therefore $0 \le r$.

Aiming for a contradiction, suppose $b \le r$.

So:

 $\ds b$ $\le$ $\ds r$ $\ds$ $<$ $\ds r + b$ as $b > 0$ $\ds \leadsto \ \$ $\ds 0$ $\le$ $\ds r - b$ $\ds$ $<$ $\ds r$ subtracting $b$ throughout

But then:

 $\ds a$ $=$ $\ds q b + r$ from above $\ds \leadsto \ \$ $\ds r - b$ $=$ $\ds \paren {a - q b} - b$ $\ds$ $=$ $\ds a - b \paren {q + 1}$ $\ds \leadsto \ \$ $\ds r - b$ $\in$ $\ds S$ as $r - b$ is of the correct form

But $r - b < r$ contradicts the choice of $r$ as the least element of $S$.

Hence $r < b$ as required.

Thus the existence of $q$ and $r$ satisfying $a = q b + r, 0 \le r < b$ has been demonstrated.

$\blacksquare$

## Proof 2

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 $r = 0$ and hence $q = a$.

$\Box$

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 $0 \le r = r_0 < b$

$\blacksquare$

## Proof 3

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.

$\blacksquare$