Division Theorem/Positive Divisor/Positive Dividend/Existence

From ProofWiki
Jump to navigation Jump to search

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:

\(\displaystyle b\) \(\le\) \(\displaystyle r\)
\(\displaystyle \) \(<\) \(\displaystyle r + b\) as $b > 0$
\(\displaystyle \leadsto \ \ \) \(\displaystyle 0\) \(\le\) \(\displaystyle r - b\)
\(\displaystyle \) \(<\) \(\displaystyle r\) subtracting $b$ throughout

But then:

\(\displaystyle a\) \(=\) \(\displaystyle q b + r\) from above
\(\displaystyle \leadsto \ \ \) \(\displaystyle r - b\) \(=\) \(\displaystyle \paren {a - q b} - b\)
\(\displaystyle \) \(=\) \(\displaystyle a - b \paren {q + 1}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle r - b\) \(\in\) \(\displaystyle S\) 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 $b$:

\(\displaystyle a\) \(=\) \(\displaystyle \sum_{k \mathop = 0}^s r_k b^k\)
\(\displaystyle \) \(=\) \(\displaystyle b \sum_{k \mathop = 0}^{s - 1} r_k b^{k - 1} + r_0\)
\(\displaystyle \) \(=\) \(\displaystyle b q + r\) 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 $k \in \left[{0 \,.\,.\, a + 1}\right]$, let $r_k = b k$.

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

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

It is also the case that:

$a + 1 \in \left[{1 \,.\,.\, b \left({a + 1}\right)}\right]$

So from Strictly Increasing Sequence induces Partition there exists $q \in \left[{0 \,.\,.\, a}\right]$ such that:

$a + 1 \in \left[{r_q + 1 \,.\,.\, r_{q + 1}}\right]$

Let $r = a - b q$.

Then $a = b q + r$.

We have:

\(\displaystyle b q + 1\) \(=\) \(\displaystyle r_q + 1\)
\(\displaystyle \) \(\le\) \(\displaystyle a + 1\)
\(\displaystyle \) \(\le\) \(\displaystyle r_{q + 1}\)
\(\displaystyle \) \(=\) \(\displaystyle b q + b\)

and so:

\(\displaystyle 0\) \(\le\) \(\displaystyle r\)
\(\displaystyle \) \(=\) \(\displaystyle \left({a + 1}\right) - \left({b q + 1}\right)\)
\(\displaystyle \) \(\le\) \(\displaystyle \left({b q + b}\right) - \left({b q + 1}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle b - 1\)
\(\displaystyle \) \(<\) \(\displaystyle b\)

Hence the result.

$\blacksquare$