Division Theorem/Positive Divisor/Positive Dividend/Existence/Proof 1
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.
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.
$\blacksquare$
Sources
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $1$: Properties of the Natural Numbers: $\S 21$
- 1982: Martin Davis: Computability and Unsolvability (2nd ed.) ... (previous) ... (next): Appendix $1$: Some Results from the Elementary Theory of Numbers: Theorem $6$
- 1996: John F. Humphreys: A Course in Group Theory ... (next): $\text{A}.1$: Number theory: Theorem $\text{A}.1$