Division Theorem/Positive Divisor/Positive Dividend/Uniqueness/Proof 1
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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.
Suppose $q_1, r_1$ and $q_2, r_2$ are two pairs of $q, r$ that satisfy $a = q b + r, 0 \le r < b$.
That is:
\(\ds a\) | \(=\) | \(\ds q_1 b + r_1, 0 \le r_1 < b\) | ||||||||||||
\(\ds a\) | \(=\) | \(\ds q_2 b + r_2, 0 \le r_2 < b\) |
This gives:
- $0 = b \paren {q_1 - q_2} + \paren {r_1 - r_2}$
Aiming for a contradiction, suppose that $q_1 \ne q_2$.
Without loss of generality, suppose that $q_1 > q_2$.
Then:
\(\ds q_1 - q_2\) | \(\ge\) | \(\ds 1\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds r_2 - r_1\) | \(=\) | \(\ds b \paren {q_1 - q_2}\) | |||||||||||
\(\ds \) | \(\ge\) | \(\ds b \times 1\) | as $b > 0$ | |||||||||||
\(\ds \) | \(=\) | \(\ds b\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds r_2\) | \(\ge\) | \(\ds r_1 + b\) | |||||||||||
\(\ds \) | \(\ge\) | \(\ds b\) |
This contradicts the assumption that $r_2 < b$.
A similar contradiction follows from the assumption that $q_1 < q_2$.
Therefore $q_1 = q_2$ and so $r_1 = r_2$.
Thus it follows that $q$ and $r$ are unique.
$\blacksquare$
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 $6$