# Division Theorem/Positive Divisor/Uniqueness/Proof 1

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## Theorem

For every pair of integers $a, b$ where $b > 0$, the integers $q, r$ such that $a = q b + r$ and $0 \le r < b$ are unique:

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

## Proof

It is given by Division Theorem: Positive Divisor: 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

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): $\S 24$: Theorem $24.1$