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:

\(\displaystyle a\) \(=\) \(\displaystyle q_1 b + r_1, 0 \le r_1 < b\) $\quad$ $\quad$
\(\displaystyle a\) \(=\) \(\displaystyle q_2 b + r_2, 0 \le r_2 < b\) $\quad$ $\quad$


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:

\(\displaystyle q_1 - q_2\) \(\ge\) \(\displaystyle 1\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle r_2 - r_1\) \(=\) \(\displaystyle b \left({q_1 - q_2}\right)\) $\quad$ $\quad$
\(\displaystyle \) \(\ge\) \(\displaystyle b \times 1\) $\quad$ as $b > 0$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle b\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle r_2\) \(\ge\) \(\displaystyle r_1 + b\) $\quad$ $\quad$
\(\displaystyle \) \(\ge\) \(\displaystyle b\) $\quad$ $\quad$

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.

$\blacksquare$


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