Division Theorem/Positive Divisor/Positive Dividend/Existence/Proof 2
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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 $a = 0$.
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 $b$:
\(\ds a\) | \(=\) | \(\ds \sum_{k \mathop = 0}^s r_k b^k\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds b \sum_{k \mathop = 0}^{s - 1} r_k b^{k - 1} + r_0\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds b q + r\) | where $0 \le r = r_0 < b$ |
$\blacksquare$
Sources
- 1971: George E. Andrews: Number Theory ... (previous) ... (next): $\text {2-1}$ Euclid's Division Lemma: Theorem $\text {2-1}$