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 $b$:

\(\displaystyle a\) \(=\) \(\displaystyle \sum_{k \mathop = 0}^s r_k b^k\)
\(\displaystyle \) \(=\) \(\displaystyle b \sum_{k \mathop = 0}^{s - 1} r_k b^{k - 1} + r_0\)
\(\displaystyle \) \(=\) \(\displaystyle b q + r\) where $0 \le r = r_0 < b$

$\blacksquare$


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