Division Theorem/Proof 1

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For every pair of integers $a, b$ where $b \ne 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $0 \le r < \size b$:

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


From Division Theorem: Positive Divisor:

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

That is, the result holds for positive $b$.


It remains to show that the result also holds for negative values of $b$.

Let $b < 0$.


$\size b = -b > 0$

where $\size b$ denotes the absolute value of $b$: by definition $\size b > 0$.

From Division Theorem: Positive Divisor, we have the existence of $\tilde q, \tilde r \in \Z$ such that:

$a = \tilde q \size b + \tilde r, 0 \le \tilde r < \size b$

Since $\size b = -b$:

$a = \tilde q \paren {-b} + \paren {\tilde r} = \paren {-\tilde q} b + \tilde r$


$q = -\tilde q, r = \tilde r$

the existence has been proved of integers $q$ and $r$ that satisfy the requirements.

The proof that they are unique is the same as that for the proof for positive $b$, but with $\size b$ replacing $b$.