Division Theorem/Positive Divisor/Existence/Proof 1

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Theorem

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

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


Proof

From Division Theorem: Positive Divisor: Positive Dividend: Existence:

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

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

$\Box$


It remains to be shown that the statement holds for $a < 0$.


From Division Theorem: Positive Divisor: Positive Dividend:

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

where $\size a$ denotes the absolute value of $a$: by definition $\size a \ge 0$.


As $a < 0$ it follows by definition of absolute value that $\size a = -a$.

This gives:

\(\displaystyle a\) \(=\) \(\displaystyle -\size a\)
\(\displaystyle \) \(=\) \(\displaystyle -\paren {\tilde q b + \tilde r}\)
\(\displaystyle \) \(=\) \(\displaystyle b \paren {-\tilde q} + \paren {-\tilde r}\)


Let $\tilde r = 0$.

Then:

$q = -\tilde q, r = \tilde r = 0$

which gives:

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

as required.


Otherwise:

$0 < \tilde r < b \implies 0 < b - \tilde r < b$

which suggests a rearrangement of the expression for $a$ above:

\(\displaystyle a\) \(=\) \(\displaystyle b \paren {-\tilde q} + \paren {-\tilde r}\)
\(\displaystyle \) \(=\) \(\displaystyle b \paren {-\tilde q} - b + b - \tilde r\)
\(\displaystyle \) \(=\) \(\displaystyle b \paren {-1 - \tilde q} + \paren {b - \tilde r}\)


Taking:

$q = \paren {-1 - \tilde q}$

and:

$r = \paren {b - \tilde r}$

the required result follows.


Uniqueness of $q$ and $r$ follow from the Uniqueness of $\tilde q$ and $\tilde r$.

$\blacksquare$


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