GCD for Negative Integers
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Theorem
- $\gcd \set {a, b} = \gcd \set {\size a, b} = \gcd \set {a, \size b} = \gcd \set {\size a, \size b}$
Alternatively, this can be put:
- $\gcd \set {a, b} = \gcd \set {-a, b} = \gcd \set {a, -b} = \gcd \set {-a, -b}$
which follows directly from the above.
Proof
Note that $\size a = \pm a$.
Suppose that:
- $u \divides a$
where $\divides$ denotes divisibility.
Then:
- $\exists q \in \Z: a = q u$
Then:
- $\size a = \pm q u = \paren {\pm q} u \implies u \divides \size a$
So every divisor of $a$ is a divisor of $\size a$.
Similarly, note that:
- $a = \pm \size a$
so every divisor of $\size a$ is a divisor of $a$.
So it follows that the common divisors of $a$ and $b$ are the same as those of $a$ and $\size b$, and so on.
In particular:
- $\gcd \set {a, b} = \gcd \set {a, \size b}$
and so on.
$\blacksquare$