Signum Function is Completely Multiplicative

Theorem
The signum function on the set of real numbers is a completely multiplicative function:


 * $\forall x, y \in \R: \operatorname{sgn} \left({x y}\right) = \operatorname{sgn} \left({x}\right) \operatorname{sgn} \left({y}\right)$

Proof
Let $x = 0$ or $y = 0$.

Then:

and either $\operatorname{sgn} \left({x}\right) = 0$ or $\operatorname{sgn} \left({y}\right) = 0$ and so:

Let $x > 0$ and $y > 0$.

Then:

and:

Let $x < 0$ and $y < 0$.

Then:

and:

Let $x < 0$ and $y > 0$.

Then:

and:

The same argument, mutatis mutandis, covers the case where $x > 0$ and $y < 0$.