Signum Function is Completely Multiplicative

Theorem

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

$\forall x, y \in \R: \map \sgn {x y} = \map \sgn x \map \sgn y$

Proof

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

Then:

\(\ds x y\)

\(=\)

\(\ds 0\)

\(\text {(1)}: \quad\)

\(\ds \leadsto \ \ \)

\(\ds \map \sgn {x y}\)

\(=\)

\(\ds 0\)

and either $\map \sgn x = 0$ or $\map \sgn y = 0$ and so:

\(\ds \map \sgn x \map \sgn y\)

\(=\)

\(\ds 0\)

\(\ds \)

\(=\)

\(\ds \map \sgn {x y}\)

from $(1)$ above

$\Box$

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

Then:

\(\ds \map \sgn x\)

\(=\)

\(\ds 1\)

\(\, \ds \land \, \)

\(\ds \map \sgn y\)

\(=\)

\(\ds 1\)

\(\ds \leadsto \ \ \)

\(\ds \map \sgn x \map \sgn y\)

\(=\)

\(\ds 1\)

and:

\(\ds x y\)

\(>\)

\(\ds 0\)

\(\ds \map \sgn {x y}\)

\(=\)

\(\ds 1\)

Definition of Signum Function

\(\ds \)

\(=\)

\(\ds \map \sgn x \map \sgn y\)

$\Box$

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

Then:

\(\ds \map \sgn x\)

\(=\)

\(\ds -1\)

\(\, \ds \land \, \)

\(\ds \map \sgn y\)

\(=\)

\(\ds -1\)

\(\ds \leadsto \ \ \)

\(\ds \map \sgn x \map \sgn y\)

\(=\)

\(\ds 1\)

and:

\(\ds x y\)

\(>\)

\(\ds 0\)

\(\ds \map \sgn {x y}\)

\(=\)

\(\ds 1\)

Definition of Signum Function

\(\ds \)

\(=\)

\(\ds \map \sgn x \map \sgn y\)

$\Box$

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

Then:

\(\ds \map \sgn x\)

\(=\)

\(\ds -1\)

\(\, \ds \land \, \)

\(\ds \map \sgn y\)

\(=\)

\(\ds 1\)

\(\ds \leadsto \ \ \)

\(\ds \map \sgn x \map \sgn y\)

\(=\)

\(\ds -1\)

and:

\(\ds x y\)

\(<\)

\(\ds 0\)

\(\ds \map \sgn {x y}\)

\(=\)

\(\ds -1\)

Definition of Signum Function

\(\ds \)

\(=\)

\(\ds \map \sgn x \map \sgn y\)

$\Box$

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

$\blacksquare$