Signum Function on Integers is Extension of Signum on Natural Numbers

Theorem

Let $\sgn_\Z: \Z \to \set {-1, 0, 1}$ be the signum function on the integers.

Let $\sgn_\N: \N \to \set {0, 1}$ be the signum function on the natural numbers.

Then $\sgn_\Z: \Z \to \Z$ is an extension of $\sgn_\N: \N \to \N$.

Proof

Let $n \in \Z: n \ge 0$.

Then by definition of the signum function:

$\map {\sgn_\Z} n = \begin {cases} 0 & : n = 0 \\ 1 & : n > 0 \end {cases}$

So by definition of the signum function on the natural numbers:

$\forall n \in \N: \map {\sgn_\Z} n = \map {\sgn_\N} n$

Hence the result, by definition of extension.

$\blacksquare$