Signum Function on Integers is Extension of Signum on Natural Numbers

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Theorem

Let $\operatorname{sgn}_\Z: \Z \to \left\{ {-1, 0, 1}\right\}$ be the signum function on the integers.

Let $\operatorname{sgn}_\N: \N \to \left\{ {0, 1}\right\}$ be the signum function on the natural numbers.


Then $\operatorname{sgn}_\Z: \Z \to \Z$ is an extension of $\operatorname{sgn}_\N: \N \to \N$.


Proof

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

Then by definition of the signum function:

$\operatorname{sgn}_\Z \left({n}\right) = \begin{cases} 0 & : n = 0 \\ 1 & : n > 0 \end{cases}$

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

$\forall n \in \N: \operatorname{sgn}_\Z \left({n}\right) = \operatorname{sgn}_\N \left({n}\right)$


Hence the result, by definition of extension.

$\blacksquare$