Signum Function is Quotient of Number with Absolute Value

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Theorem

Let $x \in \R_{\ne 0}$ be a non-zero real number.

Then:

$\map \sgn x = \dfrac x {\size x} = \dfrac {\size x} x$

where:

$\map \sgn x$ denotes the signum function of $x$
$\size x$ denotes the absolute value of $x$.


Proof

Let $x \in \R_{\ne 0}$.

Then either $x > 0$ or $x < 0$.


Let $x > 0$.

Then:

\(\ds \frac x {\size x}\) \(=\) \(\ds \frac x x\) Definition of Absolute Value, as $x > 0$
\(\ds \) \(=\) \(\ds 1\)
\(\ds \) \(=\) \(\ds \map \sgn x\) Definition of Signum Function, as $x > 0$


Similarly:

\(\ds \frac {\size x} x\) \(=\) \(\ds \frac x x\) Definition of Absolute Value, as $x > 0$
\(\ds \) \(=\) \(\ds 1\)
\(\ds \) \(=\) \(\ds \map \sgn x\) Definition of Signum Function, as $x > 0$


$\Box$


Let $x < 0$.

Then:

\(\ds \frac x {\size x}\) \(=\) \(\ds \frac x {-x}\) Definition of Absolute Value, as $x < 0$
\(\ds \) \(=\) \(\ds -1\)
\(\ds \) \(=\) \(\ds \map \sgn x\) Definition of Signum Function, as $x < 0$


Similarly:

\(\ds \frac {\size x} x\) \(=\) \(\ds \frac {-x} x\) Definition of Absolute Value, as $x < 0$
\(\ds \) \(=\) \(\ds -1\)
\(\ds \) \(=\) \(\ds \map \sgn x\) Definition of Signum Function, as $x < 0$

$\blacksquare$