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 $x < 0$ |
$\blacksquare$