Signum Function of Reciprocal

Theorem

Let $x \in \R$ such that $x \ne 0$.

Then:

$\map \sgn x = \map \sgn {\dfrac 1 x}$

where $\map \sgn x$ denotes the signum of $x$.

Proof

\(\ds \map \sgn x\)

\(=\)

\(\ds 1\)

\(\ds \leadstoandfrom \ \ \)

\(\ds x\)

\(>\)

\(\ds 0\)

Definition of Signum Function

\(\ds \leadstoandfrom \ \ \)

\(\ds \frac 1 x\)

\(>\)

\(\ds 0\)

Reciprocal of Strictly Positive Real Number is Strictly Positive

\(\ds \leadstoandfrom \ \ \)

\(\ds \map \sgn {\dfrac 1 x}\)

\(=\)

\(\ds 1\)

Definition of Signum Function

$\blacksquare$