Sign of Quotient of Factors of Difference of Squares

Theorem
Let $a, b \in \R$ such that $a \ne b$.

Then
 * $\operatorname{sgn} \left({a^2 - b^2}\right) = \operatorname{sgn} \left({\dfrac {a + b} {a - b} }\right)$

where $\operatorname{sgn}$ denotes the signum of a real number:
 * $\forall x \in \R_{\ne 0}: \operatorname{sgn} \left({x}\right) = \dfrac x {\left\vert{x}\right\vert}$

where $\left\vert{x}\right\vert$ denotes the absolute value of $x$.

Proof
Let $a^2 - b^2 > 0$.

Then: