Sign of Quotient of Factors of Difference of Squares

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Theorem

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

Then

$\map \sgn {a^2 - b^2} = \map \sgn {\dfrac {a + b} {a - b} } = \map \sgn {\dfrac {a - b} {a + b} }$

where $\sgn$ denotes the signum of a real number.


Corollary

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

Then

$-\map \sgn {\dfrac {b - a} {b + a} } = \map \sgn {a^2 - b^2} = -\map \sgn {\dfrac {b + a} {b - a} }$

where $\sgn$ denotes the signum of a real number.


Proof

\(\displaystyle \map \sgn {\frac {a - b} {a + b} }\) \(=\) \(\displaystyle \map \sgn {a - b} \frac 1 {\map \sgn {a + b} }\) Signum Function is Completely Multiplicative
\(\displaystyle \) \(=\) \(\displaystyle \map \sgn {a - b} \map \sgn {a + b}\) Signum Function of Reciprocal
\(\displaystyle \) \(=\) \(\displaystyle \map \sgn {\paren {a - b} \paren {a + b} }\) Signum Function is Completely Multiplicative
\(\displaystyle \) \(=\) \(\displaystyle \map \sgn {a^2 - b^2}\) Difference of Two Squares
\(\displaystyle \) \(=\) \(\displaystyle \map \sgn {\paren {a + b} \paren {a - b} }\) Difference of Two Squares
\(\displaystyle \) \(=\) \(\displaystyle \map \sgn {a + b} \map \sgn {a - b}\) Signum Function is Completely Multiplicative
\(\displaystyle \) \(=\) \(\displaystyle \map \sgn {a + b} \frac 1 {\map \sgn {a - b} }\) Signum Function of Reciprocal
\(\displaystyle \) \(=\) \(\displaystyle \map \sgn {\frac {a + b} {a - b} }\) Signum Function is Completely Multiplicative

$\blacksquare$