Sign of Quotient of Factors of Difference of Squares/Corollary

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Corollary to Sign of Quotient of Factors of Difference of Squares

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

\(\ds \map \sgn {\frac {b - a} {b + a} }\) \(=\) \(\ds \map \sgn {\paren {-1} \frac {a - b} {a + b} }\)
\(\ds \) \(=\) \(\ds \map \sgn {-1} \map \sgn {\frac {a - b} {a + b} }\) Signum Function is Completely Multiplicative
\(\ds \) \(=\) \(\ds \paren {-1} \map \sgn {\frac {a - b} {a + b} }\) Definition of Signum Function
\(\ds \) \(=\) \(\ds -\map \sgn {a^2 - b^2}\) Sign of Quotient of Factors of Difference of Squares
\(\ds \) \(=\) \(\ds \paren {-1} \map \sgn {\frac {a + b} {a - b} }\) Sign of Quotient of Factors of Difference of Squares
\(\ds \) \(=\) \(\ds \map \sgn {-1} \map \sgn {\frac {a + b} {a - b} }\) Definition of Signum Function
\(\ds \) \(=\) \(\ds \map \sgn {\paren {-1} \frac {a + b} {a - b} }\) Signum Function is Completely Multiplicative
\(\ds \) \(=\) \(\ds \map \sgn {\frac {b + a} {b - a} }\)

$\blacksquare$