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