# Sign of Quotient of Factors of Difference of Squares

## 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

$\blacksquare$