# Reciprocal Function is Strictly Decreasing/Proof 2

## Theorem

$\operatorname{recip}: \R \setminus \set 0 \to \R$, $x \mapsto \dfrac 1 x$
on the open interval $\openint 0 \to$
on the open interval $\openint \gets 0$

## Proof

### Strictly Increasing on $(0 \,.\,.\, \to)$

Let $a,b \in \operatorname{Dom}\left({\operatorname{recip}}\right)$ such that $a$ and $b$ are both positive.

Let $a < b$.

Then $0 < a < b$.

$0 < b^{-1} < a^{-1}$

That is, $\dfrac 1 b < \dfrac 1 a$.

$\Box$

### Strictly Increasing on $(\gets \,.\,.\, 0)$

Let $a,b \in \operatorname{Dom}\left({\operatorname{recip}}\right)$ such that $a$ and $b$ are both negative.

Let $a < b$.

Then $a < b < 0$.

$0 < -b < -a$
$0 < (-a)^{-1} < (-b)^{-1}$
$(-b)^{-1} = -b^{-1}$
$(-a)^{-1} = -a^{-1}$

Thus:

$0 < -a^{-1} < -b^{-1}$
$-(-b^{-1}) < -(-a^{-1}) < 0$
$b^{-1} < a^{-1} < 0$

Thus in particular:

$\dfrac 1 b < \dfrac 1 a$

$\blacksquare$