Reciprocal Function is Strictly Decreasing

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 1

Let $x \in \left ({0 \,.\,.\, +\infty} \right)$.

By the definition of negative powers:

$\dfrac 1 x = x^{-1}$
 $\displaystyle \frac {\mathrm d} {\mathrm d x} x^{-1}$ $=$ $\displaystyle -x^{-2}$
$-x^{-2} < 0$

for all $x$ within the domain.

Thus from Derivative of Monotone Function, $\operatorname{recip}$ is strictly decreasing.

The proof for $x \in \left ({-\infty \,.\,.\, 0} \right)$ is similar.

$\blacksquare$

Proof 2

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:

$\dfrac 1 b < \dfrac 1 a$

$\blacksquare$

Warning

Though the reciprocal function is decreasing on $\openint \gets 0$ and on $\openint 0 \to$, it is not decreasing on $\openint \gets 0 \cup \openint 0 \to$.

This is because there is a nonremovable discontinuity at the origin.