Reciprocal Function is Strictly Decreasing

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Theorem

The reciprocal function:

$\operatorname{recip}: \R \setminus \set 0 \to \R$, $x \mapsto \dfrac 1 x$

is strictly decreasing:

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:

$\dfrac 1 x = x^{-1}$

From Power Rule for Derivatives:

\(\displaystyle \frac {\mathrm d} {\mathrm d x} x^{-1}\) \(=\) \(\displaystyle -x^{-2}\)

From Square of Real Number is Non-Negative:

$-x^{-2} < 0$

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 $0 < a < b$.

By Properties of Totally Ordered Field:

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

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 $a < b < 0$.

By Group Inverse Reverses Ordering in Ordered Group:

$0 < -b < -a$

By Properties of Totally Ordered Field:

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

By Negative of Product Inverse:

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

Thus:

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

By Group Inverse Reverses Ordering in Ordered Group:

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

By Inverse of Group Inverse:

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

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.


Also see