Reciprocal Function is Strictly Decreasing/Proof 1

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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

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.

$\blacksquare$