Reciprocal Function is Strictly Decreasing/Proof 2

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

Strictly Increasing on $\openint 0 \to$

Let $a, b \in \Dom {\operatorname {recip} }$ such that $a$ and $b$ are both (strictly) 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 $\openint \gets 0$

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

Let $a < b$.

Then $a < b < 0$.

By Inversion Mapping Reverses Ordering in Ordered Group:

$0 < -b < -a$

By Properties of Totally Ordered Field:

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

By Negative of Product Inverse:

$\paren {-b}^{-1} = -b^{-1}$

$\paren {-a}^{-1} = -a^{-1}$

Thus:

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

By Inversion Mapping Reverses Ordering in Ordered Group:

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

By Inverse of Group Inverse:

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

Thus in particular:

$\dfrac 1 b < \dfrac 1 a$

$\blacksquare$