Reciprocal Function is Strictly Decreasing/Proof 2

Theorem
The reciprocal function:


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

is strictly decreasing:


 * on the open interval $\left ({0 \,.\,.\, +\infty} \right)$


 * on the open interval $\left ({-\infty \,.\,.\, 0} \right)$

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

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:
 * $\dfrac 1 b < \dfrac 1 a$