Reciprocal Function is Strictly Decreasing/Proof 2

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

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