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 $\left ({-\infty \,.\,.\, 0} \right)$

Proof
Let $a,b \in \operatorname{Dom}\left({\operatorname{recip}}\right)$ such that $a$ and $b$ are both positive or both negative.

Let $a < b$

By Ordering of Reciprocals:
 * $\dfrac 1 b \le \dfrac 1 a$

Suppose for the sake of contradiction that $\dfrac 1 b = \dfrac 1 a$.

Then $\dfrac 1 {1/b} = \dfrac 1 {1/b}$.

By Inverse of Multiplicative Inverse: $b = a$, contradicting the fact that $a < b$.

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

The result follows from the dual of Order Monomorphism iff Strictly Increasing.