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.

Sufficient Condition
From the Trichotomy Law for Real Numbers, one of the following holds:


 * $(1): \quad a > b$
 * $(2): \quad a = b$
 * $(3): \quad a < b$

If $(1)$ or $(2)$, the statement:


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

holds vacuously.

If $(3)$,the result follows from Ordering of Reciprocals.

Necessary Condition
From the Trichotomy Law for Real Numbers, one of the following holds:


 * $(1): \quad \dfrac 1 a > \dfrac 1 b$


 * $(2): \quad \dfrac 1 a = \dfrac 1 b$


 * $(3): \quad \dfrac 1 a < \dfrac 1 b$

If $(1)$ or $(2)$, the statement:


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

holds vacuously.

If $(3)$: