Ordering of Reciprocals/Proof 2

Theorem
Let $x, y \in \R$ be real numbers such that $x,y \in \left({0 \,.\,.\, \to}\right)$ or $x,y \in \left({\gets \,.\,.\, 0}\right)$

Then:
 * $x \le y \iff \dfrac 1 y \le \dfrac 1 x$

Proof
By Reciprocal Function Strictly Decreasing, the reciprocal function is strictly decreasing.

Thus
 * $x \le y \implies \dfrac 1 y \le \dfrac 1 x$

Suppose then that $\dfrac 1 y \le \dfrac 1 x$.

Since $x,y < 0$ or $x,y > 0$, we know that $\dfrac 1 y, \dfrac 1 x < 0$ or $\dfrac 1 y, \dfrac 1 x > 0$.

Thus we can apply the above to show that:
 * $\dfrac 1 {1/x} \le \dfrac 1 {1/y}$

By Inverse of Multiplicative Inverse:
 * $x \le y$.