Ordering of Reciprocals

Theorem
Let $x, y \in \R$ be real numbers such that $x > 0, y > 0$.

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

Corollary
The result holds for $x < 0, y < 0$ as well.

Proof
We have that the real numbers form a totally ordered field.

When $x = y$ it follows directly from the properties of a field that $\dfrac 1 y = \dfrac 1 x$.

Otherwise $x < y$.

It then follows from Properties of Totally Ordered Fields that $\dfrac 1 y < \dfrac 1 x$.

The result follows.