Ordering of Reciprocals

Theorem

Let $x, y \in \R$ be real numbers such that $x, y \in \openint 0 \to$ or $x, y \in \openint \gets 0$

Then:

$x \le y \iff \dfrac 1 y \le \dfrac 1 x$

Proof 1

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

By Mapping from Totally Ordered Set is Dual Order Embedding iff Strictly Decreasing, the reciprocal function is a dual order embedding.

That is:

$x \le y \iff \dfrac 1 y \le \dfrac 1 x$

$\blacksquare$

Proof 2

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

Thus

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

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

If $x, y > 0$, then from Reciprocal of Strictly Positive Real Number is Strictly Positive: $\dfrac 1 y, \dfrac 1 x > 0$

Similarly, if $x, y < 0$ or $x, y > 0$, then from Reciprocal of Strictly Negative Real Number is Strictly Negative:

$\dfrac 1 y, \dfrac 1 x < 0$

Thus we can apply the above to show that:

$\dfrac 1 {1/x} \le \dfrac 1 {1/y}$

By Inverse of Multiplicative Inverse:

$x \le y$

$\blacksquare$