Ordering of Reciprocals/Proof 1

From ProofWiki
Jump to navigation Jump to search

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

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:

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

$\blacksquare$