Inverse of Positive Real is Positive

Theorem

Let $a \in \R$ such that $a > 0$.

Then $a^{-1} = \dfrac 1 a > 0$.

It follows directly that $a < 0 \implies a^{-1} < 0$.

Suppose $a > 0$ but $a^{-1} \le 0$.

$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle a^{-1}$	$\le$	$\displaystyle $	$\displaystyle 0$	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle \implies$	$\displaystyle $	$\displaystyle a \times a^{-1}$	$\le$	$\displaystyle $	$\displaystyle a \times 0$	$\displaystyle $	$\displaystyle $	Real Number Ordering is Compatible with Multiplication
$\displaystyle $	$\displaystyle \implies$	$\displaystyle $	$\displaystyle 1$	$\le$	$\displaystyle $	$\displaystyle 0$	$\displaystyle $	$\displaystyle $

The result follows from Proof by Contradiction.

$\blacksquare$

\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle a^{-1}\)	\(\le\)	\(\displaystyle \)	\(\displaystyle 0\)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle a \times a^{-1}\)	\(\le\)	\(\displaystyle \)	\(\displaystyle a \times 0\)	\(\displaystyle \)	\(\displaystyle \)	Real Number Ordering is Compatible with Multiplication
\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle 1\)	\(\le\)	\(\displaystyle \)	\(\displaystyle 0\)	\(\displaystyle \)	\(\displaystyle \)