Inverse of Positive Real is Positive

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

The result follows from proof by contradiction.

$\blacksquare$

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