# Inverse of Positive Real is Positive

From ProofWiki

## Theorem

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

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

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

## Proof

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

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle a^{-1}\) | \(\le\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle 0\) | \(\displaystyle \) | \(\displaystyle \) | |||

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

\(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle 1\) | \(\le\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle 0\) | \(\displaystyle \) | \(\displaystyle \) |

The result follows from Proof by Contradiction.

$\blacksquare$