# 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 $a < 0 \implies a^{-1} < 0$.

## Proof

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

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

The result follows from Proof by Contradiction.

$\blacksquare$