# Reciprocal of Positive Real Number is Positive

 It has been suggested that this page or section be merged into Reciprocal of Strictly Positive Real Number is Strictly Positive. (Discuss)

## 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

Aiming for a contradiction, suppose $a > 0$ but $a^{-1} \le 0$.

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

The result follows from Proof by Contradiction.

$\blacksquare$