# Reciprocal of Positive Real Number is Positive

Jump to navigation
Jump to search

## 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.

$\blacksquare$

## Sources

- 1977: K.G. Binmore:
*Mathematical Analysis: A Straightforward Approach*... (previous) ... (next): $\S 1$: Real Numbers: $\S 1.5$:*Example*