# Reciprocal of Strictly Positive Real Number is Strictly Positive

## Contents

## Theorem

- $\forall x \in \R: x > 0 \implies \dfrac 1 x > 0$

## Proof

Let $x > 0$.

Aiming for a contradiction, suppose $\dfrac 1 x < 0$.

Then:

\(\displaystyle x\) | \(>\) | \(\displaystyle 0\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x \times \dfrac 1 x\) | \(<\) | \(\displaystyle 0 \times 0\) | Order of Real Numbers is Dual of Order Multiplied by Negative Number | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle 1\) | \(<\) | \(\displaystyle 0\) | Real Number Axioms: $\R M4$: Inverse |

But from Real Zero is Less than Real One:

- $1 > 0$

Therefore by Proof by Contradiction:

- $\dfrac 1 x > 0$

$\blacksquare$

## Also see

## Sources

- 2000: James R. Munkres:
*Topology*(2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Exercise $2 \ \text{(i)}$