# Reciprocal of Strictly Negative Real Number is Strictly Negative

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## 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:

- $\dfrac 1 x < 0$

$\blacksquare$