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:

$\dfrac 1 x < 0$

$\blacksquare$


Also see