Reciprocal of Strictly Negative Real Number is Strictly Negative
Jump to navigation
Jump to search
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:
\(\ds x\) | \(<\) | \(\ds 0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x \times \dfrac 1 x\) | \(<\) | \(\ds 0 \times 0\) | Real Number Ordering is Compatible with Multiplication: Negative Factor | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds 1\) | \(<\) | \(\ds 0\) | Real Number Axiom $\R \text A4$: Inverses for Addition |
But from Real Zero is Less than Real One:
- $1 > 0$
Therefore by Proof by Contradiction:
- $\dfrac 1 x < 0$
$\blacksquare$