# Order of Real Numbers is Dual of Order Multiplied by Negative Number

## Theorem

- $\forall x, y, z \in \R: x > y, z < 0 \implies x \times z < y \times z$

## Proof

Let $z < 0$.

Then from Real Number is Greater than Zero iff its Negative is Less than Zero:

- $-z > 0$

and so:

\(\displaystyle x\) | \(>\) | \(\displaystyle y\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x \times \paren {-z}\) | \(>\) | \(\displaystyle y \times \paren {-z}\) | Real Number Axioms: $\R O2$: compatibility with multiplication | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle -\paren {x \times z}\) | \(>\) | \(\displaystyle -\paren {y \times z}\) | Multiplication by Negative Real Number | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x \times z\) | \(<\) | \(\displaystyle y \times z\) | Order of Real Numbers is Dual of Order of their Negatives |

$\blacksquare$

## Sources

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