# Inequality of Product of Unequal Numbers

## Theorem

Let $a, b, c, d \in \R$.

Then:

$0 < a < b \land 0 < c < d \implies 0 < a c < b d$

## Proof

 $\text {(1)}: \quad$ $\ds$  $\ds 0 < a < b$ $\text {(2)}: \quad$ $\ds$ $\leadsto$ $\ds 0 < b$ Ordering is Transitive

 $\text {(3)}: \quad$ $\ds$  $\ds 0 < c < d$ $\text {(4)}: \quad$ $\ds$ $\leadsto$ $\ds 0 < c$ Ordering is Transitive

 $\ds$ $\leadsto$ $\ds 0 < a c < b c$ $(1)$ and $(4)$: Real Number Axioms: $\R \text O 2$: Usual ordering is compatible with multiplication $\ds$  $\ds 0 < b c < b d$ $(2)$ and $(3)$: Real Number Axioms: $\R \text O 2$: Usual ordering is compatible with multiplication $\ds$ $\leadsto$ $\ds 0 < a c < b d$ Ordering is Transitive

$\blacksquare$