# Inequality of Product of Unequal Numbers

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## Theorem

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

Then:

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

## Proof

\(\text {(1)}: \quad\) | \(\displaystyle \) | \(\) | \(\displaystyle 0 < a < b\) | ||||||||||

\(\text {(2)}: \quad\) | \(\displaystyle \) | \(\leadsto\) | \(\displaystyle 0 < b\) | Ordering is Transitive |

\(\text {(3)}: \quad\) | \(\displaystyle \) | \(\) | \(\displaystyle 0 < c < d\) | ||||||||||

\(\text {(4)}: \quad\) | \(\displaystyle \) | \(\leadsto\) | \(\displaystyle 0 < c\) | Ordering is Transitive |

\(\displaystyle \) | \(\leadsto\) | \(\displaystyle 0 < a c < b c\) | $(1)$ and $(4)$: Real Number Axioms: $\R O 2$: Usual ordering is compatible with multiplication | ||||||||||

\(\displaystyle \) | \(\) | \(\displaystyle 0 < b c < b d\) | $(2)$ and $(3)$: Real Number Axioms: $\R O 2$: Usual ordering is compatible with multiplication | ||||||||||

\(\displaystyle \) | \(\leadsto\) | \(\displaystyle 0 < a c < b d\) | Ordering is Transitive |

$\blacksquare$