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
\(\text {(1)}: \quad\) | \(\ds \) | \(\) | \(\ds 0 < a < b\) | |||||||||||
\(\text {(2)}: \quad\) | \(\ds \) | \(\leadsto\) | \(\ds 0 < b\) | 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 Axiom $\R \text O2$: Usual Ordering is Compatible with Multiplication | |||||||||||
\(\ds \) | \(\) | \(\ds 0 < b c < b d\) | $(2)$ and $(3)$: Real Number Axiom $\R \text O2$: Usual Ordering is Compatible with Multiplication | |||||||||||
\(\ds \) | \(\leadsto\) | \(\ds 0 < a c < b d\) | Ordering is Transitive |
$\blacksquare$