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

$\blacksquare$