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 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

$\blacksquare$