Positive Real Number Inequalities can be Multiplied

Theorem

Let $a, b, c, d \in \R$ such that $a > b$ and $c > d$.

Let $b > 0$ and $d > 0$.

Then $a c > b d$.

If $b < 0$ or $d < 0$ the inequality does not hold.

Proof

Note that as $d > 0$ it follows that $c > d > 0$ and so $c > 0$.

\(\ds a\)

\(>\)

\(\ds b\)

\(\ds \leadsto \ \ \)

\(\ds a c\)

\(>\)

\(\ds b c\)

Real Number Ordering is Compatible with Multiplication, as $c > 0$

\(\ds c\)

\(>\)

\(\ds d\)

\(\ds \leadsto \ \ \)

\(\ds b c\)

\(>\)

\(\ds b d\)

Real Number Ordering is Compatible with Multiplication, as $b > 0$

Finally:

\(\ds a c\)

\(>\)

\(\ds b c\)

\(\ds b c\)

\(>\)

\(\ds b d\)

\(\ds \leadsto \ \ \)

\(\ds a c\)

\(>\)

\(\ds b d\)

Trichotomy Law for Real Numbers

$\blacksquare$

Disproof for Negative Parameters

Proof by Counterexample:

Let $a = c = -1, b = d = -2$.

Then $a c = 1$ but $b d = 2$.

$\blacksquare$

Sources

• 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 1$: Real Numbers: Exercise $\S 1.8 \ (3)$