Positive Real Number Inequalities can be Multiplied

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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 $c > 0$.

\(\displaystyle a\) \(>\) \(\displaystyle b\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle a c\) \(>\) \(\displaystyle b c\) Real Number Ordering is Compatible with Multiplication, as $c > 0$


\(\displaystyle c\) \(>\) \(\displaystyle d\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle b c\) \(>\) \(\displaystyle b d\) Real Number Ordering is Compatible with Multiplication, as $b > 0$


Finally:

\(\displaystyle a c\) \(>\) \(\displaystyle b c\)
\(\displaystyle b c\) \(>\) \(\displaystyle b d\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle a c\) \(>\) \(\displaystyle b d\) 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