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 $c > 0$.

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

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

Finally:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle a c\)	\(>\)	\(\displaystyle b c\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle b c\)	\(>\)	\(\displaystyle b d\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle a c\)	\(>\)	\(\displaystyle b d\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Trichotomy Law for Real Numbers

$\blacksquare$

Disproof for Negative Parameters

Proof by Counterexample:

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

Then $ac = 1$ but $bd = 2$.

$\blacksquare$

Sources

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