# 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 $c > 0$.

 $\displaystyle a$ $>$ $\displaystyle b$ $\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

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

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

$\blacksquare$