Real Number Ordering is Compatible with Multiplication/Negative Factor

Theorem

 * $\forall a, b, c \in \R: a < b \land c < 0 \implies a c > b c$

where $\R$ is the set of real numbers.

Proof
From Real Numbers form Ordered Integral Domain, $\struct {\R, +, \times, \le}$ forms an ordered integral domain.

Thus: