Real Number Ordering is Compatible with Multiplication

Theorem

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


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

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

Proof
This follows from the fact that the Rational Numbers form Subfield of Real Numbers.