This is the fourth post of the Basic Algebra Theorems Series, and in this post, we show that (-*a*)(-*b*) = *ab*. In proving the theorems, please refer to the Axioms of Real Numbers.

**Theorem**

For any a, b, (-*a*)(-*b*) = *ab*

**Proof**

By the first corollary in the previous theorem, -a = (-1)a and -b = (-1)b. Using the Commutative Axiom of Multiplication (Axiom 3M) and the associative axiom (Axiom 2M) several times, results to

(-*a*)(-*b*) = (-1)(*a*)(-1)(-*b*) = (-1)(-1)(*a*)(*b*).

By the second corollary in the previous theorem, we have

(-1)(-1)(*a*)(*b*) = *ab. *

-(a+b)=(-a)+(-b)