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