Proof that (-a)(-b) = ab

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. 

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