This is the third part of the Basic Algebra Theorems Proof Series. In this post, we use the Axioms of Real Numbers to show that the product of two a negative number and positive number is negative. That is, we show that the product of –a and b is –ab. Please refer to the the preceding link to verify the axioms used below.

Theorem

For any a, b = (-a)b = –ab.

Proof

We know that –ab is a unique solution to the equation x + ab  = 0, therefore it is sufficient to show that

ab + (-a)b = 0

But

ab + (-a)b = (a + (-a))b

by the  Distributive Property of Real Numbers (Axiom 5A) and 

a + (-a) = 0

by Axiom 5A (the existence of Additive Identity).

Therefore,

ab + (-a)b = (a + (-a))b = 0b = 0

and we are done.

The theorem above give to 2 corollaries.

Corollary 1

For any number b, (-1)b = –b.

If we take a = -1, then (-1)b = – (1b) = –b by the existence of multiplicative identity (Axiom 5M).

Corollary 2

(-1)(-1) = 1

Proof: Left as an exercise.