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 = *0*b = *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* = – (1*b*) = -*b* by the existence of multiplicative identity (Axiom 5M).

**Corollary 2**

(-1)(-1) = 1

Proof: Left as an exercise.

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