The Product of Negative Number and a Positive Number is Negative

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.


For any a, b = (-a)b = –ab.


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

ab + (-a)b = 0


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).


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.


The Proof of the Cancellation Law of Addition

In the preceding posts, I have discussed the Axioms of Real Numbers and the Axioms of Equality. We will use these axioms to prove several basic theorems in Algebra. In this post, we prove the cancellation law of addition. That is, for any numbers a, b, and c,  if a + c = b + c, then a = b.


For any real numbers a, b, c,  if a + c = b + c, then a = b.


a + c = b + c

Adding  –c  to both sides, we have

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

Using Axiom 2A, the associativity of addition, we have

a + (c + (-c)) = b + (c + (-c)).

By Axiom 6A, the existence of additive inverse, we get

a + 0 = b + 0.

By Axiom 5A, the existence of additive identity

a = b.

That ends the proof.