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 x + ab = 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.
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).
(-1)(-1) = 1
Proof: Left as an exercise.