Why do we get a positive integer on multiplying two negative integers?

1 Answer
Apr 15, 2017

Answer:

Use distributivity of multiplication over addition and other properties of arithmetic to demonstrate...

Explanation:

Addition and multiplication of integers have various properties, known as axioms. I will use the shorthand #AA# "for all", #EE# "there exists", #:# "such that" as follows:

There is an additive identity #0#:

#EE 0 : AA a " " a+0 = 0+a = a#

Addition is commutative:

#AA a, b " " a+b = b+a#

Addition is associative:

#AA a, b, c " " (a+b)+c = a+(b+c)#

All integers have an inverse under addition:

#AA a EE b : a+b = b+a = 0#

There is a multiplicative identity #1#:

#EE 1 : AA a " " a*1 = 1*a = a#

Multiplication is commutative:

#AA a, b " " a*b = b*a#

Multiplication is associative:

#AA a, b, c " " (a * b) * c = a * (b * c)#

Multiplication is left and right distributive over addition:

#AA a, b, c " " a * (b + c) = (a * b) + (a * c)#

#AA a, b, c " " (a + b) * c = (a * c) + (b * c)#

We use the notation #-a# to represent the additive inverse of #a# and the notation #a-b# as a shorthand for #a+(-b)#.

Note that associativity of addition means that we can unambiguously write:

#a+b+c#

Using the PEMDAS convention that addition and subtraction are performed left to right, we can avoid writing some more brackets yet keep things unambiguous.

Then we find:

#(-a)(-b) = (-a)(-b)+0#

#color(white)((-a)(-b)) = (-a)(-b)+(-ab)+ab#

#color(white)((-a)(-b)) = ((-a)(-b)-ab)+ab#

#color(white)((-a)(-b)) = ((-a)(-b)+0-ab)+ab#

#color(white)((-a)(-b)) = ((-a)(-b)+(a)(-b)-(a)(-b)-ab)+ab#

#color(white)((-a)(-b)) = ((-a)(-b)+(a)(-b))-((a)(-b)+ab))+ab#

#color(white)((-a)(-b)) = ((-a)+a)(-b)-(a)((-b)+b))+ab#

#color(white)((-a)(-b)) = (0*(-b))-(a*0)+ab#

#color(white)((-a)(-b)) = 0-0+ab#

#color(white)((-a)(-b)) = 0+ab#

#color(white)((-a)(-b)) = ab#

So if #a, b# are positive and you are content that #ab# is also positive, then #(-a)*(-b) = ab# is also positive.