Question

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

Answer 1

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.