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

Apr 15, 2017

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 $\forall$ "for all", $\exists$ "there exists", $:$ "such that" as follows:

There is an additive identity $0$:

$\exists 0 : \forall a \text{ } a + 0 = 0 + a = a$

$\forall a , b \text{ } a + b = b + a$

$\forall a , b , c \text{ } \left(a + b\right) + c = a + \left(b + c\right)$

All integers have an inverse under addition:

$\forall a \exists b : a + b = b + a = 0$

There is a multiplicative identity $1$:

$\exists 1 : \forall a \text{ } a \cdot 1 = 1 \cdot a = a$

Multiplication is commutative:

$\forall a , b \text{ } a \cdot b = b \cdot a$

Multiplication is associative:

$\forall a , b , c \text{ } \left(a \cdot b\right) \cdot c = a \cdot \left(b \cdot c\right)$

Multiplication is left and right distributive over addition:

$\forall a , b , c \text{ } a \cdot \left(b + c\right) = \left(a \cdot b\right) + \left(a \cdot c\right)$

$\forall a , b , c \text{ } \left(a + b\right) \cdot c = \left(a \cdot c\right) + \left(b \cdot c\right)$

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

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:

$\left(- a\right) \left(- b\right) = \left(- a\right) \left(- b\right) + 0$

$\textcolor{w h i t e}{\left(- a\right) \left(- b\right)} = \left(- a\right) \left(- b\right) + \left(- a b\right) + a b$

$\textcolor{w h i t e}{\left(- a\right) \left(- b\right)} = \left(\left(- a\right) \left(- b\right) - a b\right) + a b$

$\textcolor{w h i t e}{\left(- a\right) \left(- b\right)} = \left(\left(- a\right) \left(- b\right) + 0 - a b\right) + a b$

$\textcolor{w h i t e}{\left(- a\right) \left(- b\right)} = \left(\left(- a\right) \left(- b\right) + \left(a\right) \left(- b\right) - \left(a\right) \left(- b\right) - a b\right) + a b$

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

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

$\textcolor{w h i t e}{\left(- a\right) \left(- b\right)} = \left(0 \cdot \left(- b\right)\right) - \left(a \cdot 0\right) + a b$

$\textcolor{w h i t e}{\left(- a\right) \left(- b\right)} = 0 - 0 + a b$

$\textcolor{w h i t e}{\left(- a\right) \left(- b\right)} = 0 + a b$

$\textcolor{w h i t e}{\left(- a\right) \left(- b\right)} = a b$

So if $a , b$ are positive and you are content that $a b$ is also positive, then $\left(- a\right) \cdot \left(- b\right) = a b$ is also positive.