# How can we prove that a negative number multiplied/divided/subtacted by a negative gives a positive result?

## I've always wondered if we could prove this...

Apr 15, 2018

It's not the case that a negative minus a negative is always positive. For instance, $- 3 - \left(- 2\right) = - 3 + 2 = - 1$, which is still negative.

We can, however, rigorously prove that the product of two negative numbers is always positive. From that conclusion we can additionally prove that the same holds for a negative number divided by a negative number. The proof is not out of reach for an algebra student, either, as long as you are willing to accept that the product of two positive numbers is positive.

Consider that any negative number can be written in the form $- a$, where $a$ is a positive number.

Suppose we have two negative numbers, $x$ and $y$. Because any negative number can be written $- a$, where $a > 0$, $x = - a$ and $y = - b$, where $a , b$ are positive numbers. The product $x y$ is then equal to $\left(- a\right) \left(- b\right) = \left(- 1\right) \left(- 1\right) \left(a b\right) = a b$. The product is equivalent to the product of two positive numbers, which must also be positive. $\square$

We have proved that if $x , y$ are two negative numbers, then $x y$ is positive. This can be extended to division as follows.

Suppose we have negative numbers $x , y$. The quotient $\frac{x}{y}$ is equivalent to $x \left(\frac{1}{y}\right) = x z$, where $z = \frac{1}{y}$. Because $y$ is negative, $y = - a$, where a is some positive number. Then $z = \frac{1}{y} = \frac{1}{- a} = - \frac{1}{a}$. So $z$ is negative. Thus, we again have the product of two negative numbers, which (by above) must be positive.