# Product of Negatives

By Susam Pal on 28 Feb 2010

## Negative Times Negative

A negative real number multiplied by a negative real number results in a positive real number.

We learn this fact in school after our knowledge of numbers expand to include the negative real numbers. We usually take this fact for granted. In this post, we will prove it.

In this discussion, we assume that we already know some basic properties of arithmetic operations such as the distributive property of multiplication over subtraction, existence of the additive inverse of real numbers, etc.

We also take for granted the fact the product of a positive real number and a negative real number is a negative real number, i.e., $$a \times (-b) = (-a) \times b = -(a \times b)$$ where $$a$$ and $$b$$ are arbitrary positive real numbers. We only want to show that the product of two negative real numbers must be positive, i.e., $$(-a) \times (-b) = a \times b$$ where $$a$$ and $$b$$ are arbitrary positive real numbers.

These assumptions help us to focus on the interesting parts of the arithmetic operations involved rather than getting bogged down in the details of every property of arithmetic operations we need to use.

## Intuitive Example

Let us first understand why the product of two negative numbers must be positive by taking a concrete example. We start with something very ordinary, say, $7 \times 8 = 56.$ Now let us rewrite $$7$$ as $$(10 - 3)$$ and $$8$$ as $$(10 - 2),$$ so that we get $(10 - 3) \times (10 - 2) = 56.$

Using the distributive property of multiplication over subtraction, we get \begin{align*} & (10 - 3) \times 10 + (10 - 3) \times (-2) = 56 \\ & \iff 10 \times 10 + (-3) \times 10 + 10 \times (-2) + (-3) \times (-2) = 56 \\ & \iff 100 + (-30) + (-20) + (-3) \times (-2) = 56 \\ & \iff 50 + (-3) \times (-2) = 56 \\ & \iff 50 + (-3) \times (-2) - 50 = 56 - 50 \\ & \iff (-3) \times (-2) = 6. \end{align*}

Hopefully, this builds up some intuition regarding why negative $$3$$ times negative $$2$$ must be positive $$6.$$ We see that $$(10 - 3) \times (10 - 2) = 56$$ is true if and only if $$(-3) \times (-2) = 6.$$

This example demonstrates that the fact that the product of two negative real numbers is a positive real number definitely holds good for $$-3$$ and $$-2.$$ But does it hold good for any two arbitrary real numbers?

## Algebraic Proof

To show that the fact that the product of two negative real numbers is a positive real number holds good for all real numbers, we show that for any two positive real numbers $$a$$ and $$b,$$ $$(-a) \times (-b) = a \times b.$$

We know that $a - a = 0.$ Multiplying both sides by $$-b,$$ we get $(a - a) \times (-b) = 0 \times (-b).$ Using the distributive property of multiplication over subtraction, we get $a \times (-b) + (-a) \times (-b) = 0.$ We know that $$a \times (-b) = -(a \times b).$$ We are trying to find out what $$(-a) \times (-b)$$ is. Using what we know, we get $-(a \times b) + (-a) \times (-b) = 0.$ Adding $$a \times b$$ to both sides, we get $(-a) \times (-b) = a \times b.$