Multiplying and Dividing Integers: The Sign Rules

The two rules

$$(+)(+) = + \qquad (-)(-) = + \qquad (+)(-) = - \qquad (-)(+) = -$$

Same signs give a positive. Different signs give a negative.

$$(-4)(-3) = 12 \qquad (-4)(3) = -12$$

Division behaves identically:

$$-12 \div -3 = 4 \qquad -12 \div 3 = -4$$

One set of rules covers both operations, which is no accident — dividing by $3$ is multiplying by $\tfrac{1}{3}$, and the sign of a reciprocal matches the sign of the number.

Operation	Signs	Result
$(-6)(-2)$	same	$12$
$(-6)(2)$	different	$-12$
$-20 \div -5$	same	$4$
$-20 \div 5$	different	$-4$

Why a positive times a negative is negative

Multiplication by a whole number is repeated addition:

$$3 \times (-4) = (-4) + (-4) + (-4) = -12$$

Three losses of four is a loss of twelve. Nothing surprising there — the negative carries through.

Why two negatives make a positive

This one deserves a real reason rather than a rule to memorise, and there are two.

The pattern. Step down through the multiplier:

Product	Value
$3 \times (-4)$	$-12$
$2 \times (-4)$	$-8$
$1 \times (-4)$	$-4$
$0 \times (-4)$	$0$
$-1 \times (-4)$	$?$

Each step down the table adds $4$. Continuing that pattern one more row gives $4$:

$$-1 \times (-4) = 4$$

Nothing else would keep the column consistent, and there is no reason for a pattern that has held for four rows to break at the fifth.

The distributive property. This is the argument that settles it. Start from something certainly true:

$$-4 \times 0 = 0$$

Write $0$ as $3 + (-3)$:

$$-4 \times (3 + (-3)) = 0$$

Distribute:

$$(-4)(3) + (-4)(-3) = 0$$

The first product is $-12$, so:

$$-12 + (-4)(-3) = 0$$

The only number that makes that true is $12$:

$$(-4)(-3) = 12$$

So two negatives multiplying to a positive is forced. Any other answer would break the distributive property, and that property is what makes all of algebra work.

Several factors at once

Negatives pair off, and each pair produces a positive. So only the count matters:

Negative factors	Sign of the product
even	positive
odd	negative

$$(-2)(-3)(-4) = -24 \qquad \text{three negatives, so negative}$$ $$(-2)(-3)(-4)(-1) = 24 \qquad \text{four negatives, so positive}$$

Work out the size from the digits, then apply the sign separately. That is quicker and less error-prone than tracking the sign through each step.

The same logic explains why an even power of a negative is positive: $(-2)^4$ is four negative factors, which pair off completely.

Division and zero

Every integer division works except dividing by zero:

$$\frac{-12}{-3} = 4 \qquad \frac{0}{-5} = 0 \qquad \frac{-5}{0} \text{ is undefined}$$

Division asks: what multiplied by the divisor gives the dividend? For $\tfrac{-5}{0}$ that is asking what times zero gives $-5$, and nothing does — every product with zero is zero.

Note also that an integer divided by an integer need not be an integer. $\tfrac{-7}{2}$ is $-3.5$, which is a perfectly good rational number and not a whole one.

Worked examples

Common mistakes

Practice problems