Multiplying and Dividing Negative Numbers

Two minus signs multiply to a plus.

A negative number, several times

Multiplying is repeated adding, so 3 × (−2) is three lots of −2: (−2) + (−2) + (−2) = −6. Three falls of 2 make a fall of 6.

A quiz that gives −3 marks for each wrong answer takes 4 × (−3) = −12 marks for 4 wrong answers. A positive number times a negative number is negative.

A column of multiplications

Now write the multiplications by −2 in a column: 3 × (−2) = −6, 2 × (−2) = −4, 1 × (−2) = −2 and 0 × (−2) = 0.

Down the column, the left number drops by 1 each time and the answer climbs by 2.

× (−2)3−62−41−200

Each row down, the number multiplied by −2 is 1 less, and the answer is 2 more.

Carry the column below zero

The next number down is −1. The pattern has no reason to stop at zero, so the answer climbs by 2 once more, from 0 to 2. So (−1) × (−2) = 2, and one more row gives (−2) × (−2) = 4.

A negative number times a negative number is positive.

× (−2)3−62−41−200−12−24

Below zero the answer keeps climbing by 2: (−1) × (−2) = 2 and (−2) × (−2) = 4.

The sign rules

Work out the size of the answer as usual, then decide its sign. Two numbers with the same sign multiply to a positive answer, and two numbers with different signs multiply to a negative answer.

So 4 × 3 = 12 and (−4) × (−3) = 12, while (−4) × 3 = −12 and 4 × (−3) = −12.

Two reversals

A walk shows why two negatives make a positive. A walker moves along a straight road, and position = velocity × time, measured from where the walker is now. A velocity is a speed with a direction: walking east is a positive velocity and walking west a negative one. A time in the future is positive and a time in the past negative.

Walking west at 4 meters a second, where was the walker 3 seconds ago? They were heading west, so 3 seconds ago they were 12 meters east of where they are now: (−4) × (−3) = 12. Reversing the direction makes the answer negative, and reversing the clock as well turns it back to positive.

−20−15−10−505101520position 6velocity 3 m/s · time 2 s3 × 2 = 6vt−5 west+5 east−4 s (ago)+4 s (ahead)

east at v and t seconds ahead: position = v × t > 0, east of the start

Set the velocity to −4 and the time to −3 and find the walker

Set the velocity to −4 and the time to −3. Making one of them negative puts the walker west of the start; making both negative puts the walker east again, at 12.

Division follows the same rules

Division undoes multiplication, so it follows the same sign rules. (−3) × (−2) = 6, so 6 ÷ (−2) = −3. And 3 × (−2) = −6, so (−6) ÷ (−2) = 3.

To divide, ask what times the divisor gives the number being divided. For (−16) ÷ (−4), ask what times −4 gives −16. 4 × (−4) = −16, so (−16) ÷ (−4) = 4.

Same signs give a positive answer and different signs give a negative answer, for division just as for multiplication.

÷ (−2)6−34−22−100−21−42−63

Each row down, the number divided by −2 drops by 2 and the answer climbs by 1. Below zero the answers are positive: (−6) ÷ (−2) = 3.

The usual mistakes

Keeping the minus signs. (−3) × (−4) is not −12. Two negatives multiplied give a positive, so (−3) × (−4) = 12.

Adding instead of multiplying. (−3) × (−4) is not −7. Multiplying asks for 3 lots of 4, which is 12, and then the signs decide: same signs, so positive.

Two negatives only make a positive when they are multiplied or divided. (−3) + (−4) is still −7, because two falls stack.

Worked example: Temperature Readings on Either Side of Zero

Question At a mountain hut the temperature at 6 a.m. was −6 °C. By noon it had risen by 10 °C. By 6 p.m. it had fallen by 7 °C from the noon reading, and by midnight it had fallen by a further 8 °C. (a) Find the temperature at midnight. (b) Find the mean of the four readings.

  1. 1.Start at −6 on the number line. A rise of 10 °C is a move of 10 to the right: −6 + 10 = 4. The noon reading is 4 °C.

    0−6+104
    0−6+104
    A rise of 10 °C is a move of 10 to the right: −6 + 10 = 4 °C at noon.
  2. 2.A fall of 7 °C is a move of 7 to the left. From 4 it takes 4 to reach zero and 3 more to pass it: 4 − 7 = −3. The 6 p.m. reading is −3 °C.

    0−6+104−7−3
    0−6+104−7−3
    A fall of 7 °C passes through zero: 4 − 7 = −3 °C at 6 p.m.
  3. 3.A further fall of 8 °C gives −3 − 8 = −11. (a) The temperature at midnight is −11 °C.

    0−6+104−7−3−8−11
    0−6+104−7−3−8−11
    (a) A further fall of 8 °C gives −3 − 8 = −11 °C at midnight.
  4. 4.Add the four readings. The one positive reading is 4, and the negative readings add up to −6 + (−3) + (−11) = −20. The sum is 4 + (−20) = −16.

    0−6+104−7−3−8−11−6 + 4 + (−3) + (−11) = −16
    0−6+104−7−3−8−11−6 + 4 + (−3) + (−11) = −16
    The four readings add up to 4 + (−20) = −16.
  5. 5.(b) The mean is −16 ÷ 4 = −4 °C, because a negative number divided by a positive number is negative. Check: −4 lies between the lowest reading, −11, and the highest, 4.

    0−6+104−7−3−8−11the sum −16, divided by 4, is −4mean −4
    0−6+104−7−3−8−11the sum −16, divided by 4, is −4mean −4
    (b) The mean is −16 ÷ 4 = −4 °C.

Answer: (a) −11 °C; (b) −4 °C

Common mistakes

  • Writing −3 − 8 = 5 or −3 − 8 = −5. The temperature is already below zero and it falls again, so the move is further to the left: −3 − 8 = −11.
  • Finding the mean of 6, 4, 3 and 11 and ignoring the signs. The signs are part of the readings, so the sum is −16, not 24.

More negative numbers and linear equations problems, worked step by step →

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