Primary & Middle School Mathematics · Grades 1–9

54Multiplying Negative Numbers

Grade 7 added and subtracted relative numbers (Chapter 46); it remains to multiply and divide them. One question dominates the chapter: why should “minus times minus” be plus? We give a reason, not just a rule.

54.1 The sign rules

Theorem 54.1 (Sign of a product)

The product of two relative numbers has distance to zero equal to the product of the distances, and its sign is given by:

$\times$	$+$	$-$
$+$	$+$	$-$
$-$	$-$	$+$

Same signs: positive product. Opposite signs: negative product. The same table governs quotients.

Why $(-1) \times (-1) = +1$ . First, $3 \times (-4)$ means $(-4) + (-4) + (-4) = -12$ : a positive times a negative is negative. Now watch the pattern:

3 \times (-4) = -12, \quad 2 \times (-4) = -8, \quad 1 \times (-4) = -4, \quad 0 \times (-4) = 0 .

Each step, the first factor drops by $1$ and the result rises by $4$ . Continuing the pattern one more step:

(-1) \times (-4) = 0 + 4 = +4 .

Any other choice would break the regularity of arithmetic (distributivity). So minus times minus must be plus. ∎

The pattern argument in a picture: the points (k,\ k × (-4)) line up, and continuing the line past k = 0 forces (-1)×(-4) = +4 and (-2)×(-4) = +8. — The pattern argument in a picture: the points $(k,\ k \times (-4))$ line up, and continuing the line past $k = 0$ forces $(-1)\times(-4) = +4$ and $(-2)\times(-4) = +8$ .

Example 54.2

\begin{align*} (-7) \times (+6) &= -42 && \text{(opposite signs)} \\ (-5) \times (-8) &= +40 && \text{(same signs)} \\ (+2.5) \times (-4) &= -10 \\ (-36) \div (-9) &= +4 \\ (+15) \div (-2) &= -7.5 . \end{align*}

54.2 Products of several factors

Proposition 54.3 (Sign of a long product)

The sign of a product of several nonzero factors depends only on the number of negative factors:

an even number of negative factors gives a positive product;
an odd number gives a negative product.

Proof. Negative factors pair up, and each pair contributes a positive sign (Theorem 54.1); an odd count leaves one unpaired negative factor, which makes the whole product negative. ∎

Example 54.4

$(-2) \times (+3) \times (-5) \times (-1)$ : three negative factors (odd), so the product is negative; distances: $2 \times 3 \times 5 \times 1 = 30$ . Result: $-30$ . Decide the sign first, then multiply the distances — two easy tasks instead of one error-prone one.

Example 54.5 (Powers of negatives)

$(-2)^2 = (-2) \times (-2) = +4$ , and $(-2)^3 = (-2)^2 \times (-2) = -8$ : even exponents give positive values, odd exponents keep the sign. Careful with notation: $(-3)^2 = 9$ but $-3^2 = -(3 \times 3) = -9$ .

54.3 Computing with all four operations

Method 54.6 (Safe computation with relative numbers)

Respect the priorities of Chapter 45: brackets, then $\times$ and $\div$ , then $+$ and $-$ ;
at each multiplication or division, find the sign first, then the distance;
rewrite one step per line.

Example 54.7

\begin{align*} 5 - 3 \times (-4) &= 5 - (-12) && \text{(product first: opposite signs)}\\ &= 5 + 12 = 17 . \end{align*}

\begin{align*} \frac{(-6) \times (-10)}{-4} &= \frac{+60}{-4} && \text{(numerator: same signs)}\\ &= -15 && \text{(quotient: opposite signs).} \end{align*}

Example 54.8 (Substituting negative values)

Evaluate $E = 3x^2 - 2x$ for $x = -5$ , brackets around the value:

E = 3 \times (-5)^2 - 2 \times (-5) = 3 \times 25 - (-10) = 75 + 10 = 85 .

The brackets in $(-5)^2$ are essential: without them the square would apply to $5$ only.

54.4 Exercises

Exercise 54.1 ★

Compute:

(-8) \times (+7), \qquad (-9) \times (-6), \qquad (+12) \times (-0.5), \qquad (-1) \times (-1) \times (-1).

Solution

Solution of Exercise 54.1.

$(-8) \times (+7) = -56$ ; $(-9) \times (-6) = +54$ ; $(+12) \times (-0.5) = -6$ ; $(-1) \times (-1) \times (-1) = -1$ (three negative factors: odd).

Exercise 54.2 ★

Compute:

(-63) \div (+9), \qquad (-8) \div (-16), \qquad \frac{45}{-5}, \qquad \frac{-7.2}{-8} .

Solution

Solution of Exercise 54.2.

$(-63) \div (+9) = -7$ ; $(-8) \div (-16) = +0.5$ ; $\dfrac{45}{-5} = -9$ ; $\dfrac{-7.2}{-8} = +0.9$ .

Exercise 54.3 ★

Give only the sign of each product, without computing it:

(-3) \times (-7) \times (+2) \times (-5); \qquad (-1)^{10}; \qquad (-2)^{7}; \qquad (-4) \times 0 \times (-6).

Solution

Solution of Exercise 54.3.

Three negative factors: negative. $(-1)^{10}$ : ten negative factors, even: positive (it equals $1$ ). $(-2)^7$ : seven factors, odd: negative. $(-4) \times 0 \times (-6) = 0$ : neither positive nor negative.

Exercise 54.4 ★

Compute:

(-2)^4, \qquad -2^4, \qquad (-3)^3, \qquad (-1)^{2026} .

Solution

Solution of Exercise 54.4.

$(-2)^4 = +16$ ; $-2^4 = -16$ (the minus is not squared); $(-3)^3 = -27$ ; $(-1)^{2026} = +1$ (even exponent).

Exercise 54.5 ★

Compute step by step, respecting priorities:

7 + 2 \times (-6), \qquad (-4) \times (5 - 9), \qquad 18 \div (-3) - (-2) .

Solution

Solution of Exercise 54.5.

$7 + 2 \times (-6) = 7 + (-12) = -5$ .

$(-4) \times (5 - 9) = (-4) \times (-4) = +16$ .

$18 \div (-3) - (-2) = -6 + 2 = -4$ .

Exercise 54.6 ★

Compute:

\frac{(-5) \times (+8)}{-4}, \qquad \frac{(-3) \times (-6)}{(-2) \times (+9)} .

Solution

Solution of Exercise 54.6.

$\dfrac{(-5) \times (+8)}{-4} = \dfrac{-40}{-4} = +10$ .

$\dfrac{(-3) \times (-6)}{(-2) \times (+9)} = \dfrac{+18}{-18} = -1$ .

Exercise 54.7 ★

Evaluate for $x = -3$ :

5x, \qquad x^2, \qquad -x^2, \qquad 2x^2 + 4x, \qquad (2x)^2 .

Solution

Solution of Exercise 54.7.

For $x = -3$ : $5x = -15$ ; $x^2 = (-3)^2 = 9$ ; $-x^2 = -9$ ; $2x^2 + 4x = 18 - 12 = 6$ ; $(2x)^2 = (-6)^2 = 36$ .

Exercise 54.8 ★★

Complete each equality:

(-6) \times \;?\; = 42, \qquad \;?\; \div (-4) = -2.5, \qquad (-5) \times \;?\; = -1 .

Solution

Solution of Exercise 54.8.

$(-6) \times (-7) = 42$ ; $10 \div (-4) = -2.5$ ; $(-5) \times \frac15 = -1$ , so the missing number is $+\frac15 = 0.2$ .

Exercise 54.9 ★★

True or false? Justify or give a counterexample.

The product of two relative numbers is always at least as large as their sum.
The square of a relative number is never negative.
If a product of three factors is positive, all three factors are positive.

Solution

Solution of Exercise 54.9.

1. False: $0.5 \times 0.5 = 0.25$ is smaller than $0.5 + 0.5 = 1$ (or: $2 \times 3 = 6 > 5$ , but the claim must hold always).

2. True: a square is a product of two numbers of the same sign, hence positive or zero.

3. False: $(-2) \times (-3) \times 4 = +24$ is positive with two negative factors.

Exercise 54.10 ★★

Each wrong answer at a quiz counts $-3$ points, each right answer $+5$ . Zoe answered $20$ questions and scored $36$ points. How many answers were right? (Try values, or set up the computation $5r - 3(20 - r) = 36$ .)

Solution

Solution of Exercise 54.10.

Let $r$ be the number of right answers; then $20 - r$ are wrong, and

5r - 3(20 - r) = 36 \ \Longrightarrow\ 5r - 60 + 3r = 36 \ \Longrightarrow\ 8r = 96 \ \Longrightarrow\ r = 12 .

Twelve right answers (and eight wrong: check, $60 - 24 = 36$ ).

Exercise 54.11 ★★★

Using distributivity (as in the proof of Theorem 54.1), expand and justify each step of

0 = (-1) \times 0 = (-1) \times \bigl(1 + (-1)\bigr) = (-1) \times 1 + (-1) \times (-1),

and conclude that $(-1) \times (-1)$ must equal $+1$ .

Solution

Solution of Exercise 54.11.

Zero times anything is zero, so $(-1) \times 0 = 0$ . Writing $0 = 1 + (-1)$ and distributing:

0 = (-1) \times \bigl(1 + (-1)\bigr) = (-1) \times 1 + (-1) \times (-1) = -1 + (-1) \times (-1).

The number $(-1) \times (-1)$ added to $-1$ gives $0$ : it must be the opposite of $-1$ , that is $+1$ . Distributivity leaves no other choice.

54.5 Problem: Why minus times minus is plus

Problem 54.1

Weekend problem — the sign rules deduced from distributivity, and the alternating sum $1 - 2 + 3 - 4 + \dots$

The proof of Theorem 54.1 continued a pattern and claimed that “any other choice would break distributivity”. This problem makes the claim exact: starting from distributivity alone (Theorem 45.6), you will prove the sign rules — no picture, no pattern, the way algebra does it — then put them to work on powers of $-1$ and on a famous sum with alternating signs. Throughout, $a$ and $b$ are relative numbers; we take for granted the addition rules of Chapter 46, that $1 \times a = a$ , and that a product may be computed in any order (so distributivity applies on either side of a product).

Part I — The sign rules are theorems.

Prove that $0 \times a = 0$ for every relative number $a$ . (Write $0 = 0 + 0$ , expand $(0 + 0) \times a$ , and ask which number, added to $0 \times a$ , gives $0 \times a$ back.)
Expand $\bigl(1 + (-1)\bigr) \times a$ to show that $(-1) \times a + a = 0$ , and conclude that

$(-1) \times a = -a :$

multiplying by $-1$ gives the opposite. Compare with Exercise 54.11, which is the case $a = -1$ .
Deduce that $(-a) \times b = -(a \times b)$ : one minus sign comes out of a product unchanged.
Deduce that $(-a) \times (-b) = a \times b$ : two minus signs cancel.
Every relative number is its distance to zero with a sign in front: $a = +d$ or $a = -d$ . Explain how questions 3 and 4 prove all four cases of the sign table of Theorem 54.1, distances included.

Part II — Powers of $-1$ .

Compute $(-1)^2$ , $(-1)^3$ , $(-1)^4$ , $(-1)^5$ , then give, with justification from Proposition 54.3, the value of $(-1)^n$ for every whole number $n \geq 1$ .
Give (without computing any distance) the sign of

$(-1) \times (-2) \times (-3) \times \dots \times (-10),$

and write the distance of this product as a product of whole numbers, without working it out.
Show that $(-a)^2 = a^2$ and that $(-a)^3 = -a^3$ . For which exponents does the minus sign survive?
Use the sign rules to show that the square of a relative number is never negative, and deduce that no relative number $x$ satisfies $x^2 = -4$ .
Zoe claims: “ $(-1)^{2026} + (-1)^{2027} = 0$ , and more generally two consecutive powers of $-1$ always cancel.” Is she right? Justify.

Part III — The alternating sum. With the sign rules secured, we can compute sums that mix both signs. For a whole number $n \geq 1$ , let

A_n = 1 - 2 + 3 - 4 + \dots

be the sum of the whole numbers from $1$ to $n$ with alternating signs: the last term is $+n$ when $n$ is odd, $-n$ when $n$ is even. So $A_1 = 1$ and $A_2 = 1 - 2 = -1$ .

Compute $A_3$ , $A_4$ , $A_5$ , $A_6$ and $A_7$ . What do you conjecture?
Suppose $n$ is even. Group the terms two by two, $(1 - 2) + (3 - 4) + \dots$ , count the pairs, and prove that $A_n = -\frac{n}{2}$ .
Suppose $n$ is odd. Explain why $A_n = A_{n-1} + n$ , and deduce from the previous question that $A_n = \frac{n + 1}{2}$ .
Compute $A_{100}$ , $A_{101}$ and $A_{2026}$ .
Alma computed $A_{101} - A_{100} = 51 - (-50) = 101$ and was surprised to find exactly $101$ . Should she have been? Explain in one sentence, and state the value of $A_{2027} - A_{2026}$ without any further computation.

Solution

Solution of Problem 54.1.

1. Since $0 = 0 + 0$ , distributivity (Theorem 45.6) gives

0 \times a = (0 + 0) \times a = 0 \times a + 0 \times a .

So adding $0 \times a$ to itself changes nothing — and the only number one can add without changing anything is $0$ . (Directly: subtract $0 \times a$ from both sides.) Hence $0 \times a = 0$ .

2. Expanding, then using $1 + (-1) = 0$ and question 1:

a + (-1) \times a = 1 \times a + (-1) \times a = \bigl(1 + (-1)\bigr) \times a = 0 \times a = 0 .

So $(-1) \times a$ is the number which, added to $a$ , gives $0$ : that is exactly the opposite of $a$ , and $(-1) \times a = -a$ . Exercise 54.11 is the special case $a = -1$ : there, $(-1) \times (-1) = -(-1) = +1$ .

3. Write $-a = (-1) \times a$ (question 2) and regroup the factors:

(-a) \times b = \bigl((-1) \times a\bigr) \times b = (-1) \times (a \times b) = -(a \times b) .

4. Apply question 3 twice (once on each factor):

(-a) \times (-b) = -\bigl(a \times (-b)\bigr) = -\bigl(-(a \times b)\bigr) = a \times b ,

since the opposite of the opposite of a number is the number itself.

5. Write $a = +d$ or $a = -d$ and $b = +e$ or $b = -e$ , where $d$ and $e$ are the distances to zero. The four cases:

(+d) \times (+e) = de, \quad (+d) \times (-e) = -de, \quad (-d) \times (+e) = -de, \quad (-d) \times (-e) = +de,

the middle two by question 3, the last by question 4. In every case the distance of the result is $d \times e$ , the product of the distances, and the sign is the one announced by the table of Theorem 54.1: same signs positive, opposite signs negative.

6. $(-1)^2 = +1$ , $(-1)^3 = -1$ , $(-1)^4 = +1$ , $(-1)^5 = -1$ . In general $(-1)^n$ is a product of $n$ negative factors, each of distance $1$ : by Proposition 54.3 it is $+1$ when $n$ is even and $-1$ when $n$ is odd.

7. There are ten negative factors — an even number — so the product is positive (Proposition 54.3). Its distance is $1 \times 2 \times 3 \times \dots \times 10$ .

8. By question 4, $(-a)^2 = (-a) \times (-a) = a \times a = a^2$ . Then by question 3,

(-a)^3 = (-a)^2 \times (-a) = a^2 \times (-a) = -(a^2 \times a) = -a^3 .

The minus sign disappears for even exponents and survives for odd ones, in line with Example 54.5.

9. If $a = 0$ then $a^2 = 0$ . Otherwise $a$ and $a$ have the same sign, so $a^2 = a \times a$ is positive (Theorem 54.1). A square is therefore never negative; since $-4$ is negative, no relative number $x$ satisfies $x^2 = -4$ .

10. Zoe is right. The exponents $2026$ and $2027$ have opposite parities, so $(-1)^{2026} = +1$ and $(-1)^{2027} = -1$ , which add to $0$ . In general, of two consecutive exponents one is even and one is odd, so the two powers are $+1$ and $-1$ in some order: opposite numbers, whose sum is always $0$ .

11. $A_3 = 1 - 2 + 3 = 2$ , then $A_4 = A_3 - 4 = -2$ , $A_5 = A_4 + 5 = 3$ , $A_6 = A_5 - 6 = -3$ , $A_7 = A_6 + 7 = 4$ . Conjecture: $A_n = -\frac{n}{2}$ for even $n$ , and $A_n = \frac{n+1}{2}$ for odd $n$ .

12. For even $n$ the terms pair up completely:

A_n = (1 - 2) + (3 - 4) + \dots + \bigl((n-1) - n\bigr),

and each pair equals $-1$ . There are $\frac{n}{2}$ pairs (two terms per pair, $n$ terms in all), so $A_n = \frac{n}{2} \times (-1) = -\frac{n}{2}$ .

13. For odd $n$ , the sum $A_n$ is the sum $A_{n-1}$ of the first $n - 1$ terms, plus the last term, which is $+n$ since $n$ is odd. As $n - 1$ is even, question 12 gives

A_n = A_{n-1} + n = -\frac{n-1}{2} + n = \frac{-(n-1) + 2n}{2} = \frac{n+1}{2} .

14. $A_{100} = -\frac{100}{2} = -50$ , $A_{101} = \frac{102}{2} = 51$ , $A_{2026} = -\frac{2026}{2} = -1013$ .

15. No surprise: passing from $A_{100}$ to $A_{101}$ adds exactly one term, namely $+101$ , so the difference had to be $101$ . For the same reason, $A_{2027} - A_{2026} = +2027$ — without computing $A_{2027}$ at all.