---
title: "Multiplying Negative Numbers"
book: "Primary & Middle School Mathematics"
subject: math
language: en
chapter: 54
exercises: 11
source: https://one-course.com/books/math/1/en/chapter/54-multiplying-negative-numbers
---

# Chapter 54 — Multiplying Negative Numbers

Grade 7 added and subtracted [relative numbers](https://one-course.com/books/math/1/en/chapter/46-negative-numbers#def-g7-negatives-def) ([Chapter 46](https://one-course.com/books/math/1/en/chapter/46-negative-numbers#ch-g7-negatives)); 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](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) of two [relative numbers](https://one-course.com/books/math/1/en/chapter/46-negative-numbers#def-g7-negatives-def) has distance to [zero](https://one-course.com/books/math/1/en/chapter/1-counting-to-20#def-g1-counting-zero) equal to the [product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) of the distances, and its sign is given by:

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

*Same signs: positive [product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def). Opposite signs: negative [product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def).* The same table governs [quotients](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder).

**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.](https://one-course.com/images/onecourse/chapters/math-1/g8-negprod/fig-364e71576001.svg)

*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](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) of several nonzero factors depends only on the number of negative factors:

- an *[even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd)* number of negative factors gives a positive [product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) ;
- an *[odd](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd)* number gives a negative [product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) .

**Proof.** Negative factors pair up, and each pair contributes a positive sign ([Theorem 54.1](#thm-g8-negprod-rules)); an [odd](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) count leaves one unpaired negative factor, which makes the whole [product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) negative. ∎

**Example 54.4.**

$(-2) \times (+3) \times (-5) \times (-1)$: three negative factors ([odd](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd)), so the [product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) 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](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) exponents give positive values, [odd](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) 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).**

1. Respect the priorities of [Chapter 45](https://one-course.com/books/math/1/en/chapter/45-priorities-of-operations#ch-g7-priorities) : brackets, then $\times$ and $\div$ , then $+$ and $-$ ;
2. at each multiplication or division, find the *sign first* , then the distance;
3. 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 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](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd)).

**Exercise 54.2 ★.**

Compute:

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

**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](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def), without computing it:

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

**Solution of Exercise 54.3.**

Three negative factors: *negative*. $(-1)^{10}$: ten negative factors, [even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd): *positive* (it equals $1$). $(-2)^7$: seven factors, [odd](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd): *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 of Exercise 54.4.**

$(-2)^4 = +16$; $-2^4 = -16$ (the minus is not squared); $(-3)^3 = -27$; $(-1)^{2026} = +1$ ([even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) 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 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 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 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 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.

1. The [product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) of two [relative numbers](https://one-course.com/books/math/1/en/chapter/46-negative-numbers#def-g7-negatives-def) is always at least as large as their [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) .
2. The square of a relative number is never negative.
3. If a [product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) of three factors is positive, all three factors are positive.

**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](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) of two numbers of the same sign, hence positive or [zero](https://one-course.com/books/math/1/en/chapter/1-counting-to-20#def-g1-counting-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 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](#thm-g8-negprod-rules)), 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 of Exercise 54.11.**

[Zero](https://one-course.com/books/math/1/en/chapter/1-counting-to-20#def-g1-counting-zero) times anything is [zero](https://one-course.com/books/math/1/en/chapter/1-counting-to-20#def-g1-counting-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](#thm-g8-negprod-rules) 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](https://one-course.com/books/math/1/en/chapter/45-priorities-of-operations#thm-g7-priorities-distributivity)), 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](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) with alternating signs. Throughout, $a$ and $b$ are [relative numbers](https://one-course.com/books/math/1/en/chapter/46-negative-numbers#def-g7-negatives-def); we take for granted the addition rules of [Chapter 46](https://one-course.com/books/math/1/en/chapter/46-negative-numbers#ch-g7-negatives), that $1 \times a = a$, and that a [product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) may be computed in any order (so distributivity applies on either side of a [product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def)).

**Part I — The sign rules are theorems.**

1. Prove that $0 \times a = 0$ for every [relative number](https://one-course.com/books/math/1/en/chapter/46-negative-numbers#def-g7-negatives-def) $a$ . (Write $0 = 0 + 0$ , expand $(0 + 0) \times a$ , and ask which number, added to $0 \times a$ , gives $0 \times a$ back.)
2. 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](#exo-g8-negprod-11), which is the case $a = -1$.
3. Deduce that $(-a) \times b = -(a \times b)$ : one minus sign comes out of a [product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) unchanged.
4. Deduce that $(-a) \times (-b) = a \times b$ : two minus signs cancel.
5. Every [relative number](https://one-course.com/books/math/1/en/chapter/46-negative-numbers#def-g7-negatives-def) is its distance to [zero](https://one-course.com/books/math/1/en/chapter/1-counting-to-20#def-g1-counting-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](#thm-g8-negprod-rules) , distances included.

**Part II — Powers of $-1$.**

6. Compute $(-1)^2$ , $(-1)^3$ , $(-1)^4$ , $(-1)^5$ , then give, with justification from [Proposition 54.3](#prop-g8-negprod-many) , the value of $(-1)^n$ for every whole number $n \geq 1$ .
7. Give (without computing any distance) the sign of $$(-1) \times (-2) \times (-3) \times \dots \times (-10),$$ and write the distance of this [product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) as a [product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) of whole numbers, without working it out.
8. Show that $(-a)^2 = a^2$ and that $(-a)^3 = -a^3$ . For which exponents does the minus sign survive?
9. Use the sign rules to show that the square of a relative number is never negative, and deduce that no [relative number](https://one-course.com/books/math/1/en/chapter/46-negative-numbers#def-g7-negatives-def) $x$ satisfies $x^2 = -4$ .
10. 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](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def).** With the sign rules secured, we can compute [sums](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) that mix both signs. For a whole number $n \geq 1$, let

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

be the [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) of the whole numbers from $1$ to $n$ with alternating signs: the last term is $+n$ when $n$ is [odd](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd), $-n$ when $n$ is [even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd). So $A_1 = 1$ and $A_2 = 1 - 2 = -1$.

11. Compute $A_3$ , $A_4$ , $A_5$ , $A_6$ and $A_7$ . What do you conjecture?
12. Suppose $n$ is [even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) . Group the terms two by two, $(1 - 2) + (3 - 4) + \dots$ , count the pairs, and prove that $A_n = -\frac{n}{2}$ .
13. Suppose $n$ is [odd](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) . Explain why $A_n = A_{n-1} + n$ , and deduce from the previous question that $A_n = \frac{n + 1}{2}$ .
14. Compute $A_{100}$ , $A_{101}$ and $A_{2026}$ .
15. 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 of Problem 54.1.**

**1.** Since $0 = 0 + 0$, distributivity ([Theorem 45.6](https://one-course.com/books/math/1/en/chapter/45-priorities-of-operations#thm-g7-priorities-distributivity)) 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](#exo-g8-negprod-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](https://one-course.com/books/math/1/en/chapter/1-counting-to-20#def-g1-counting-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](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) of the distances, and the sign is the one announced by the table of [Theorem 54.1](#thm-g8-negprod-rules): 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](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) of $n$ negative factors, each of distance $1$: by [Proposition 54.3](#prop-g8-negprod-many) it is $+1$ when $n$ is [even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) and $-1$ when $n$ is [odd](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd).

**7.** There are ten negative factors — an [even number](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) — so the [product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) is *positive* ([Proposition 54.3](#prop-g8-negprod-many)). 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](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) exponents and survives for [odd](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) ones, in line with [Example 54.5](#ex-g8-negprod-powers).

**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](#thm-g8-negprod-rules)). A square is therefore never negative; since $-4$ is negative, no [relative number](https://one-course.com/books/math/1/en/chapter/46-negative-numbers#def-g7-negatives-def) $x$ 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](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) and one is [odd](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd), so the two powers are $+1$ and $-1$ in some order: opposite numbers, whose [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) 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](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) $n$, and $A_n = \frac{n+1}{2}$ for [odd](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) $n$.

**12.** For [even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) $n$ 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](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) $n$, the [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) $A_n$ is the [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) $A_{n-1}$ of the first $n - 1$ terms, plus the last term, which is $+n$ since $n$ is [odd](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd). As $n - 1$ is [even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd), 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](https://one-course.com/books/math/1/en/chapter/3-subtraction-first-steps#ex-g1-subtraction-difference) had to be $101$. For the same reason, $A_{2027} - A_{2026} = +2027$ — without computing $A_{2027}$ at all.