---
title: "Negative Numbers"
book: "Primary & Middle School Mathematics"
subject: math
language: en
chapter: 46
exercises: 12
source: https://one-course.com/books/math/1/en/chapter/46-negative-numbers
---

# Chapter 46 — Negative Numbers

Temperatures below [zero](https://one-course.com/books/math/1/en/chapter/1-counting-to-20#def-g1-counting-zero), floors below ground, money owed: many quantities go below $0$. Negative numbers extend the number line to the left of [zero](https://one-course.com/books/math/1/en/chapter/1-counting-to-20#def-g1-counting-zero). This chapter teaches how to compare them and how to add and subtract them — their multiplication waits for [Chapter 54](https://one-course.com/books/math/1/en/chapter/54-multiplying-negative-numbers#ch-g8-negprod).

## 46.1 Relative numbers

**Definition 46.1 (Relative numbers).**

A *relative number* is made of a sign ($+$ or $-$) and a *distance to [zero](https://one-course.com/books/math/1/en/chapter/1-counting-to-20#def-g1-counting-zero)*: $+3$ and $-3$ are both at distance $3$ from $0$, on opposite sides. Positive numbers are usually written without their sign: $+3 = 3$. The numbers $+3$ and $-3$ are *opposites* of each other.

![The number line extended to the left: -3 and +3 are opposite numbers, at the same distance from 0.](https://one-course.com/images/onecourse/chapters/math-1/g7-negatives/fig-d11f9c05fe48.svg)

*The number line extended to the left: $-3$ and $+3$ are opposite numbers, at the same distance from $0$.*

**Method 46.2 (Comparing relative numbers).**

On the number line, smaller means further left. In practice:

1. a negative number is always smaller than a positive one: $-7 < 2$ ;
2. between two positives, the usual order;
3. between two negatives, the one with the *larger* distance to [zero](https://one-course.com/books/math/1/en/chapter/1-counting-to-20#def-g1-counting-zero) is the *smaller* : $-7 < -2$ .

**Example 46.3.**

Order from smallest to largest: $3$; $-5$; $0$; $-1.5$; $2.8$. Negatives first, the most negative leading:

$$
-5 < -1.5 < 0 < 2.8 < 3 .
$$

## 46.2 Adding relative numbers

Think of a [relative number](#def-g7-negatives-def) as a movement along the line: $+5$ means “walk $5$ steps right”, $-3$ means “walk $3$ steps left”. Adding numbers means chaining the walks.

**Method 46.4 (Adding two relative numbers).**

1. *Same sign* : add the distances, keep the common sign: $(-3) + (-4) = -7$ .
2. *Opposite signs* : subtract the smaller distance from the larger, take the sign of the larger: $(-7) + (+2) = -5$ and $(+7) + (-2) = +5$ .
3. [Opposites](#def-g7-negatives-def) cancel: $(+4) + (-4) = 0$ .

![Adding as walking: starting at 0, the walk -3.5 then the walk +2 lands at (-3.5) + 2 = -1.5.](https://one-course.com/images/onecourse/chapters/math-1/g7-negatives/fig-c6e3d0f5b921.svg)

*Adding as walking: starting at $0$, the walk $-3.5$ then the walk $+2$ lands at $(-3.5) + 2 = -1.5$.*

**Example 46.5.**

Step by step, grouping as convenient:

$$
\begin{align*}
(-8) + (+3) &= -5 && \text{(opposite signs: } 8 - 3 = 5,
\text{ sign of } 8)\\
(-2.5) + (-1.5) &= -4 && \text{(same sign: add distances)}\\
(+6) + (-9) + (+3) &= (-3) + (+3) = 0 .
\end{align*}
$$

## 46.3 Subtracting relative numbers

**Theorem 46.6 (Subtraction rule).**

Subtracting a [relative number](#def-g7-negatives-def) is the same as *adding its opposite*:

$$
a - b = a + (-b) .
$$

**Why it works.** $a - b$ is the number that added to $b$ gives $a$. Check that $a + (-b)$ does the job: $\bigl(a + (-b)\bigr) + b = a + \bigl((-b) + b\bigr) = a + 0 = a$. ∎

**Example 46.7.**

Every subtraction becomes an addition, then [Method 46.4](#met-g7-negatives-add) applies:

$$
\begin{align*}
5 - 9 &= 5 + (-9) = -4, \\
(-3) - (+4) &= (-3) + (-4) = -7, \\
(-2) - (-6) &= (-2) + (+6) = +4 .
\end{align*}
$$

The last line is the famous one: *subtracting a negative means adding*. Removing a debt makes you richer!

**Example 46.8 (Temperature differences).**

In one day the temperature went from $-6$ degrees to $+9$ degrees. The rise is the [difference](https://one-course.com/books/math/1/en/chapter/3-subtraction-first-steps#ex-g1-subtraction-difference)

$$
9 - (-6) = 9 + 6 = 15 \text{ degrees}.
$$

On the number line: from $-6$ up to $0$ is $6$ degrees, from $0$ to $9$ is $9$ more.

## 46.4 Coordinates in the whole plane

**Definition 46.9 (Coordinates with signs).**

With two [perpendicular](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) graduated axes crossing at the origin $O$, every point of the plane gets two relative coordinates $(x, y)$: negative $x$ means left of the vertical axis, negative $y$ means below the horizontal axis.

![One point in each quarter of the plane. Always read the horizontal coordinate first.](https://one-course.com/images/onecourse/chapters/math-1/g7-negatives/fig-e9a527c0178e.svg)

*One point in each [quarter](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) of the plane. Always read the horizontal coordinate first.*

## 46.5 Exercises

**Exercise 46.1 ★.**

Give the opposite of: $7$; $-3.5$; $0$; $+12$. Then complete: “two numbers are [opposites](#def-g7-negatives-def) when their [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) is ?”.

**Solution of Exercise 46.1.**

[Opposites](#def-g7-negatives-def): $-7$; $+3.5$; $0$ (its own opposite); $-12$. Two numbers are [opposites](#def-g7-negatives-def) when their [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) is $0$.

**Exercise 46.2 ★.**

Copy and complete with $<$ or $>$:

$$
-4 \;?\; 1, \qquad
-4 \;?\; -7, \qquad
-2.5 \;?\; -2.4, \qquad
0 \;?\; -0.1 .
$$

**Solution of Exercise 46.2.**

$-4 < 1$; $-4 > -7$; $-2.5 < -2.4$; $0 > -0.1$.

**Exercise 46.3 ★.**

Order from smallest to largest: $-3$; $2.5$; $-0.5$; $-3.1$; $0$; $1$.

**Solution of Exercise 46.3.**

$-3.1 < -3 < -0.5 < 0 < 1 < 2.5$.

**Exercise 46.4 ★.**

Compute:

$$
(-5) + (-9), \qquad
(-12) + (+7), \qquad
(+3.5) + (-1.5), \qquad
(-6) + (+6).
$$

**Solution of Exercise 46.4.**

$(-5) + (-9) = -14$; $(-12) + (+7) = -5$; $(+3.5) + (-1.5) = +2$; $(-6) + (+6) = 0$.

**Exercise 46.5 ★.**

Transform into additions, then compute:

$$
4 - 11, \qquad
(-5) - (+3), \qquad
(-7) - (-10), \qquad
0 - (-8).
$$

**Solution of Exercise 46.5.**

$4 - 11 = 4 + (-11) = -7$.

$(-5) - (+3) = (-5) + (-3) = -8$.

$(-7) - (-10) = (-7) + (+10) = +3$.

$0 - (-8) = 0 + (+8) = +8$.

**Exercise 46.6 ★.**

Compute by grouping cleverly ([opposites](#def-g7-negatives-def) first):

$$
(+7) + (-12) + (+3) + (-8), \qquad
(-5.5) + (+9) + (-4.5) + (+1).
$$

**Solution of Exercise 46.6.**

$(+7) + (-12) + (+3) + (-8) = (+10) + (-20) = -10$ (positives together, negatives together).

$(-5.5) + (+9) + (-4.5) + (+1) = (-10) + (+10) = 0$.

**Exercise 46.7 ★.**

The temperature was $-3$ degrees this morning; it rose by $8$ degrees during the day, then dropped by $12$ degrees at night. Write a single computation giving the final temperature, and compute it.

**Solution of Exercise 46.7.**

$(-3) + 8 - 12 = 5 - 12 = -7$: it is $-7$ degrees at night.

**Exercise 46.8 ★.**

Plot in a coordinate system: $A(2, 3)$; $B(-3, 1)$; $C(-2, -2)$; $D(3, -1)$; $E(0, -3)$. Which point lies on an axis?

**Solution of Exercise 46.8.**

Free plot; $E(0, -3)$ lies on the vertical axis (its first coordinate is $0$).

**Exercise 46.9 ★★.**

Compute the [difference](https://one-course.com/books/math/1/en/chapter/3-subtraction-first-steps#ex-g1-subtraction-difference) between the highest and lowest temperature: Moscow in January, max $-4$ degrees, min $-13$ degrees; Helsinki, max $2$, min $-9$.

**Solution of Exercise 46.9.**

Moscow: $(-4) - (-13) = -4 + 13 = 9$ degrees of [difference](https://one-course.com/books/math/1/en/chapter/3-subtraction-first-steps#ex-g1-subtraction-difference). Helsinki: $2 - (-9) = 11$ degrees.

**Exercise 46.10 ★★.**

A diver is at altitude $-12$ m (twelve meters below the surface). She rises $7$ m, then descends $4$ m. At what altitude is she? How far must she still rise to reach the surface?

**Solution of Exercise 46.10.**

$(-12) + 7 - 4 = -5 - 4 = -9$: she is at $-9$ m. To reach the surface (altitude $0$) she must rise $0 - (-9) = 9$ m.

**Exercise 46.11 ★★.**

Plot $A(-2, 1)$, $B(2, 1)$, $C(2, -2)$. Find the coordinates of the point $D$ so that $ABCD$ is a rectangle, and compute its [perimeter](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-perimeter) (count the grid squares for the side lengths).

**Solution of Exercise 46.11.**

$D(-2, -2)$: same first coordinate as $A$, same second coordinate as $C$. Sides: $AB = 4$ (from $-2$ to $2$) and $BC = 3$ (from $1$ down to $-2$), so the [perimeter](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-perimeter) is $2 \times (4 + 3) = 14$.

**Exercise 46.12 ★★★.**

In the magic square below, every row, column and diagonal must have the same [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def). Copy and complete it:

| $?$ | $-7$ | $0$ |
| --- | --- | --- |
| $-3$ | $-2$ | $?$ |
| $-4$ | $?$ | $?$ |

(Start by finding the magic [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) using the diagonal that is almost complete.)

**Solution of Exercise 46.12.**

The complete diagonal $0, -2, -4$ gives the magic [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) $0 + (-2) + (-4) = -6$. Then, cell by cell: top-left $= -6 - (-7) - 0 = 1$; middle-right $= -6 - (-3) - (-2) =
-1$; bottom-right $= -6 - 0 - (-1) = -5$; bottom-middle $= -6 - (-7) - (-2) = 3$.

| $1$ | $-7$ | $0$ |
| --- | --- | --- |
| $-3$ | $-2$ | $-1$ |
| $-4$ | $3$ | $-5$ |

Every row, column and diagonal sums to $-6$.

## 46.6 Problem: The year zero that never was

**Problem 46.1.**

Weekend problem — relative numbers on the timeline of history: durations across the era boundary, the astronomers’ trick, and the net-change principle

The number line of this chapter has a famous real-world twin: the timeline of history, with the years BC stretching left and the years AD stretching right. But the historians’ line hides a trap that has spoiled many a computation: *there is no year [zero](https://one-course.com/books/math/1/en/chapter/1-counting-to-20#def-g1-counting-zero)* — the year 1 BC is followed immediately by AD 1. This problem computes with temperatures, lifts and bank accounts, then repairs history’s broken [zero](https://one-course.com/books/math/1/en/chapter/1-counting-to-20#def-g1-counting-zero) the way astronomers do: with [relative numbers](#def-g7-negatives-def).

**Part I — Warm-ups, and a first trap.**

1. The summit of Mont Blanc is at altitude $+4\,808$ m; the shore of the Dead Sea at $-430$ m. Compute the [difference](https://one-course.com/books/math/1/en/chapter/3-subtraction-first-steps#ex-g1-subtraction-difference) in altitude between them.
2. A bank account holds $-45$ euros (an overdraft). Its owner deposits $120$ euros, pays a bill of $60$ euros, then deposits $15$ euros. Write the balance as a single chain of additions and compute it. What single deposit would have brought the original $-45$ back to exactly $0$ ?
3. Julius Caesar was born in $100$ BC and died in $44$ BC. Counting BC years as negative numbers ( $100$ BC as $-100$ , $44$ BC as $-44$ ), compute $(-44) - (-100)$ : how many years did Caesar live?
4. Emperor Augustus was born in $63$ BC and died in AD $14$ . The same recipe gives $14 - (-63) = 77$ years — but historians insist he died at $76$ . Their objection: the count $77$ walks through a year that never existed. Which one?
5. Astronomers repair the timeline by relabelling: AD years keep their number ( $+14$ stays $+14$ ), but $1$ BC becomes $0$ , $2$ BC becomes $-1$ , and generally the year $n$ BC becomes $-(n - 1)$ . Convert $63$ BC, and redo the computation of question 4 with the astronomers’ numbers. Does the answer now satisfy the historians?

**Part II — Computing across the eras.** From now on, use the astronomers’ relabelling for every computation, converting back to BC/AD at the end (a year labelled $0$ or negative $-m$ is the historians’ year $m + 1$ BC).

6. Convert to astronomers’ numbers: $1$ BC; $10$ BC; AD $476$ ; and $753$ BC, the legendary founding of Rome.
7. How many years passed from the founding of Rome ( $753$ BC) to the fall of the Western Empire (AD $476$ )?
8. A comet returns every $76$ years and appeared in $12$ BC. Give the historians’ years of its next three appearances, and of the appearance just before $12$ BC.
9. How many years passed from $5$ BC to AD $5$ ? (The tempting answer is $10$ ; the timeline says otherwise.)
10. A child born in $3$ BC turned ten years old in which year (BC or AD)?

**Part III — The net-change principle.**

11. The *distance* between two numbers on the line is the larger minus the smaller — always positive. Compute the distance between $-7$ and $-2$ , then between $-3.5$ and $4.5$ .
12. A lift starts at floor $0$ and makes the moves $+3$ , $-5$ , $+2$ , $-1$ , $+4$ . Compute the final floor by grouping cleverly ( [Method 46.4](#met-g7-negatives-add) , [opposites](#def-g7-negatives-def) first where possible). Would performing the moves in a different order change the destination? Why?
13. Day by day, the temperature changes by $+2$ , $-4$ , $+1$ , $0$ , $-3$ , $+5$ , $-2$ degrees over a week. Compute the *[net](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-net) change* . If the week started at $-2$ degrees, where did it end? State the principle: final value $=$ starting value $+$ ( [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) of all changes).
14. A hiker’s walk brings her back exactly to her starting altitude. What must the [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) of all her altitude changes be, and why? Of her six recorded changes, five are $+4$ , $-7$ , $+2$ , $+3$ , $-5$ (in hundreds of meters): find the sixth.
15. The finale, back in Rome. A historian computes: “the Republic ran from $509$ BC to $27$ BC, then the Empire to AD $476$ ; in all, from $509$ BC to AD $476$ , that is $509 + 476 = 985$ years of Roman state.” Redo the computation with astronomers’ numbers. What is the correct total — and which year, once again, caused the error?

**Solution of Problem 46.1.**

**1.** $4\,808 - (-430) = 4\,808 + 430 = 5\,238$ m ([Theorem 46.6](#thm-g7-negatives-sub): subtracting a negative means adding).

**2.** Balance:

$$
(-45) + 120 + (-60) + 15 = 75 + (-60) + 15 = 15 + 15 = 30
\text{ euros}.
$$

To return the initial $-45$ to $0$, deposit its opposite: $45$ euros.

**3.** $(-44) - (-100) = (-44) + 100 = 56$: Caesar lived $56$ years. (No trap here: both years are on the same side of the era boundary.)

**4.** The year $0$: counting from $-63$ up to $+14$ passes through $0$, but the historians’ calendar jumps straight from $1$ BC to AD $1$ — the year $0$ never existed, so the naive count is one year too long.

**5.** $63$ BC becomes $-(63 - 1) = -62$. Then $14 - (-62) = 14 + 62 = 76$ years: exactly the historians’ figure. On the astronomers’ relabelled line there are no gaps, so ordinary subtraction gives true durations.

**6.** $1$ BC $\to 0$; $10$ BC $\to -9$; AD $476 \to +476$; $753$ BC $\to -752$.

**7.** $476 - (-752) = 476 + 752 = 1\,228$ years.

**8.** $12$ BC is $-11$. Next appearances: $-11 + 76 = 65$, then $141$, then $217$: the years AD $65$, AD $141$ and AD $217$. Before: $-11 - 76 = -87$, which is the historians’ year $87 + 1 = 88$ BC.

**9.** $5$ BC is $-4$, so $5 - (-4) = 9$ years — not $10$. The missing year [zero](https://one-course.com/books/math/1/en/chapter/1-counting-to-20#def-g1-counting-zero) strikes whenever a duration crosses the era boundary.

**10.** $3$ BC is $-2$; ten years later is $-2 + 10 = +8$: the child turned ten in AD $8$.

**11.** Between $-7$ and $-2$: $(-2) - (-7) = 5$. Between $-3.5$ and $4.5$: $4.5 - (-3.5) = 8$.

**12.** Grouping [opposites](#def-g7-negatives-def) and friends:

$$
3 + (-5) + 2 + (-1) + 4
= \bigl(3 + 2\bigr) + (-5) + (-1) + 4
= 0 + (-1) + 4 = +3 :
$$

the lift ends at floor $3$. The order does not matter: a [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) of [relative numbers](#def-g7-negatives-def) can be reorganized freely (each move is walked along the same line, and the total walk is the same).

**13.** [Net](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-net) change: $2 + (-4) + 1 + 0 + (-3) + 5 + (-2) = -1$ degree. Starting at $-2$: final temperature $-2 + (-1) = -3$ degrees. In general, the final value is the starting value plus the [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) of all the changes — no need to follow the ups and downs one by one.

**14.** Back to the start means [net](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-net) change $0$: the ups and downs must cancel exactly. The five known changes sum to $4 + (-7) + 2 + 3 + (-5) = -3$, so the sixth change is the opposite: $+3$ (three hundred meters up).

**15.** Astronomers: $509$ BC $\to -508$, and $476 - (-508) = 984$ years. The historian’s $509 + 476 = 985$ counts one year too many — once more the phantom year [zero](https://one-course.com/books/math/1/en/chapter/1-counting-to-20#def-g1-counting-zero), silently included in the addition but absent from history. (As a check, the two eras: Republic $-508$ to $-26$, $482$ years; Empire $-26$ to $+476$, $502$ years; $482 + 502 = 984$.)