Primary & Middle School Mathematics · Grades 1–9

46Negative Numbers

Temperatures below zero, floors below ground, money owed: many quantities go below $0$ . Negative numbers extend the number line to the left of zero. This chapter teaches how to compare them and how to add and subtract them — their multiplication waits for Chapter 54.

46.1 Relative numbers

Definition 46.1 (Relative numbers)

A relative number is made of a sign ( $+$ or $-$ ) and a distance to 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. — 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:

a negative number is always smaller than a positive one: $-7 < 2$ ;
between two positives, the usual order;
between two negatives, the one with the larger distance to 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 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)

Same sign: add the distances, keep the common sign: $(-3) + (-4) = -7$ .
Opposite signs: subtract the smaller distance from the larger, take the sign of the larger: $(-7) + (+2) = -5$ and $(+7) + (-2) = +5$ .
Opposites 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. — 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 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 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

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 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.

46.5 Exercises

Exercise 46.1 ★

Give the opposite of: $7$ ; $-3.5$ ; $0$ ; $+12$ . Then complete: “two numbers are opposites when their sum is ?”.

Solution

Solution of Exercise 46.1.

Opposites: $-7$ ; $+3.5$ ; $0$ (its own opposite); $-12$ . Two numbers are opposites when their sum 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

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

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

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

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 first):

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

Solution

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

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

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 between the highest and lowest temperature: Moscow in January, max $-4$ degrees, min $-13$ degrees; Helsinki, max $2$ , min $-9$ .

Solution

Solution of Exercise 46.9.

Moscow: $(-4) - (-13) = -4 + 13 = 9$ degrees of 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

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 (count the grid squares for the side lengths).

Solution

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 is $2 \times (4 + 3) = 14$ .

Exercise 46.12 ★★★

In the magic square below, every row, column and diagonal must have the same sum. Copy and complete it:

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

(Start by finding the magic sum using the diagonal that is almost complete.)

Solution

Solution of Exercise 46.12.

The complete diagonal $0, -2, -4$ gives the magic sum $0 + (-2) + (-4) = -6$ . 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 — 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 the way astronomers do: with relative numbers.

Part I — Warm-ups, and a first trap.

The summit of Mont Blanc is at altitude $+4\,808$ m; the shore of the Dead Sea at $-430$ m. Compute the difference in altitude between them.
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 $0$ ?
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?
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?
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).

Convert to astronomers’ numbers: $1$ BC; $10$ BC; AD $476$ ; and $753$ BC, the legendary founding of Rome.
How many years passed from the founding of Rome ( $753$ BC) to the fall of the Western Empire (AD $476$ )?
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.
How many years passed from $5$ BC to AD $5$ ? (The tempting answer is $10$ ; the timeline says otherwise.)
A child born in $3$ BC turned ten years old in which year (BC or AD)?

Part III — The net-change principle.

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 $4.5$ .
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, opposites first where possible). Would performing the moves in a different order change the destination? Why?
Day by day, the temperature changes by $+2$ , $-4$ , $+1$ , $0$ , $-3$ , $+5$ , $-2$ degrees over a week. Compute the net change. If the week started at $-2$ degrees, where did it end? State the principle: final value $=$ starting value $+$ (sum of all changes).
A hiker’s walk brings her back exactly to her starting altitude. What must the sum 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.
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

Solution of Problem 46.1.

1. $4\,808 - (-430) = 4\,808 + 430 = 5\,238$ m (Theorem 46.6: 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 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 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 of relative numbers can be reorganized freely (each move is walked along the same line, and the total walk is the same).

13. 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 of all the changes — no need to follow the ups and downs one by one.

14. Back to the start means 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, 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; $482 + 502 = 984$ .)