Chapter 10: The Other Side of Zero (Integers)

Four floors up: press + four times, i.e. a movement of +4. Three floors down: press − three times, i.e. −3.

+4 (+ + + +) to go up four floors, and −3 (− − −) to go down three floors.

With the Welcome Hall as Floor 0: Food Court +1, Art Centre +2, Book Store +3, Ice Cream +4, Sports Centre +5, Space Centre +6; and below: Toy Store −1, Video Games −2, Cinema −3, the next two floors −4 and −5 (the bottom floor, Dinosaurs).

Welcome Hall 0; above: +1 (Food Court) to +6 (Space); below: −1 (Toys), −2 (Video Games), −3 (Cinema), −4, −5.

$(+2)+(-3)=-1$: you reach Floor −1, the Toy Store.

$(+2)+(-3)=-1$, the Toy Store.

a. +5 b. +5 c. +1 d. +1 e. 0 f. +2 g. −2

a. +5 b. +5 c. +1 d. +1 e. 0 f. +2 g. −2

$$(+1)+(-6)=-5,0+(-5)=-5,(-2)+(-3)=-5,(+4)+(-9)=-5,(-5)+0=-5$$

(The movement is always $-5$ minus the starting floor.)

e.g. (+1) + (−6), 0 + (−5), (−2) + (−3), (+4) + (−9)

a. +5 b. +5 c. $4-3-2=$ −1 d. $-1+2-3=$ −2

a. +5 b. +5 c. −1 d. −2

The inverse undoes the movement, so it has the same size and the opposite sign:

• +4 → −4, −4 → +4, −3 → +3, 0 → 0, +2 → −2, −1 → +1.
• Pairs: (+5, −5), (−7, +7), (−8, +8), (+9, −9).

−4, +4, +3, 0, −2, +1; pairs (+5, −5), (−7, +7), (−8, +8), (+9, −9).

Asin: +5; Binnu: −3; Aman: −1. Binnu (Floor −3) is on the lowest floor.

Asin +5, Binnu −3, Aman −1; Binnu is on the lowest floor.

A lower floor is the smaller number.

a. $-2<+5$ b. $-5<+4$ c. $-5<-3$ d. $+6>-6$ e. $0>-4$ f. $0<+4$

a. < b. < c. < d. > e. > f. <

a. $-10>-12$ b. $+17>-10$ c. $0>-20$ d. $+9>-9$ e. $-25<-7$ f. $+15>-17$

a. > b. > c. > d. > e. < f. >

Counting the marks (one mark per floor) from A = −12 upward:

B = −9, C = −6, F = +2, G = +6, H = +11. The floors to be marked: −7 (P), −4 (Q), +3 (R), −10 (S).

B = −9, C = −6, F = +2, G = +6, H = +11; −7, −4, +3 and −10 are marked as P, Q, R and S.

• $15-5$: $5+?=15$, so 10.
• $100-10$: $10+?=100$, so 90.
• $74-34$: $34+?=74$, so 40.

10, 90 and 40

a. −3 b. −2 c. +3 d. +2 e. +7 f. −1 g. −3 h. 0 i. −2 j. +6

(For example, in e: from Floor −3 to Floor +4 you go up 7 floors.)

a. −3 b. −2 c. +3 d. +2 e. +7 f. −1 g. −3 h. 0 i. −2 j. +6

a. +160 b. −240 c. +250 d. −150 e. −160 f. +160 g. −240

a. +160 b. −240 c. +250 d. −150 e. −160 f. +160 g. −240

a. −155 b. +160 c. +160 d. +230 e. +230 f. +101 g. +101 h. +3000

Notice that b and c, d and e, f and g give the same answers: subtracting a negative number is the same as adding the corresponding positive number.

a. −155 b. 160 c. 160 d. 230 e. 230 f. 101 g. 101 h. 3000

• 5 → 9: 4 steps forward (+4), $9-5=4$.
• 9 → 3: 6 steps backward (−6), $3-9=-6$.
• 3 → −2: 5 steps backward (−5), $-2-3=-5$.

+4, −6 and −5

For example, A = 2, B = 5, C = 8 and P = −1, Q = −3, R = −7. In increasing order: $-7<-3<-1$.

For example 2, 5, 8 and −1, −3, −7; −7 < −3 < −1.

Yes, $2>-3$: on the number line, 2 is to the right of −3. Yes, $-2<3$: −2 is to the left of 3. (Every positive number is greater than every negative number.)

Yes to both: numbers to the right on the number line are greater.

(i) −5 (ii) 0 (iii) 10 (iv) −10 (v) 14 (vi) 2

(i) −5 (ii) 0 (iii) 10 (iv) −10 (v) 14 (vi) 2

a. 10 positive tokens: +10. b. 5 negative tokens: −5. c. 5 zero pairs cancel, 2 negatives are left: −2. d. 2 zero pairs cancel, 4 positives are left: +4.

a. +10 b. −5 c. −2 d. +4

a. 3 zero pairs cancel, 2 negatives are left: $(+3)+(-5)=-2$; the attendant is on Floor −2.

b. 3 zero pairs cancel, 3 positives are left: $(+6)+(-3)=+3$; the attendant is on Floor +3.

a. (+3) + (−5) = −2, Floor −2 b. (+6) + (−3) = +3, Floor +3

a. Take 7 positives from 10: +3.

b. Take 4 negatives from 8: −4.

c. Take 4 negatives from 9: −5.

d. Only 9 positives: add 3 zero pairs, then take 12 positives; 3 negatives remain: −3.

e. Only 5 negatives: add 2 zero pairs, take 7 negatives; 2 positives remain: +2.

f. Add 4 zero pairs, take 6 negatives; 4 positives remain: +4.

a. +3 b. −4 c. −5 d. −3 e. +2 f. +4

a. +2 b. −3 c. +2 d. −5 e. +5 f. −12

a. +2 b. −3 c. +2 d. −5 e. +5 f. −12

There are 3 negatives and no positives, so put in 5 zero pairs and take away the 5 positives. 3 + 5 = 8 negatives remain: $-3-(+5)=-8$.

5 zero pairs; the result is −8.

a. −13 b. +15 c. −14 d. −19 e. +10 f. −9

a. −13 b. +15 c. −14 d. −19 e. +10 f. −9

• After the credit: $100+60=$ ₹160.
• After the bill: $160-30=$ ₹130.
• After the purchase: $130-150=$ −₹20. Yes, some banks allow a negative balance for a short time (usually with a fee or interest).
• After earning ₹200: $-20+200=$ ₹180.

₹160, ₹130, −₹20 (possible in some banks), then ₹180.

$$(+30)+(+40)+(+50)+(-40)+(-50)+(-60)=120-150=-30$$

The balance is −₹30.

−₹30

Debits: $1+2+4+8+16+32+64+128=255$ (one less than the next power of 2, as in Chapter 1). Balance $=-255+256=$ ₹1.

₹1

Positive balance: you have money available for needs and emergencies; you do not pay extra fees or interest; cheques and payments do not bounce; and the bank may pay you interest on your savings.

Temporary negative balance: when an important payment cannot wait (a medical emergency, school fees), or when a purchase for a business will earn more money soon, like the purchase in the example that let you earn ₹200 the next day. It is worthwhile only if you can repay quickly and the fees are small.

A positive balance keeps money ready and avoids fees; a short negative balance may be worth it for an emergency or a purchase that will soon earn more than it costs.

• Highest point: A; lowest point: D.
• Decreasing order: A, E, C, G, F, B, D.
• Increasing order: D, B, F, G, C, E, A.

A +1500, B −500, C +300, D −1200, E +1200, F −200, G +100 (m); highest A, lowest D; decreasing A, E, C, G, F, B, D.

• Highest: Mount Everest, about +8849 m (8848.86 m).
• Lowest on the ocean floor: the Challenger Deep in the Mariana Trench (Pacific Ocean), about −10,935 m (some surveys give about −10,994 m; nearly 11 km below sea level). The lowest point on land is the shore of the Dead Sea, about −430 m.

Mount Everest, about +8849 m; the Challenger Deep, about −10,935 m (lowest on land: the Dead Sea shore, about −430 m).

Examples: Leh, Kargil and Dras (Ladakh), Gulmarg and Srinagar (Jammu and Kashmir), Keylong, Manali and Shimla (Himachal Pradesh), Tawang (Arunachal Pradesh), and Lachung (Sikkim). Dras is one of the coldest inhabited places, with winter nights below −20°C.

Common feature: they are in the Himalayas, at high altitudes, in the north of India. Air becomes colder as we go higher, and in winter the Sun's rays are slanting and the nights are long, so temperatures fall below freezing.

Leh, Kargil, Dras, Gulmarg, Shimla, Tawang and others: all are high in the Himalayas, where higher altitude and winter make it very cold.

The day is warmest in the early afternoon and coldest before dawn:

14°C at 2 p.m., 8°C at 11 a.m., −2°C at 11 p.m., −4°C at 2 a.m.

• Top row: $5+(-3)+(-5)=-3$
• Bottom row: $(-8)+(-2)+7=-3$
• Left column: $5+0+(-8)=-3$
• Right column: $(-5)+(-5)+7=-3$

The border sum is −3.

−3

Fill the corner cells first, then each row and column has only one empty cell left.

More ways for grid (iii):

• 7, −11, 0 / −11, ☐, −5 / 0, −5, 1
• 7, −12, 1 / −12, ☐, −5 / 1, −5, 0

(Check: every row and column of each grid adds up to the required sum.)

For example (i) −10, 10, 4 / 5, −5 / 9, −10, 5 (ii) 6, 8, −16 / 11, −5 / −19, −2, 19 (iii) 7, −2, −9 / −3, −5 / −8, −6, 10 (and many more).

All three grids can be filled in many ways. Each grid has 8 border cells and only 4 conditions (the 4 border sums). When fewer cells are given, more cells can be chosen freely: once you choose a value for one free cell, the others in that row or column adjust to match the sum. Grid (iii) has the fewest given numbers, so it has the most freedom.

Puzzle for classmates: border sum −6 with 4 in the top-left corner and −3 in the bottom-right corner. (One answer: 4, −5, −5 / −2, ☐, 2 / −8, 5, −3.)

All of them, because there are more empty border cells than conditions, so some cells can be chosen freely. Puzzle example: border sum −6 with corners 4 and −3.

1. The sum is always −1 for the first grid (3, 4, 0, 9 / −2, −1, −5, 4 / 1, 2, −2, 7 / −7, −6, −10, −1), whatever numbers you choose.

2. For the grid 7, 10, 13, 16 / −2, 1, 4, 7 / −11, −8, −5, −2 / −20, −17, −14, −11 the sum is always −8; for the grid −11, −10, −9, −8 / … / 1, 2, 3, 4 it is always −14. (The textbook prints −7 in the last row of the first grid; following the pattern of the rows it should be −17, which gives the constant sum −8.)

3. What is special: each grid is like an addition table. In the second grid, for example, every row is the row above it plus 4, and in each row the numbers go up by 1. So every number = (a number for its row) + (a number for its column). Choosing one number from each row and each column always uses every row number once and every column number once, so the total never changes. The magic is in both the numbers and the way they are arranged. To make your own: write numbers along the top and side of a 4 × 4 grid and fill each cell with the sum.

1. Always −1 2. −8 and −14 3. Each cell is (row number) + (column number), like an addition table, so any choice of one cell per row and column gives the same sum.

a. −6, −5, −4, −3, −2, −1

b. −3, −2, −1, 0, 1, 2, 3

c. −14, −13, −12, −11, −10, −9

d. −29, −28, −27, −26, −25, −24

a. −6 to −1 b. −3 to 3 c. −14 to −9 d. −29 to −24

For example $(-5)+7+(-10)=-8$, $(-2)+(-3)+(-3)=-8$, or $10+(-20)+2=-8$.

e.g. −5, 7 and −10 (or −2, −3 and −3)

The faces are three odd negatives (−1, −3, −5) and three even positives (2, 4, 6). Listing all sums:

• two negatives: −2, −4, −6, −8, −10;
• two positives: 4, 6, 8, 10, 12;
• one of each: 1, 3, 5, −1, −3, 1, 3, −1 (all odd, from −3 to 5).

The numbers not possible are −9, −7, −5, 0, 2, 7, 9 and 11.

−9, −7, −5, 0, 2, 7, 9 and 11

−5, −21, −5, −21, −5, 5, 5, 21

Taking the present year as 2026:

a. $2026-150=$ 1876 CE.

b. $2026-2200=-174$. Since there is no year 0 (1 BCE is followed directly by 1 CE), we must go one more year back: 175 BCE.

c. 680 BCE is like −680; $-680+320=-360$, so 360 BCE (we do not pass year 0, so no adjustment is needed).

a. 1876 b. 175 BCE c. 360 BCE (for the year 2026)

a. Add 6 each time: −16, −10, −4.

b. Two patterns alternate: 3, 2, 1, 0, … (down by 1) and 4, 5, 6, 7, … (up by 1): −1, 8, −2.

c. The differences are −6, −5, −4, −3, so the earlier ones were −8 and −7, and the next ones are −2, −1, 0: 27, 19, 12, 6, 1, −3, −6, −8, −9, −9.

a. −16, −10, −4 b. −1, 8, −2 c. 27, 19 … −8, −9, −9

$$(-2)+(-9)-(+18)-(+1)=-30$$

exactly. (Another: $(-5)+(-9)-(+18)+(+1)$ … gives −31.)

(−2) + (−9) − (+18) − (+1) = −30

a. Always positive (e.g. 3 − (−2) = 5).

b. Positive, negative or zero (e.g. 5 + (−2) = 3, 2 + (−5) = −3).

c. Always negative (e.g. −3 + (−4) = −7).

d. Positive, negative or zero (e.g. −2 − (−5) = 3, −5 − (−2) = −3).

e. Always negative (e.g. −3 − 4 = −7).

f. Positive, negative or zero (e.g. −2 + 5 = 3, −5 + 2 = −3).

a. always positive b. either c. always negative d. either e. always negative f. either

Each group of 5 tokens (3 positive, 2 negative) has value $3-2=+1$. There are $100÷5=20$ groups, so the value is +20.

+20

Think of $a-b$ as "the movement needed to go from Floor $b$ to Floor $a$":

Treat a − b as the movement from Floor b to Floor a; e.g. 7 − 4 = 3, 4 − 7 = −3, 2 − (−3) = 5, (−6) − (−6) = 0, 0 − (−2) = 2.