NCERT Solutions Class 9 Maths Chapter 3 the World of Numbers

F R E E S T U D Y M AT E R I A L F O R E V E R Y S T U D E N T

C L A S S 9 · M AT H S

NCERT Solutions

Chapter 3: The World of
Numbers

NCERT Textbook — Ganita Manjari

BOOK PAGES SECTIONS QUESTIONS MEDIUM

41 – 67 19 57 English

Solutions, notes, sample papers & more at 55 pages

Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

CLASS 9 · MATHS · GANITA MANJARI

NCERT Solutions — Chapter 3: The World of Numbers
This chapter walks the whole road from tally notches on the Ishango bone to the unbroken real number line
— Brahmagupta's śhūnya and negative numbers, the rational numbers and their density, the proof that √2 is
irrational, and the decimal signature that separates rational from irrational.

TEXTBOOK BOOK PAGES

Ganita Manjari (Class 9) 41 – 67

SECTIONS QUESTIONS

19 57

MEDIUM

English

Exercise Set 3.1 — Page 43
3.1 The Dawn of Mathematics: The Human Need to Count

Q1 A merchant in the port city of Lothal is exchanging bags of spices for copper ingots.
He receives 15 ingots for every 2 bags of spices. If he brings 12 bags of spices to the
market, how many copper ingots will he leave with?

He leaves with 90 copper ingots.

12 bags = 6 lots of 2 bags

each lot of 2 bags → 15 ingots

ingots = 6 × 15 = 90

The same answer through the unit rate:

1 bag → 15 ÷ 2 = 7.5 ingots

12 bags → 12 × 7.5 = 90 ingots

Page 1 of 55

Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

Why it happens: the exchange is a fixed ratio 15 : 2. Multiplying both parts of a ratio
by the same number does not change it, and 12 = 6 × 2, so the ingots must be 6 × 15.
The unit rate 7.5 is not a whole number, yet the answer is — because 12 is a multiple
of 2.

Q2 Look at the sequence of numbers on one column of the Ishango bone: 11, 13, 17, 19.
What do these numbers have in common? List the next three numbers that fit this
pattern.

13 19

11 17

Fig. 3.1, page 42 — redrawn sketch of the tally groups notched on the Ishango bone.

All four are prime numbers — in fact they are exactly the primes between 10 and 20.

11 = 1 × 11 only

13 = 1 × 13 only

17 = 1 × 17 only

19 = 1 × 19 only

Continuing past 19, we test each number for a factor other than 1 and itself:

Page 2 of 55

Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

NUMBER FACTORS PRIME?

20, 21, 22 2·10, 3·7, 2·11 no

23 none but 1 and 23 yes

24 … 28 all composite no

29 none but 1 and 29 yes

30 2·15 no

31 none but 1 and 31 yes

The next three are 23, 29 and 31.

Did you know? To test whether n is prime you need only try prime divisors up to √n.
For 31, √31 < 6, so testing 2, 3 and 5 is enough.

Q3 We know that Natural Numbers are closed under addition (the sum of any two
natural numbers is always a natural number). Are they closed under subtraction?
Provide a couple of examples to justify your answer.

No — the natural numbers are not closed under subtraction.

3 − 5 = −2, and −2 is not a natural number

7 − 7 = 0, and 0 is not a natural number either

Why it happens: closure means the operation can never take you outside the set.
Addition cannot: a + b is always at least as large as a, so it stays in {1, 2, 3, …}.
Subtraction can: a − b lands outside whenever b ≥ a. One counter-example is enough
to destroy a closure claim; a hundred successful cases prove nothing.

This failure is exactly what pushed mathematics forward. To make subtraction always possible,
Brahmagupta added zero and the negative numbers — and the enlarged set, the integers ℤ, is
closed under subtraction.

Page 3 of 55

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Class 9 Maths Chapter 3 The World of Numbers
a g l AglaSem · NCERT Solutions

co m
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Ancient Indians used the joints of their fingers to count, a practice still seen today.
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Q4*

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Each finger has 3 joints, and the thumb is used to count them. How many can you
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On one hand you can count up to 12.
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total joints = 4 × 3 = 12

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Why it happens: a counting base is simply the size of the natural group you finish
before starting again. Here the hand .fills
a s emdozens gives 12 × 12 = 144 (a gross), which is
agl to a foot and 12 months to a year all survive.
the other hand to record completed
why dozens, grosses, 12 inches

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Think and Reflect — Page 46
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3.3.1 The Arithmetic of Integers

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and debt. If a negative number represents a debt, then multiplying by a negative

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agare effectively ₹12 richer!
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number represents the removal of that debt. (Hint: If someone

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Because removing a debt makes you richer — and “removing” is what the second negative sign
does.
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m ase
.co


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Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

one debt of ₹3 → −3

taking away 4 such debts → (−3) × (−4)

your position improves by ₹12 → +12

The book's picture is convincing, but the rule is also forced by the arithmetic we already accept.
Watch what the distributive law demands:

(−3) × [4 + (−4)] = (−3) × 0 = 0

so (−3) × 4 + (−3) × (−4) = 0

we already know (−3) × 4 = −12

therefore −12 + (−3) × (−4) = 0

hence (−3) × (−4) = +12

Why it happens: once we insist that a × 0 = 0 and that p(q + r) = pq + pr must keep
holding for negative numbers, the value of (−3) × (−4) is no longer a matter of choice.
Any other answer would break the distributive law. So “minus times minus is plus” is
not a convention invented to be neat — it is the only value that keeps the whole
system consistent.

Check it yourself: run the same argument on (−1) × (−1). From (−1)(1 + (−1)) = 0 you
get −1 + (−1)(−1) = 0, so (−1)(−1) = 1.

Exercise Set 3.2 — Page 46
3.3 Integers: Expanding the Horizon

Q1 The temperature in the high-altitude desert of Ladakh is recorded as 4 °C at noon.
By midnight, it drops by 15 °C. What is the midnight temperature?

The midnight temperature is −11 °C.

Page 5 of 55

Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

midnight = noon temperature − fall

= 4 − 15

= −11 °C

On the number line, start at 4 and move 15 units to the left. You pass 0 after 4 steps and
continue 11 more steps into the negative side.

a fall of 15
−11 4

−15 −12 −9 −6 −3 0 3 6 9

A drop of 15 °C from 4 °C lands 11 units to the left of zero.

Tip: a “drop” always means subtract. A drop of 15 from 4 is written 4 − 15, never 15 −
4.

Q2 A spice trader takes a loan (debt) of ₹850. The next day, he makes a profit (fortune)
of ₹1,200. The following week, he incurs a loss of ₹450. Write this sequence as an
equation using integers and calculate his final financial standing.

Writing a debt as a negative integer and a fortune as a positive one:

(−850) + 1200 + (−450)

= 350 + (−450)

= −100

His final standing is −₹100, that is, he is ₹100 in debt.

Why it happens: Brahmagupta's rules let all three transactions live on one number
line. A debt of 850 and a fortune of 1200 partly cancel — 850 of the profit clears the
loan and 350 is left over. The later loss of 450 is larger than that 350, so it pushes the
trader back across zero by 100.

Page 6 of 55

Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

Check it yourself: the order does not matter. 1200 + (−850) + (−450) = 1200 − 1300 =
−100, the same answer, because addition of integers is commutative and associative.

Q3 Calculate the following using Brahmagupta’s laws: (i) (–12) × 5 (ii) (–8) × (–7) (iii) 0 – (–
14) (iv) (–20) ÷ 4

PART BRAHMAGUPTA'S LAW USED WORKING ANSWER

(i) (−12) × 5 debt × fortune = debt 12 × 5 = 60, sign negative −60

(ii) (−8) × (−7) debt × debt = fortune 8 × 7 = 56, sign positive 56

(iii) 0 − (−14) removing a debt is a fortune 0 + 14 14

(iv) (−20) ÷ 4 debt shared into 4 equal parts 20 ÷ 4 = 5, sign negative −5

Why it happens: in each case the size of the answer comes from ordinary whole-
number arithmetic; only the sign comes from the debt–fortune rule. Part (iv) is a
division, and it obeys the same sign rule as multiplication because (−5) × 4 = −20.

Check it yourself: (iii) reads “zero minus a debt of 14”. If you owe nothing and
someone cancels a debt of ₹14 on your behalf, you are ₹14 better off — so the
answer must be +14.

Q4 Explain, using a real-world example of debt, why subtracting a negative number is
the same as adding a positive number (e.g., 10 – (–5) = 15).

Suppose you have ₹10 in hand and your account also carries a debt of ₹5 written as −5. If a
relative takes that debt away — subtracts the −5 — nothing is removed from your pocket, yet
you are ₹5 better off.

Page 7 of 55

Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

10 − (−5)

= 10 + 5

= 15

The algebraic reason is that subtracting means adding the additive inverse, and the additive
inverse of −5 is +5:

a − b = a + (−b)

10 − (−5) = 10 + [−(−5)] = 10 + 5 = 15

Why it happens: −(−5) must equal 5, because −5 and 5 add to 0 and a number has
only one additive inverse. So the two minus signs undo one another. In the debt
picture: a debt is a negative amount, and cancelling a negative amount is a gain.

Tip: read “−(−5)” as “remove a debt of 5”, not as “two minus signs somehow become
plus”. The meaning makes the rule impossible to forget.

Think and Reflect — Page 47
3.4 Filling the Spaces: Fractions and Rational Numbers

Q1 Can you explain why we need q ≠ 0 in the definition of a rational number?

Because division by zero has no answer — and the symbol p/q means “the number which,
multiplied by q, gives p”.

let p/q = x, so x × q = p

if q = 0 then x × 0 = p

but x × 0 = 0 for every x

Now two separate disasters appear:

If p ≠ 0, say 5/0, we would need x × 0 = 5. No number does that, so no value exists.

Page 8 of 55

as e
Class 9 Maths Chapter 3 The World of Numbers
a g l AglaSem · NCERT Solutions

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If p = 0, that is 0/0, then x × 0 = 0 is true for every x. So every number would qualify, and the
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symbol names nothing in particular.
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Why it happens: either there is no candidate or there are infinitely many, and in
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also wreck arithmetic: from 1 × 0 = 2 × 0 we could “cancel” the 0 and conclude 1 = 2.

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Tip: the same restriction reappears in the division rule a/b ÷ c/d = ad/bc, which the
g 0 and c ≠ 0 — the extra c ≠ 0 is there because c is
chapter states only for b ≠ 0, da≠
about to become a denominator.

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Think eand a g
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3.4 Filling the Spaces: Fractions and Rational Numbers

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Q1

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Replace each rational number by an equivalent one whose denominator is a common multiple

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a/b = (a × m/b)/m and c/d = (c × m/d)/m

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then a/b + c/d = (a·m/b + c·m/d)/m
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For example, to add 7/12 and 5/8: agl

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12 = 2² × 3, 8 = 2³ → LCM = 2³ × 3 = 24
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7/12 = 14/24 and 5/8 = 15/24
a g l
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Why it happens: multiplying numerator and denominator by the same non-zero
a g la
number does not change the
s e mvalue, because (ka)/(kb) = a/b whenever k ≠ 0. Once
a gla of the same size, the counts can simply be added — which
both fractions count parts
is exactly the rule a/b + c/b = (a + c)/b that the chapter states.

co m
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m ase
.co


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Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

Tip: the product b × d always works as a common denominator, but the LCM keeps
the numbers small and often saves the final reduction step.

Q2 Verify the distributive law for rational numbers.

The law claims p(q + r) = pq + pr for all rational p, q, r. Take p = 2/3, q = 1/4, r = 5/6.

Left side
q + r = 1/4 + 5/6 = 3/12 + 10/12 = 13/12

p(q + r) = 2/3 × 13/12 = 26/36 = 13/18

Right side

pq = 2/3 × 1/4 = 2/12 = 1/6

pr = 2/3 × 5/6 = 10/18 = 5/9

pq + pr = 1/6 + 5/9 = 3/18 + 10/18 = 13/18

Both sides give 13/18, so the law holds for this triple.
A single example only illustrates. Here is the general proof, using p = a/b, q = c/d, r = e/d with a
common denominator d:

p(q + r) = (a/b) × (c + e)/d = a(c + e)/(bd)

= (ac + ae)/(bd) (integers distribute)

= ac/(bd) + ae/(bd)

= (a/b)(c/d) + (a/b)(e/d) = pq + pr

Why it happens: the whole argument rests on one line — a(c + e) = ac + ae for
integers. Rational arithmetic is built out of integer arithmetic, so every law the
integers obey is inherited by ℚ.

Exercise Set 3.3 — Page 49

Page 10 of 55

Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

3.4 Filling the Spaces: Fractions and Rational Numbers

Q1 Prove that the following rational numbers are equal: (i) 2/3 and 4/6 (ii) 5/4 and 10/8
(iii) −3/5 and −6/10 (iv) 9/3 and 3

The chapter's test is: a/b = c/d exactly when ad = bc. Apply it to each pair.

PART CROSS PRODUCTS AD AND BC EQUAL?

(i) 2/3, 4/6 2 × 6 = 12 and 3 × 4 = 12 yes

(ii) 5/4, 10/8 5 × 8 = 40 and 4 × 10 = 40 yes

(iii) −3/5, −6/10 (−3) × 10 = −30 and 5 × (−6) = −30 yes

(iv) 9/3, 3/1 9 × 1 = 9 and 3 × 3 = 9 yes

The same four facts seen through equivalent fractions:

4/6 = (2 × 2)/(2 × 3) = 2/3

10/8 = (2 × 5)/(2 × 4) = 5/4

−6/10 = (2 × −3)/(2 × 5) = −3/5

9/3 = (3 × 3)/(3 × 1) = 3

Why it happens: ad = bc is just the cleared-denominator form of a/b = c/d —
multiply both sides by bd and the fractions disappear. That is why the test needs no
division and works even when the fractions are negative.

Tip: in (iv), a whole number is written as 3/1 before the test is applied. Every integer
is a rational number with denominator 1.

Page 11 of 55

Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

Q2 Find the sum: (i) 2/5 + 3/10 (ii) 7/12 + 5/8 (iii) −4/7 + 3/14

(i) LCM(5, 10) = 10

2/5 + 3/10 = 4/10 + 3/10 = 7/10

7 and 10 are co-prime → 7/10

(ii) 12 = 2² × 3, 8 = 2³ → LCM = 24

7/12 + 5/8 = 14/24 + 15/24 = 29/24 (= 1 5/24)

(iii) LCM(7, 14) = 14
−4/7 + 3/14 = −8/14 + 3/14 = −5/14

5 and 14 are co-prime → −5/14

Why it happens: the rule a/b + c/b = (a + c)/b only works when the parts are of equal
size, so every sum begins by rewriting both fractions over a common denominator.
After that, only the numerators are added — the denominator names the size of the
part and does not change.

Check it yourself: in (iii) the answer must be negative, since 4/7 = 8/14 is larger than
3/14.

Q3 Find the difference: (i) 5/6 − 1/4 (ii) 11/8 − 3/4 (iii) −7/9 − (−2/3)

(i) LCM(6, 4) = 12

5/6 − 1/4 = 10/12 − 3/12 = 7/12

Page 12 of 55

Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

(ii) LCM(8, 4) = 8

11/8 − 3/4 = 11/8 − 6/8 = 5/8

(iii) subtracting a negative is adding its opposite

−7/9 − (−2/3) = −7/9 + 2/3

= −7/9 + 6/9 = −1/9

Why it happens: part (iii) uses the integer rule of Section 3.3 inside a fraction. Since
2/3 = 6/9 is smaller than 7/9, adding it does not quite reach 0, so the answer stays
negative — a useful check before you compute anything.

Q4 Find the product: (i) 2/3 × 3/10 (ii) 7/11 × 5/8 (iii) −4/7 × 5/14

The rule is a/b × c/d = ac/bd. Cancelling common factors before multiplying keeps the numbers
small.

(i) 2/3 × 3/10 = (2 × 3)/(3 × 10)

cancel 3, cancel 2 → 1/5

= 1/5

(ii) 7/11 × 5/8 = 35/88

88 = 2³ × 11, 35 = 5 × 7 — no common factor

= 35/88

(iii) −4/7 × 5/14 = −20/98

divide numerator and denominator by 2

= −10/49

Page 13 of 55

as e
Class 9 Maths Chapter 3 The World of Numbers
a g l AglaSem · NCERT Solutions

co m
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Why it happens: a product of a positive and a negative rational is negative — the

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same debt-and-fortune rule as for integers, since the sign travels with the
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numerator.
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Find the quotient: (i) 2/3 ÷ 3/10 (ii) 7/11 ÷ 5/8 (iii) −4/7 ÷ 5/14
.
Q5

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Dividing by c/d means multiplying by its reciprocal d/c, which needs c ≠ 0.

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(i) 2/3 ÷ 3/10 = 2/3 × 10/3 = 20/9
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a (ii) 7/11 ÷ 5/8 = 7/11 × 8/5 = 56/55

m a s
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(iii) −4/7 ÷ 5/14 = −4/7 × 14/5

= −56/35, divide both by 7
a

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= −8/5
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m it happens: the reciprocal of c/d is the number that multiplies it to 1, since (c/d)
a s eWhy
agl × (d/c) = cd/dc = 1. So dividing by c/d and multiplying by d/c must give the same
result. Note in (i) and (ii) the quotient is larger than the first number — dividing by a
se m
fraction less than 1 makes things bigger.
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Show that: (1/2 + 3/4) × 8/3 = 1/2 × 8/3 + 3/4 × 8/3.
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ag Left side
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1/2 + 3/4 = 2/4 + 3/4 = 5/4
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(5/4) × (8/3) = 40/12 = 10/3
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co m
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m ase
.co


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Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

Right side

1/2 × 8/3 = 8/6 = 4/3

3/4 × 8/3 = 24/12 = 2 = 6/3

4/3 + 6/3 = 10/3

Both sides equal 10/3, so the statement is proved.

Why it happens: this is the distributive law (p + q)r = pr + qr with p = 1/2, q = 3/4, r =
8/3. It is not a coincidence about these particular fractions — the law holds for every
triple of rational numbers, and the equality would still be true if 8/3 were replaced by
any rational number at all.

Q7 Simplify the following using the distributive property: 7/9 (6/7 − 3/4).

Distribute 7/9 across the bracket:

7/9 × 6/7 − 7/9 × 3/4

= 42/63 − 21/36
= 2/3 − 7/12

= 8/12 − 7/12

= 1/12

Doing the bracket first must give the same answer — a useful check:

6/7 − 3/4 = 24/28 − 21/28 = 3/28

7/9 × 3/28 = 21/252 = 1/12

Why it happens: distributivity is stated in the chapter for addition, p(q + r) = pq + pr;
it extends to subtraction because q − r means q + (−r). Here the first route is lighter,
since 7/9 × 6/7 cancels the 7 immediately.

Page 15 of 55

Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

Q8 Find the rational number x such that: 5/6 (x + 3/5) = 5/6 x + 1/2.

Expand the left side using the distributive law:

5/6 (x + 3/5) = 5/6 · x + 5/6 · 3/5

5/6 × 3/5 = 15/30 = 1/2

so the left side is 5x/6 + 1/2

5x/6 + 1/2 = 5x/6 + 1/2

subtract 5x/6 from both sides → 1/2 = 1/2

The equation reduces to a true statement with no x left in it. So it is an identity: every rational
number x satisfies it.

Why it happens: the equation is nothing but the distributive law itself, written out
for p = 5/6, q = x, r = 3/5, with the product 5/6 × 3/5 already simplified to 1/2. A law
that holds for all rationals cannot single out one value of x. Had the right side been
5x/6 + 1/3, say, we would have reached 1/2 = 1/3, a false statement, and then no x
would work.

Check it yourself: try x = 2. Left side: 5/6 (2 + 3/5) = 5/6 × 13/5 = 13/6. Right side: 5/6
× 2 + 1/2 = 5/3 + 1/2 = 10/6 + 3/6 = 13/6. Equal — as it must be for any x.

Think and Reflect — Page 51
3.4.1 Representation of Rational Numbers on the Number Line

Q1 Try and represent 8/5 and – 7/4 on a number line.

Follow the chapter's rule: to mark p/q, cut the unit interval into q equal parts and count p of
them from 0 — right if the number is positive, left if it is negative.

Page 16 of 55

Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

8/5 = 1 3/5, so it lies between 1 and 2

cut the interval from 1 to 2 into 5 equal parts

move 3 parts to the right of 1

−7/4 = −1 3/4, so it lies between −2 and −1

cut the interval from −1 to −2 into 4 equal parts

move 3 parts to the left of −1

−7/4 8/5

−2 −1 0 1 2
quarters fifths

−7/4 sits three quarter-steps to the left of −1; 8/5 sits three fifth-steps to the right of 1.

Why it happens: the two numbers need different rulers. Fifths and quarters do not
share tick marks, so each one is located inside its own unit interval. If you wanted
both on the same set of ticks you would use twentieths: 8/5 = 32/20 and −7/4 =
−35/20.

In-text Questions — Page 52
3.4.2 The Density of Rational Numbers

Q1 Try to explain why the average of two rational numbers a and b, which equals (a +
b)/2, is always a rational number between a and b.

Two things have to be shown: that the average is rational, and that it lies between a and b.
1. It is rational. ℚ is closed under addition, and closed under division except by zero.

Page 17 of 55

Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

a = p/q and b = r/s with q, s ≠ 0

a + b = (ps + qr)/(qs)

(a + b)/2 = (ps + qr)/(2qs)

The numerator ps + qr and the denominator 2qs are integers, and 2qs ≠ 0. So (a + b)/2 has the
form of a rational number.
2. It lies between. Suppose a < b. Then

(a + b)/2 − a = (a + b − 2a)/2 = (b − a)/2 > 0

b − (a + b)/2 = (2b − a − b)/2 = (b − a)/2 > 0

Both gaps are positive and, in fact, equal — so a < (a + b)/2 < b, with the average exactly halfway.

Why it happens: this single fact is what makes ℚ dense. The average of a and b can
be averaged with a again, and again, without end — so between any two rational
numbers, however close, there are infinitely many more. Between 1 and 3/2 the
chapter finds 5/4; between 1 and 5/4 you would find 9/8, then 17/16, and so on for
ever.

Tip: the argument needs a ≠ b. If a = b the “average” is a itself and there is nothing
strictly between them.

Exercise Set 3.4 — Page 52
3.4.2 The Density of Rational Numbers

Q1 Represent the rational numbers 2/3, –5/4 and 1 1/2 on a single number line.

Each number is placed by cutting its own unit interval into the number of parts named by the
denominator.

Page 18 of 55

as e
Class 9 Maths Chapter 3 The World of Numbers
a g l AglaSem · NCERT Solutions

co m
m.
NUMBER LIES BETWEEN CONSTRUCTION

m 0 and 1 as e
2/3
.co a g
cut 0–1 into 3 parts, take 2 to the rightl
se m
g l a −2 and −1 cut −1 to −2 into 4 parts, take 1 to the left of −1
a
−5/4 = −1 1/4

1 1/2 = 3/2 1 and 2 halfway between 1 and 2

co m
e m . ag
g l as
−5/4
a 2/3 1½

. com2
−2 −1 0 1
e m
shalf
mquarters l a
.co ag
thirds

se m
g l a All three rational numbers marked on one line between −2 and 2.

a
m a s
c o agl
Tip: to draw all three on a single ruler, use twelfths — 2/3 = 8/12, −5/4 = −15/12 and
3/2 = 18/12, since 12 = LCM(3, 4, 2).m .
l a se
ag

co m
Q2 Find three distinct rational numbers that lie strictly between – 1/2 and 1/4.
m .
o m l a se
.c a g
se m

g l a
a
Rewrite both endpoints over a common denominator so that the gap contains enough whole-

m
numbered steps.

a se
. com a g l
m
ase
LCM(2, 4) = 4, so −1/2 = −2/4 and 1/4 = 1/4

agl
only 0/4 lies strictly between — one number, not three

m
so enlarge the denominator: use eighths
. co
em
as
−1/2 = −4/8 and 1/4 = 2/8
m l
.co a g
a s emNow the integers strictly between −4 and 2 are −3, −2, −1, 0, 1 — five choices. Take any three:
agl
.c
s e m
m a
−3/8, −1/8, 1/8

em . co agl
as
check: −4/8 < −3/8 < −1/8 < 1/8 < 2/8 ✓

a g l

co m
m .
m as e
.co


a g l Page 19 of 55

Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

Why it happens: with a common denominator d, the numbers between the two
endpoints are exactly the fractions k/d whose numerators k lie between the two
numerators. So the number of candidates is controlled entirely by how far apart the
numerators are — and multiplying the denominator by 2 doubles that gap without
moving the endpoints.

Check it yourself: repeated averaging works too. (−1/2 + 1/4)/2 = −1/8; then average
−1/8 with each endpoint to get −5/16 and 1/16.

Q3 Simplify the expression: (−1/4) + (5/12).

LCM(4, 12) = 12

−1/4 = −3/12

−3/12 + 5/12 = 2/12

divide numerator and denominator by 2

= 1/6

Why it happens: 5/12 is larger than 1/4 = 3/12, so the sum lands on the positive side
of 0. Reducing 2/12 to 1/6 is allowed because 2/12 and 1/6 are equivalent rational
numbers, and the chapter asks us to name a rational number by its co-prime form.

Q4 A tailor has 15 3/4 metres of fine silk. If making one kurta requires 2 1/4 metres of
silk, exactly how many kurtas can he make?

Convert both mixed numbers to improper fractions and divide.

15 3/4 = (15 × 4 + 3)/4 = 63/4

2 1/4 = (2 × 4 + 1)/4 = 9/4

63/4 ÷ 9/4 = 63/4 × 4/9 = 63/9 = 7

Page 20 of 55

Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

He can make exactly 7 kurtas, with no silk left over.

Why it happens: both lengths are whole numbers of quarter-metres — 63 quarters
and 9 quarters. Dividing them is just 63 ÷ 9, and since 9 divides 63 exactly, nothing
remains. The common denominator has done all the work.

Check it yourself: 7 × 2 1/4 = 7 × 9/4 = 63/4 = 15 3/4 metres. ✓

Q5 Find three rational numbers between 3.1415 and 3.1416.

Add one more decimal place. Between the fourth-place numbers 3.1415 and 3.1416 there is
room for nine fifth-place numbers.

3.1415 = 31415/10000 = 314150/100000

3.1416 = 31416/10000 = 314160/100000

Any numerator strictly between 314150 and 314160 works, for example:

3.14152 = 314152/100000 = 39269/12500
3.14155 = 314155/100000 = 62831/20000

3.14158 = 314158/100000 = 157079/50000

Why it happens: every terminating decimal is a rational number, and inserting a
further decimal place multiplies the denominator by 10 while multiplying both
numerators by 10 as well. The gap between the numerators widens from 1 to 10,
creating nine new rationals — and the trick can be repeated for ever. This is the
density of ℚ seen in decimal clothing.

Did you know? π = 3.14159265… lies inside this very interval, but π is not one of the
rationals you can list this way — its decimal never terminates and never repeats.

Page 21 of 55

Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

Q6* Can you think of other way(s) to find a rational number between any two rational
numbers?

Yes — the average is only one recipe. Here are three more, each with its own proof.

Common denominator, bigger denominator. Write a = p/m and b = r/m with r − p ≥ 2
(enlarge m if needed). Then (p + 1)/m lies strictly between them.
Weighted average. For any rational t with 0 < t < 1, the number ta + (1 − t)b lies between a
and b. Taking t = 1/2 gives the ordinary average; t = 1/3 gives a point one-third of the way
along.
The mediator of two fractions. For positive a = p/q and b = r/s with p/q < r/s, the fraction (p
+ r)/(q + s) always lies strictly between them.

Proof of the third, the least obvious one:

p/q < r/s means ps < qr

(p + r)/(q + s) − p/q = (q(p + r) − p(q + s))/(q(q + s))

= (qr − ps)/(q(q + s)) > 0

similarly r/s − (p + r)/(q + s) = (qr − ps)/(s(q + s)) > 0

So p/q < (p + r)/(q + s) < r/s.

Why it happens: in every method the answer is built from a, b and integers using
only +, −, × and ÷ by non-zero numbers. Since ℚ is closed under all four, the result is
bound to be rational; the inequalities then pin it inside the interval. Try the mediator
on 1/2 and 2/3: it gives (1 + 2)/(2 + 3) = 3/5, and indeed 1/2 < 3/5 < 2/3.

Tip: the mediator is not the same as adding fractions. (p + r)/(q + s) is a well-known
construction called the mediant, and it is the reason the fractions on a Farey
sequence line up so neatly.

Think and Reflect — Page 53

Page 22 of 55

Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

3.5 Irrational Numbers

Q1 Can √2 be written as a rational number p/q?

No. There is no pair of integers p, q with q ≠ 0 for which √2 = p/q.
A first look at why the answer must be no, before the formal proof of Section 3.5.1:

if √2 = p/q then 2q² = p²

count the factor 2 on each side

p² has an even number of 2s (each 2 in p appears twice)

q² also has an even number of 2s

so 2q² has an odd number of 2s

an even count can never equal an odd count

Why it happens: squaring doubles every exponent in a prime factorisation, so a
perfect square always carries each prime an even number of times. Multiplying by a
single extra 2 breaks that parity. Since 2 is not a perfect square, its square root
cannot be a ratio of integers. The number √2 is therefore irrational — the very first
number ever proved to be so.

Did you know? The chapter reaches √2 through geometry: the diagonal d of a unit
square satisfies 1² + 1² = d², so d² = 2. A perfectly ordinary length turns out to be a
number no fraction can name.

Think and Reflect — Page 55
3.5.1 The Proof of Irrationality of √2

Q1 Try to prove the irrationality of √3 using the approach of proof by contradiction. Will
the same approach work for √5, √7, or √10?

Claim: √3 is irrational. Follow the eight steps of Section 3.5.1, replacing “even” by “divisible by 3”.

Page 23 of 55

as e
Class 9 Maths Chapter 3 The World of Numbers
a g l AglaSem · NCERT Solutions

co m
em.
Step 1. Suppose √3 = p/q in lowest terms, q ≠ 0, gcd(p, q) = 1
m l as
.co
Step 2. square: 3 = p²/q²
m a g
l a se
g
Step 3. 3q² = p²
aStep 4. so 3 divides p². Since 3 is prime, 3 divides p

co m
. ag
write p = 3k
e m
Step 5. 3q² = (3k)² = 9k²
g l as
Step 6. divide by 3: q² = 3k² a

co m
m.
Step 7. so 3 divides q², hence 3 divides q

m as e
.co l
Step 8. p and q share the factor 3 — contradicting gcd(p, q) = 1
a g
se m
g l a
a Does it work for √5, √7, √10? Yes, all three.
The assumption is the only thing that can be wrong, so √3 is irrational.

m a s
m .co agl
ase
NUMBER KEY STEP VERDICT

√5
agl
5 prime: 5 | p² ⇒ 5 | p irrational

m
.co
√7 7 prime: 7 | p² ⇒ 7 | p irrational

se m
com l a
√10 10 not prime, but 2 | p² ⇒ 2 | p works irrational

. a g
m
ase
agl = 4k², i.e. 5q² = 2k², so q² is even and q is even — contradiction.
For √10, run the argument with the prime 2 alone: from 10q² = p² we get p even, p = 2k, so 10q²

se m
com g l a
.
Why it happens: the engine of the whole proof is the step “n divides p² implies n
m a
ase
divides p”, and that step is valid precisely when n is prime (or, more generally,

agl
square-free). It fails for n = 4: 4 divides 6² = 36 but 4 does not divide 6 — which is
exactly why √4 = 2 is not irrational. So the method proves √n irrational for every
positive integer n that is not a perfect square, and honestly refuses to prove it when
co m
m .
n is one.
m as e
.co a g l
sem
g l a
a c
Think and Reflect — Page 55
m .
m a s e
em . co agl
g l as
a

com
m .
m ase
.co


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Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

3.5.2 Construction of Length √n

Q1 We have seen how to obtain a line whose length is a rational number. How do we
obtain lines whose lengths are irrational?

By using the Baudhāyana–Pythagoras Theorem instead of division. Cutting a unit into q equal
parts can only ever produce rational lengths p/q; a right triangle produces lengths that are
square roots.

right triangle with legs 1 and 1

hypotenuse² = 1² + 1² = 2

hypotenuse = √2 — an irrational length

The construction of Fig. 3.11 then transfers that length onto the number line:

Step 1. Mark OA = 1 unit on the number line and erect a perpendicular at A.
Step 2. On that perpendicular mark B with AB = 1 unit and join OB. Then OB = √2.
Step 3. With centre O and radius OB draw an arc cutting the number line at P. Then OP = √2,
so P is the point √2.

B

√2
1

O 1 A P

0 1 2 3

OA = AB = 1, so OB = √2; swinging OB down with a compass marks √2 on the line at P.

Why it happens: a compass moves a length without changing it. The theorem
manufactures an irrational length out of two rational ones, and the compass carries
it to the axis. This is how the number line gets filled at points no fraction can reach.

Page 25 of 55

Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

Think and Reflect — Page 56
3.5.2 Construction of Length √n

Q1 Try to extend this method for constructing line segments of lengths √3 and √5 using
a ruler and a compass. Generalise this method to construct a line segment of any
length of the form √n, where n is a positive integer.

Length √3. Start from the √2 already built at P.

erect a perpendicular at P and mark C with PC = 1

OC² = OP² + PC² = (√2)² + 1² = 2 + 1 = 3

so OC = √3

swing OC onto the number line with the compass

Length √5. Repeat twice more, or take a shortcut with legs 1 and 2:

from √3, add a unit perpendicular: 3 + 1 = 4, giving √4 = 2

from 2, add a unit perpendicular: 4 + 1 = 5, giving √5

shortcut: legs 2 and 1 give 2² + 1² = 5 directly

The general rule. Having constructed √n, erect a perpendicular of length 1 at its far end. The
new hypotenuse satisfies

(new)² = (√n)² + 1² = n + 1

new length = √(n + 1)

So from OA = 1 = √1 the construction climbs one step at a time: √1 → √2 → √3 → √4 → … → √n
for any positive integer n, in exactly n − 1 steps. Repeating this endlessly draws the square root
spiral of Fig. 3.14.

Why it happens: each step adds exactly 1 to the square of the length, never to the
length itself. That is why the outer edge of the spiral grows more and more slowly —
the gap between √n and √(n + 1) shrinks as n grows, since √(n+1) − √n = 1/(√(n+1) +
√n).

Page 26 of 55

Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

Check it yourself: the construction never needs a protractor. A ruler and compass
suffice, because a perpendicular at a point can be drawn with a compass alone.

In-text Questions — Page 57
3.6.1 Rational Decimals: Terminating and Repeating

Q1 Can you tell for which rational numbers the decimal will be terminating?

Exactly those p/q which, in lowest terms, have a denominator whose only prime factors are 2
and 5.

q = 2^m × 5^n → decimal terminates

q has any other prime factor → decimal repeats

RATIONAL NUMBER DENOMINATOR FACTORISED DECIMAL

3/8 2³ 0.375, terminates

3/20 2² × 5 0.15, terminates

13/250 2 × 5³ 0.052, terminates

4/15 3×5 0.2666…, repeats

5/11 11 0.4545…, repeats

Why it happens: a decimal that stops after k places is a fraction with denominator
10^k = 2^k × 5^k. So p/q terminates only if p/q can be rewritten with a denominator
that is a power of 10 — and multiplying q by anything can never remove a prime
factor of 3, 7, 11, … already sitting inside it. Conversely, if q = 2^m 5^n, multiplying
top and bottom by the missing power of 2 or 5 turns the denominator into
10^max(m, n).

Tip: reduce to lowest terms first. 6/15 looks as if it has the prime 3, but 6/15 = 2/5 =
0.4, which terminates.

Page 27 of 55

Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

Think and Reflect — Page 57
3.6.1 Rational Decimals: Terminating and Repeating

Q1 Try to find the decimal expansions of 10/3 and 11/12. What do you observe about
the repetition of the digits after the decimal point?

10/3 by long division

10 ÷ 3 = 3, remainder 1

10 ÷ 3 = 3, remainder 1 (again)

10/3 = 3.3333… = 3.3

11/12 by long division
110 ÷ 12 = 9, remainder 2

20 ÷ 12 = 1, remainder 8

80 ÷ 12 = 6, remainder 8

80 ÷ 12 = 6, remainder 8 (repeats)

11/12 = 0.91666… = 0.916

What we observe. Both are non-terminating and repeating, but the repetition begins in
different places:

In 10/3 the repeating block starts immediately after the decimal point — a pure repeating
decimal. Its denominator 3 has no factor of 2 or 5.
In 11/12 two digits, 9 and 1, come first and only then does 6 repeat — a general repeating
decimal. Its denominator 12 = 2² × 3 mixes a 3 with powers of 2.

Why it happens: the 2² in 12 can be cleared into the power of 10 (that produces the
non-repeating digits), but the leftover 3 can never be cleared, so a remainder must
recur and the tail loops for ever. When the denominator has no 2s or 5s at all, as in 3,
there are no non-repeating digits to make, and the loop starts at once.

Page 28 of 55

as e
Class 9 Maths Chapter 3 The World of Numbers
a g l AglaSem · NCERT Solutions

co m
m.
Check it yourself: the number of non-repeating digits equals the larger of the

as e
co1.m l
powers of 2 and 5 in the denominator. For 12 = 2² × 5⁰ that is 2 — exactly the two
. a g
em
digits 9 and
a s
agl

co m
ag
Think and Reflect — Page 58
.
a s em
3.6.1 Rational Decimals: Terminating and Repeating

agl
The decimal expansion of p/q will be terminating precisely when the prime factors
m
Q1

co
m.
of q are only 2, only 5 or both 2 and 5. Can you explain why?

o m l a se
ANSWER c
. a g
m
e “precisely” means two statements must be proved, one in each direction. Throughout,
sword
l a
agp/q is in lowest terms.
The

m
If q = 2^m × 5^n, the decimal terminates. Let k be the larger of m and n. Multiply top and
a s
bottom by whatever is missing:
m.co agl
l a se
ag
2^m 5^n × 2^(k−m) 5^(k−n) = 2^k 5^k = 10^k

so p/q = (p × 2^(k−m) 5^(k−n))/10^k
co m
m .
m as e
.co a g l
m
A whole number divided by 10^k is a decimal that stops after k places. The chapter's own

l a se
example is 3/20 with 20 = 2² × 5, k = 2:

ag
se m
com a
3/20 = (3 × 5)/(2² × 5 × 5) = 15/100 = 0.15

. a g l
e m
g l asonly be 2s and 5s. A decimal stopping after k places is
If the decimal terminates, q can
a
p/q = N/10^k for some integer N
co m
m .
m
so p × 10^k = N × q
as e
.qcodivides p × 10^k a g l
se m
g l a
a but gcd(p, q) = 1, so q divides 10^k = 2^k 5^k
c
m .
m a s e
. co agl
A divisor of 2^k 5^k can contain no prime other than 2 and 5. Hence q has exactly the stated

e m
as
form.

a g l

co m
m .
m ase
.co


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Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

Why it happens: our numerals are written in base 10, and 10 = 2 × 5. A denominator
can be absorbed into a power of ten only if it is built from ten's own prime bricks.
Any other prime — 3, 7, 11, 13 — survives every multiplication, the division never
reaches remainder 0, and since only finitely many remainders are possible one of
them must return, sending the digits into a loop.

Did you know? In base 3, the fraction 1/3 would terminate (it is 0.1 in that base)
while 1/2 would repeat. Terminating is a property of the number and the base, not of
the number alone.

Exercise Set 3.5 — Page 61
3.6.3 Irrational Decimals: Chaos and Infinity

Q1 Without performing long division, determine which of the following rational
numbers will have terminating decimals and which will be repeating: 7/20, 4/15 and
13/250. Then check your answers by explicitly performing the long divisions and
expressing these rational numbers as decimals.

Prediction from the denominators. Each fraction is already in lowest terms, so factorise the
denominator and look for primes other than 2 and 5.

NUMBER DENOMINATOR OTHER PRIMES? PREDICTION

7/20 20 = 2² × 5 none terminating

4/15 15 = 3 × 5 3 repeating

13/250 250 = 2 × 5³ none terminating

Now the long divisions.

7/20: 70 ÷ 20 = 3, remainder 10

100 ÷ 20 = 5, remainder 0 — stop

7/20 = 0.35 (2 places, as 2² × 5 predicts max(2, 1) = 2)

Page 30 of 55

Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

4/15: 40 ÷ 15 = 2, remainder 10

100 ÷ 15 = 6, remainder 10

100 ÷ 15 = 6, remainder 10 — the remainder 10 has returned

4/15 = 0.2666… = 0.26

13/250: 130 ÷ 250 = 0, remainder 130

1300 ÷ 250 = 5, remainder 50

500 ÷ 250 = 2, remainder 0 — stop

13/250 = 0.052 (3 places, as 2 × 5³ predicts max(1, 3) = 3)

All three predictions are confirmed.

Why it happens: for the two terminating cases you can even skip the division. 7/20 =
35/100 (multiply by 5) and 13/250 = 52/1000 (multiply by 4). For 4/15 the factor 3
cannot be turned into a power of 10, so a remainder must repeat — and it does, at
the very first step.

Q2 Perform the long division for 1/13. Identify the repeating block of digits. Does it
show cyclic properties if you evaluate 2/13? Now compute 3/13, 4/13, etc. What do
you notice?

1/13 by long division

10 ÷ 13 = 0, r 10 → 100 ÷ 13 = 7, r 9

90 ÷ 13 = 6, r 12 → 120 ÷ 13 = 9, r 3

30 ÷ 13 = 2, r 4 → 40 ÷ 13 = 3, r 1

remainder 1 has returned → the digits now repeat

1/13 = 0.076923

The repeating block is 076923, of length 6. Now the other thirteenths:

Page 31 of 55

Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

FRACTION REPEATING BLOCK FRACTION REPEATING BLOCK

1/13 076923 2/13 153846

3/13 230769 4/13 307692

5/13 384615 6/13 461538

7/13 538461 8/13 615384

9/13 692307 10/13 769230

11/13 846153 12/13 923076

What we notice. The twelve blocks are not twelve different sets of digits — they fall into two
families, each a cyclic rotation of one block:

Family A (rotations of 076923): 1/13, 3/13, 4/13, 9/13, 10/13, 12/13
Family B (rotations of 153846): 2/13, 5/13, 6/13, 7/13, 8/13, 11/13

So 2/13 does not give a rotation of 1/13's block — it starts a second cycle. This is the difference
from 1/7.

Why it happens: the repeating block of 1/n has length equal to the number of
remainders that appear in the long division. For 1/7 all six possible non-zero
remainders 1…6 occur, so multiplying by 1…6 merely restarts the same cycle at a
different point — 142857 is a true cyclic number. For 1/13 only six of the twelve
possible remainders {1, 3, 4, 9, 10, 12} occur, so the twelve fractions split into two
cycles of six. The block of 1/n is fully cyclic exactly when its length is n − 1.

Check it yourself: 076923 × 3 = 230769 and 076923 × 9 = 692307 — both rotations.
But 076923 × 2 = 153846, which is not.

Page 32 of 55

Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

Q3 Classify the following numbers as rational or irrational: (i) √81 (ii) √12 (iii) 0.33333 …
(iv) 0.123451234512345 … (v) 1.01001000100001 … (Notice the pattern: Is it repeating
a single block?) (vi) 23.560185612239874790120. Find the explicit fractions in case
they are rational.

NUMBER REASON TYPE FRACTION

(i) √81 81 = 9² is a perfect square rational 9 = 9/1

(ii) √12 12 = 2² × 3 is not a perfect square irrational —

(iii) 0.33333… pure repeating, block 3 rational 1/3

(iv) 0.123451234512345… pure repeating, block 12345 rational 4115/33333

(v) 1.01001000100001… never repeats a fixed block irrational —

(vi) 23.560185612239874790120 terminating decimal rational see below

The working for the rational ones.

(i) √81 = 9, because 9 × 9 = 81 → 9/1

(iii) x = 0.3; 10x = 3.3

9x = 3 → x = 3/9 = 1/3

(iv) x = 0.12345; 5 digits repeat, so multiply by 10⁵

100000x = 12345.12345

99999x = 12345 → x = 12345/99999

99999 = 3² × 41 × 271 and 12345 = 3 × 5 × 823, common factor 3

x = 4115/33333

Page 33 of 55

as e
Class 9 Maths Chapter 3 The World of Numbers
a g l AglaSem · NCERT Solutions

co m
e m.
(vi) 21 digits after the point, so
m l as
.co
= 23560185612239874790120/10²¹
m a g
l a se
g
divide top and bottom by 40
a= 589004640305996869753/25000000000000000000

com
m . ag
l a se
Why it happens: (ii) is irrational by the parity argument of Section 3.5.1 — √12 = 2√3,
ag be a ratio of integers. (v) is the interesting one: the
and √3 is irrational, so √12 cannot
digits do follow a pattern, but the gaps of zeros grow 1, 2, 3, 4, … and never settle

co m
m.
into a block of fixed length that repeats for ever. A rational number must repeat a

as e
com block. l
block of constant length, so (v) is irrational. A visible pattern is not the same thing as
. a g
em
a repeating

l a s
ag Tip: (vi) simply stops — it has 21 decimal places and no “…”. Every decimal that stops
m a s
.co agl
is a fraction over a power of 10, hence rational, no matter how ugly the digits look.

se m
g l a
a
The number 0.9 (which means 0.99999 …) is a rational number. Using algebra (let x =
m
Q4
0.9, multiply by 10, and subtract), explain why 0.9 is exactly equal to 1.
. co
e m
m l as
m
.co a g
l a se
ag let x = 0.9 = 0.9999…

se m
one digit repeats, so multiply by 10¹ = 10
com g l a
m . a
ase
10x = 9.9999… = 9.9

agl
subtract: 10x − x = 9.9999… − 0.9999…

9x = 9
co m
m .
x=1
m as e
.co a g l
a s emSo 0.9 and 1 are two decimal names for the same number.
agl
.c
s e m
m a
e m . co agl
g l as
a

co m
m .
m ase
.co


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Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

Why it happens: the subtraction is exact only because the tail of 9s is infinite; every 9
after the point in 10x is matched by a 9 in x and cancels, leaving 9 exactly. If the 9s
stopped anywhere — say at 0.9999 — the subtraction would leave 8.9999 − 0.0000…
and x would fall short of 1. The equality is a statement about a completed infinite
expansion, not about any finite approximation.

Check it yourself: the same fact follows from arithmetic you already trust: 1/3 = 0.3,
so 3 × 1/3 = 3 × 0.3, that is 1 = 0.9. Or note that 1 − 0.9 would have to be a positive
number smaller than every 1/10^k, and no such rational number exists. This is
exactly the “non-uniqueness of decimal representations” the chapter points out:
2.47000… = 2.46999… as well.

Q5* We have seen that the repeating block of 1/7 is a cyclic number. Try to find more
numbers (n) whose reciprocals (1/n) produce decimals with repeating blocks that
are cyclic.

The reciprocal 1/n gives a fully cyclic block exactly when the repeating block has the greatest
possible length, n − 1. That forces n to be prime.

N BLOCK LENGTH REPEATING BLOCK CYCLIC?

7 6=7−1 142857 yes

11 2 09 no

13 6, not 12 076923 no

17 16 = 17 − 1 0588235294117647 yes

19 18 = 19 − 1 052631578947368421 yes

23 22 = 23 − 1 0434782608695652173913 yes

29 28 = 29 − 1 0344827586206896551724137931 yes

So n = 7, 17, 19, 23, 29 all work, and the list continues with 47, 59, 61, 97, …

check for 17: 0588235294117647 × 2 = 1176470588235294

which is the same digits rotated — a cycle ✓

Page 35 of 55

Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

Why it happens: in the long division of 1/n the remainders can only be 1, 2, …, n − 1,
so the block can never be longer than n − 1 digits. When the division actually visits all
n − 1 remainders before returning to 1, every fraction k/n restarts the same loop at a
different remainder — which on paper looks like the same digits rotated. When only
some remainders appear, as with 13, the fractions split into several separate cycles
and the block is not cyclic. Primes with this all-remainders property are called full
reptend primes.

Tip: n must be co-prime to 10. Any n divisible by 2 or 5 has some terminating part in
its decimal, so it cannot produce a full-length pure repeating block.

Think and Reflect — Page 64
3.7 Conclusion: The Never-Ending Journey

Q1 Consider this puzzle: What is the square root of –1? We know that 1 × 1 = 1. We also
know that (–1) × (–1) = 1. There is no Real Number that, when multiplied by itself,
results in a negative number. Thus, √(–1) cannot exist on number line.

There is no real number whose square is −1. The reason is Brahmagupta's own sign rule.

if x > 0 then x × x > 0 (fortune × fortune = fortune)
if x < 0 then x × x > 0 (debt × debt = fortune)

if x = 0 then x × x = 0

Every real number falls into one of these three cases, so x² ≥ 0 for every real x. Since −1 < 0, the
equation x² = −1 has no solution anywhere on the number line.

Why it happens: the number line was built by filling gaps — fractions filled the gaps
between integers, irrationals filled the gaps the fractions left. But √(−1) is not sitting
in a gap; it is excluded by the arithmetic of signs itself. No amount of further filling
can produce it, so mathematicians had to leave the line altogether and add a new
dimension, writing i for a quantity with i² = −1.

Page 36 of 55

Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

Did you know? The pattern of this chapter repeats once more here. Every time an
operation could not be carried out — subtraction inside ℕ, division inside ℤ, square
roots inside ℚ — the number system was enlarged. Imaginary numbers are the next
enlargement, and they are what make alternating-current circuits and quantum
mechanics calculable.

End-of-Chapter Exercises — Page 64
The World of Numbers

Q1 Convert the following rational numbers in the form of a terminating decimal or
non-terminating and repeating decimal, whichever the case may be, by the process
of long division: (i) 3/50 (ii) 2/9

(i) 3/50. 50 = 2 × 5², only 2s and 5s → expect a terminating decimal

30 ÷ 50 = 0, remainder 30

300 ÷ 50 = 6, remainder 0 — stop

3/50 = 0.06, a terminating decimal with 2 places

(ii) 2/9. 9 = 3², a prime other than 2 or 5 → expect repetition

20 ÷ 9 = 2, remainder 2

20 ÷ 9 = 2, remainder 2 (the remainder returns at once)

2/9 = 0.2222… = 0.2

Why it happens: the predicted number of decimal places for 3/50 is max(1, 2) = 2,
and indeed the answer stops after two places. In 2/9 the remainder can only be 1…8,
and it lands back on 2 immediately, so the single digit 2 loops for ever.

Check it yourself: 3/50 = 6/100 = 0.06 without any division; and 0.2 = 2/9 by the
pure-repeating rule, since 9x = 2.

Page 37 of 55

Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

Q2 Prove that √5 is an irrational number.

Proof by contradiction, following the pattern of Section 3.5.1.

Step 1. Suppose √5 is rational.

Then √5 = p/q with p, q integers, q ≠ 0 and gcd(p, q) = 1.

Step 2. Square both sides: 5 = p²/q²

Step 3. Multiply by q²: 5q² = p²

Step 4. So 5 divides p². Since 5 is prime, 5 must divide p.

Write p = 5k with k an integer.

Step 5. Substitute: 5q² = (5k)² = 25k²

Step 6. Divide by 5: q² = 5k²

Step 7. So 5 divides q², and therefore 5 divides q.

Step 8. Now 5 divides both p and q, so they share the factor 5.

But Step 1 said gcd(p, q) = 1. Contradiction.

Every step after the assumption is forced, so the assumption itself must be false. Hence √5
cannot be written as p/q — it is irrational.

Why it happens: Step 4 is the heart of the argument and it needs 5 to be prime. If a
prime divides a product, it must divide one of the factors; here the product is p × p,
so the prime divides p. The same proof would collapse for √4, since 4 divides 6²
without dividing 6 — and rightly so, because √4 = 2 is rational.

Page 38 of 55

as e
Class 9 Maths Chapter 3 The World of Numbers
a g l AglaSem · NCERT Solutions

co m
m.
Tip: notice that Step 3 gives another quick argument. In p² every prime appears an

as e
com be equal. l
even number of times; in 5q² the prime 5 appears an odd number of times. The two
sides can.never a g
a s em
agl

co m
ag
Convert the following decimal numbers in the form of p/q. (i) 12.6 (ii) 0.0120 (iii)
.
Q3

e m
as
3.052 (iv) 1.235 (v) 0.23 (vi) 2.05 (vii) 2.125 (viii) 3.125 (ix) 2.1625

a g l

The first two are terminating; the rest use the shifting-and-subtracting method of Cases 2 and 3.
co m
em.
m = 63/5
(i) 12.6.c=o126/10 g l as
m a
ase
agl
s
(ii) 0.0120 = 120/10000 = 12/1000 = 3/250
m a
em
.co agl
a s
gl 2 repeating
(iii) 3.052 — 1 non-repeatingadigit,

m
x = 3.0525252…
. co
e m
as
10x = 30.52 and 1000x = 3052.52
m l
m .co
1000x − 10x = 3052.5252… − 30.5252… = 3022
a g
a s e990x
agl = 3022 → x = 3022/990 = 1511/495

se m
com
(iv) 1.235 — 1 non-repeating digit, 2.repeating g l a
em a
a s
agl
10x = 12.35, 1000x = 1235.35

990x = 1223 → x = 1223/990

co m
m .
m as e
.co a g l
m
(v) 0.23 — pure repeating, 2 digits

l a se
ag 100x = 23.23, so 99x = 23 → x = 23/99

.c
s e m
m a
em . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 39 of 55

Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

(vi) 2.05 — 1 non-repeating, 1 repeating

10x = 20.5, 100x = 205.5

90x = 185 → x = 185/90 = 37/18

(vii) 2.125 — 2 non-repeating, 1 repeating

100x = 212.5, 1000x = 2125.5

900x = 1913 → x = 1913/900

(viii) 3.125 — same shape as (vii)

1000x − 100x = 3125.5… − 312.5… = 2813

900x = 2813 → x = 2813/900

(ix) 2.1625 — pure repeating, 4 digits

10000x = 21625.1625

9999x = 21623 → x = 21623/9999

DECIMAL FRACTION DECIMAL FRACTION

12.6 63/5 2.05 37/18

0.0120 3/250 2.125 1913/900

3.052 1511/495 3.125 2813/900

1.235 1223/990 2.1625 21623/9999

0.23 23/99

Why it happens: multiplying by 10^m pushes the m non-repeating digits to the left
of the point; multiplying again by 10^n slides the decimal on by exactly one full
repeating cycle. The two numbers then have identical infinite tails, so the subtraction
wipes the tail out completely and leaves a whole number. That is the whole trick, and
it is why the denominator always comes out as a string of 9s followed by a string of
0s: 990 = 99 × 10, 900 = 9 × 100, 9999 = 9999 × 1.

Page 40 of 55

Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

Check it yourself: 1511 ÷ 495 = 3.0525252…, 1223 ÷ 990 = 1.2353535…, 1913 ÷ 900 =
2.12555… — each returns the decimal we started from.

Q4 Locate the following rational numbers on the number line. (i) 0.532 (ii) 1.15

(i) 0.532 is located by successive magnification: each decimal place cuts the interval into ten
equal parts.

0.532 lies between 0.5 and 0.6 (first place is 5)

then between 0.53 and 0.54 (second place is 3)

then it is the 2nd of the ten parts of 0.53 to 0.54

0 1

0.5 0.6

0.53 0.54

0.532

Zooming in three times: 0.5–0.6, then 0.53–0.54, then the point 0.532.

(ii) 1.15 is a repeating decimal, so first turn it into a fraction and then place it exactly.

x = 1.1555…

10x = 11.5, 100x = 115.5
90x = 104 → x = 104/90 = 52/45

So 1.15 = 52/45 = 1 7/45. To mark it exactly, divide the interval from 1 to 2 into 45 equal parts
and count 7 of them from 1. By magnification it lies between 1.15 and 1.16, just past one half of
that gap.

Page 41 of 55

Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

Why it happens: a terminating decimal is located by repeated tenfold zooming,
which is precisely reading off one decimal place at a time. A repeating decimal never
finishes, so zooming alone can only ever approximate it — but its p/q form gives an
exact position, since dividing a unit into q equal parts is an exact ruler-and-compass
operation.

Q5 Find 6 rational numbers between 3 and 4.

Write both endpoints over a denominator large enough to leave six whole numerators in
between. Since we want 6 numbers, use 6 + 1 = 7.

3 = 21/7 and 4 = 28/7

the integers strictly between 21 and 28 are 22, 23, 24, 25, 26, 27 — exactly six

So six rational numbers between 3 and 4 are

22/7, 23/7, 24/7, 25/7, 26/7, 27/7

Why it happens: with a common denominator d, every number k/d with 3d < k < 4d
lies between 3 and 4, and there are exactly d − 1 such k. Choosing d = 7 gives 6 of
them — just enough. Choosing d = 10 would give nine candidates (3.1 to 3.9) and
you could pick any six.

Tip: the answer is not unique. 3.1, 3.2, 3.3, 3.4, 3.5, 3.6 is equally correct, and so is
repeated averaging.

Page 42 of 55

Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

Q6 Find 5 rational numbers between 2/5 and 3/5.

the numerators 2 and 3 are consecutive — no room yet

multiply top and bottom of both by 6

2/5 = 12/30 and 3/5 = 18/30

integers strictly between 12 and 18: 13, 14, 15, 16, 17 — exactly five

Five rational numbers between 2/5 and 3/5:

13/30, 14/30 = 7/15, 15/30 = 1/2, 16/30 = 8/15, 17/30

Why it happens: the endpoints did not move — 12/30 and 18/30 are the same
numbers as 2/5 and 3/5 — but the gap between the numerators grew from 1 to 6,
which is what creates the five slots. To place n numbers this way the numerator gap
must be at least n + 1.

Check it yourself: 15/30 = 1/2, and indeed 0.4 < 0.5 < 0.6. ✓

Q7 Find 5 rational numbers between 1/6 and 2/5.

LCM(6, 5) = 30

1/6 = 5/30 and 2/5 = 12/30

integers strictly between 5 and 12: 6, 7, 8, 9, 10, 11 — six candidates, more than enough

Choosing five of them:

6/30 = 1/5, 7/30, 8/30 = 4/15, 9/30 = 3/10, 10/30 = 1/3

Check the order in decimals: 0.16 < 0.2 < 0.23 < 0.26 < 0.3 < 0.3 < 0.4 ✓

Page 43 of 55

as e
Class 9 Maths Chapter 3 The World of Numbers
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: here the LCM alone already opens a numerator gap of 7, which is

m l a se
more than the 6 needed for five numbers. Unlike Q6, no extra enlargement was
o g denominator
.cWhenever the gap is too small, simply multiply the common
a
m
necessary.
by a e
assuitable
a g l whole number.

o m
e
. c
mnumber x. ag
s
If x/3 + x/5 = 16/15, find the rational
a
Q8

agl

co m
em.
m l as
LCM(3, 5) = 15
. c o a g
s e
x/3 + mx/5 = 5x/15 + 3x/15 = 8x/15
a
agl 8x/15 = 16/15
multiply both sides by 15: 8x = 16
m a s
m.co agl
se
x=2

g l a
a
Why it happens: the two terms share the unknown x, so the distributive law lets us
co m
collect them: x(1/3 + 1/5) = x × 8/15. Since the denominators on the two sides are
m .
m as e
l
already the same, the equation reduces to a comparison of numerators alone.
.co a g
a s em
a gl Check it yourself: 2/3 + 2/5 = 10/15 + 6/15 = 16/15 ✓
se m
com g l a
m . a
ase
agl
Q9 Let a and b be two non-zero rational numbers such that a + 1/b = 0. Without
assigning any numerical values, determine whether ab is positive or negative.
Justify your answer.

co m
m .
m
as e
ab.cisonegative — in fact ab = −1 exactly.
a g l
a sem
agl c
a + 1/b = 0 (b ≠ 0, so 1/b is defined)
m .
m a s e
. co agl
a = −1/b
e m
multiply both sides by b
g l as
ab = −1 a

co m
m .
m ase
.co


a g l Page 44 of 55

Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

Since −1 < 0, the product ab is negative for every such pair a, b.

Why it happens: the condition says a is the negative of the reciprocal of b. A
number and its reciprocal always have the same sign — if b > 0 then 1/b > 0, and if b
< 0 then 1/b < 0. Putting a minus sign in front therefore guarantees that a and b
have opposite signs, and a product of a fortune and a debt is a debt. No numerical
value was needed anywhere.

Check it yourself: b = 4 gives a = −1/4 and ab = −1. b = −2/3 gives a = 3/2 and ab =
−1. Always −1.

Q10 A rational number has a terminating decimal expansion whose last non-zero digit
occurs in the 4th decimal place. Show that such a number can be written in the
form p/10⁴, where p is an integer not divisible by 10. Is it necessary that the
denominator of this rational number, when written in the lowest form, is divisible
by 2⁴ or 5⁴? Give reasons.

Part 1: the form p/10⁴. A decimal that stops at the 4th place has the shape

N = d₀.d₁d₂d₃d₄

multiply by 10⁴ to clear the point

N × 10⁴ = an integer, call it p

so N = p/10⁴

Here d₄, the 4th decimal digit, is the last non-zero digit, and d₄ is exactly the units digit of p.
Since d₄ ≠ 0, the integer p does not end in 0, so 10 does not divide p. ∎
Part 2: yes, it is necessary. Write N = p/10⁴ = p/(2⁴ · 5⁴) and reduce to lowest terms.
Because 10 ∤ p, the integer p cannot be divisible by 2 and 5 at the same time. So at least one of
these two cases holds:

2 ∤ p. Then no factor of 2 cancels, and the reduced denominator still contains the whole 2⁴ —
so it is divisible by 2⁴.
5 ∤ p. Then no factor of 5 cancels, and the reduced denominator still contains the whole 5⁴ —
so it is divisible by 5⁴.

Hence the lowest-form denominator is always divisible by 2⁴ or by 5⁴ (possibly by both).

Page 45 of 55

Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

NUMBER P/10⁴ LOWEST FORM DENOMINATOR

0.0005 5/10⁴ 1/2000 2⁴ × 5³ — divisible by 2⁴

0.0002 2/10⁴ 1/5000 2³ × 5⁴ — divisible by 5⁴

0.1233 1233/10⁴ 1233/10000 2⁴ × 5⁴ — divisible by both

Why it happens: cancellation can only remove a prime from the denominator if the
numerator supplies it. Since p is missing at least one of 2 and 5 entirely, the
corresponding fourth power survives untouched. This is the converse side of the
terminating-decimal rule: the number of decimal places equals the larger of the two
exponents in the lowest-form denominator, and here that larger exponent must be
4.

Q11 Without performing division, determine whether the decimal expansion of 18/125
is terminating or non-terminating. If it terminates, state the number of decimal
places.

125 = 5³, and 18 = 2 × 3²

gcd(18, 125) = 1, so 18/125 is already in lowest terms

the only prime in the denominator is 5

The denominator has the form 2⁰ × 5³, so the decimal terminates. The number of places is the
larger of the two exponents:

places = max(0, 3) = 3

Confirming without long division:

18/125 = (18 × 2³)/(5³ × 2³) = 144/1000 = 0.144 — 3 places ✓

Page 46 of 55

Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

Why it happens: to turn 5³ into a power of ten we must supply the missing 2³. That
fixes the denominator at 10³, so the decimal cannot need more than three places —
and it cannot need fewer either, because 144 does not end in 0.

Q12 A rational number in its lowest form has denominator 2³ × 5. How many decimal
places will its decimal expansion have? Explain your answer.

The expansion terminates after 3 decimal places.

q = 2³ × 5¹, so m = 3 and n = 1

k = max(m, n) = 3

multiply top and bottom by 5^(3−1) = 5² = 25

p/(2³ × 5) = 25p/(2³ × 5³) = 25p/1000

A whole number over 1000 stops after three places. For example, 3/40:

3/40 = (3 × 25)/1000 = 75/1000 = 0.075 — 3 places ✓

Why it happens: to make the denominator a power of 10 both primes must appear
to the same exponent, and the exponent has to be at least as large as each of them
— so it is the larger one, max(m, n). It is never more, because supplying extra factors
of 10 only appends zeros. And it is never fewer, because in lowest terms the
numerator shares no factor with q, so nothing can cancel and shorten the
expansion.

Check it yourself: 1/40 = 0.025 and 7/40 = 0.175 — every fraction with this
denominator uses exactly three places.

Page 47 of 55

Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

Q13* Let a = 7/12 and b = 5/6. Express both a and b in the form k₁/m and k₂/m where k₁,
k₂ and m are integers and k₂ – k₁ > 6. Using the same denominator m, write
exactly five distinct rational numbers lying between a and b keeping an integer
numerator. Explain why the condition k₂ – k₁ > n + 1 is necessary to find n such
rational numbers between the two rational numbers a and b using this method.

Step 1: a common denominator that is large enough.

a = 7/12, b = 5/6 = 10/12

here k₂ − k₁ = 10 − 7 = 3, which is not > 6

so enlarge: multiply numerator and denominator by 3

a = 21/36 and b = 30/36

now m = 36, k₁ = 21, k₂ = 30 and k₂ − k₁ = 9 > 6 ✓

Step 2: five rational numbers between them. The integers strictly between 21 and 30 are 22,
23, 24, 25, 26, 27, 28, 29 — eight candidates. Choosing five:

22/36 = 11/18, 23/36, 24/36 = 2/3, 25/36, 26/36 = 13/18

Check: 21/36 < 22/36 < 23/36 < 24/36 < 25/36 < 26/36 < 30/36 ✓
Step 3: why the size condition is needed.

the numbers this method produces are k/m with k₁ < k < k₂

the number of such integers k is k₂ − k₁ − 1

to obtain n of them we need k₂ − k₁ − 1 ≥ n

that is, k₂ − k₁ ≥ n + 1

The book's condition k₂ − k₁ > n + 1 is the same requirement with a margin to spare: it
guarantees k₂ − k₁ − 1 > n, so there are strictly more than n candidates and you can choose
freely. With n = 5 it demands k₂ − k₁ > 6, and our m = 36 gives 9, comfortably enough.

Page 48 of 55

as e
Class 9 Maths Chapter 3 The World of Numbers
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: once the denominator is fixed at m, the rationals available

m l a se
between a and b are locked to a grid of step 1/m. Widening the gap between the
o
endpoints.con that grid is the only way to create more grid points,
a gand multiplying m
by ats e m
g l multiplies the gap k₂ − k₁ by t while leaving a and b unmoved. So any number of
arationals can be squeezed between two distinct rationals — the density of ℚ again,
this time counted exactly.
co m
e m . ag
g l as
a
Tip: strictly, k₂ − k₁ = n + 1 already yields exactly n numbers with no choice at all. The
book's strict inequality is a safe sufficient condition rather than the sharpest one.

co m
se m.
o m l a
m .c rational numbers x, y, z satisfy x + y + z = 0 and xya+gyz + zx = 0. Show that all
Three
se the rational numbers x, y, z must be simultaneously zero.
Q14*

l a
ag
s
m a
Square the first condition and use the second.
m .co agl
l a se
ag
(x + y + z)² = x² + y² + z² + 2(xy + yz + zx)

substitute x + y + z = 0 and xy + yz + zx = 0
com
m .
as e
cy²om+ z² = 0 l
0² = x² + y² + z² + 2 × 0
. a g
m
ase
x² +

agl Now x², y², z² are each squares of rational numbers, so each is ≥ 0. Three non-negative numbers
se m
add to 0 only if every one of them is 0:
com g l a
m . a
ase
agl
if, say, x² > 0 then x² + y² + z² ≥ x² > 0, a contradiction

so x² = 0, y² = 0, z² = 0
co m
m .
e
hence x = y = z = 0 ∎
m l as
.co a g
m Why it happens: the identity (x + y + z)² = x² + y² + z² + 2(xy + yz + zx) converts the
l a se
ag two given conditions into one statement about a sum of squares, and squares of real
.c
s e m
m a
numbers can never be negative — the very fact used in the Think and Reflect about

m . co
√(−1). Note how essential that is: over the imaginary numbers the conclusion fails,
e agl
l as
since x = 1, y = ω, z = ω² (complex cube roots of 1) satisfy both conditions without
g
being zero. a

co m
m .
m ase
.co


a g l Page 49 of 55

Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

Check it yourself: try to find a non-zero example by hand. Take x = 1, y = −1; then z =
0 from the first condition, but xy + yz + zx = −1 ≠ 0. The second condition always
blocks you.

Q15* Show that the rational number (a + b)/2 lies between the rational numbers a and
b.

Take a ≠ b and, without loss of generality, let a < b (otherwise swap the names).

Above a:

(a + b)/2 − a = (a + b − 2a)/2 = (b − a)/2

since a < b, b − a > 0, so (b − a)/2 > 0

therefore (a + b)/2 > a

Below b:

b − (a + b)/2 = (2b − a − b)/2 = (b − a)/2

again (b − a)/2 > 0

therefore (a + b)/2 < b

Combining, a < (a + b)/2 < b. ∎
It is also a rational number: a + b is rational because ℚ is closed under addition, and dividing by
2 ≠ 0 keeps it rational.

Why it happens: the two gaps computed above are identical, both equal to (b − a)/2.
That is why the average is not merely somewhere in between but exactly at the
midpoint — the same distance from each end. Geometrically it is the middle of the
segment joining a and b on the number line.

Tip: the statement needs a ≠ b. If a = b then (a + b)/2 = a, and there is no number
strictly between a and b to find.

Page 50 of 55

Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

Q16 Find the lengths of the hypotenuses of all the right triangles in Fig. 3.14 which is
referred to as the square root spiral.

1

1

1

1

1 1

1
1

1
1
1

Fig. 3.14, page 66 — the square root spiral.

Fig. 3.14 is built by the rule of Section 3.5.2: each new right triangle takes the previous
hypotenuse as one leg and a fresh segment of length 1 as the other.

Page 51 of 55

Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

1st triangle: legs 1 and 1 → h₁² = 1 + 1 = 2 → h₁ = √2

2nd triangle: legs √2 and 1 → h₂² = 2 + 1 = 3 → h₂ = √3

3rd triangle: legs √3 and 1 → h₃² = 3 + 1 = 4 → h₃ = 2
in general hₙ = √(n + 1)

The figure prints ten right triangles, so their hypotenuses are:

TRIANGLE LEGS HYPOTENUSE² HYPOTENUSE

1 1, 1 2 √2 ≈ 1.414

2 √2, 1 3 √3 ≈ 1.732

3 √3, 1 4 √4 = 2

4 2, 1 5 √5 ≈ 2.236

5 √5, 1 6 √6 ≈ 2.449

6 √6, 1 7 √7 ≈ 2.646

7 √7, 1 8 √8 = 2√2 ≈ 2.828

8 √8, 1 9 √9 = 3

9 3, 1 10 √10 ≈ 3.162

10 √10, 1 11 √11 ≈ 3.317

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Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

2
√5 √3

√6
√2
√7
O 1

√8

3

√10
√11

Each blue segment has length 1; each grey spoke from O is the hypotenuse of the previous triangle,
giving √2, √3, 2, √5, … , √11.

So the ten hypotenuses are √2, √3, 2, √5, √6, √7, 2√2, 3, √10 and √11.

Why it happens: every step adds exactly 1 to the square of the length, so the
squares run through the whole numbers 2, 3, 4, 5, … in order. Three of these squares
are perfect — 4, 9 — giving the rational lengths 2 and 3; all the others give irrational
lengths. The spiral is therefore a picture of the fact that irrational and rational points
sit side by side on the number line, and it also shows how to construct √n for every
positive integer n with only a ruler and compass.

Did you know? The turn at each step is arctan(1/√n), which shrinks as n grows. After
ten triangles the spiral has swept about 259°, and it takes seventeen triangles to
complete a full turn.

Page 53 of 55

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Class 9 Maths Chapter 3 The World of Numbers
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a g l Page 54 of 55

Class 9 Maths Chapter 3 The World of Numbers AglaSem · NCERT Solutions

Quick revision

IDEA SYMBOL / TEST OR RULE EXAMPLE FROM THE
FORM CHAPTER

Natural numbers ℕ = {1, 2, 3, …} closed under + and ×, not under 3 − 5 is not a natural
− number

Integers ℤ = {…, −1, 0, 1, fortunes and debts; (−) × (−) = (−3) × (−4) = 12
…} (+)

Rational number p/q, p, q ∈ ℤ, q ≠ q ≠ 0; take p, q co-prime 12/30 = 2/5
0

Equality of rationals a/b = c/d true exactly when ad = bc 2/3 = 4/6 since 2·6 = 3·4

Absolute value |x| distance from 0; |a − b| = |−4 − 3| = 7
distance a to b

Density (a + b)/2 always rational, always between between 1 and 3/2 lies 5/4
a and b

Terminating decimal p/q in lowest terms q = 2^m · 5^n only 3/20 = 15/100 = 0.15

Repeating decimal p/q in lowest terms q has a prime other than 2, 5 5/11 = 0.4545… repeating

Number of decimal q = 2^m · 5^n larger of m and n 1/40 = 0.025, 3 places
places

Irrational number cannot be written decimal never ends, never √2, √3, π
p/q repeats

Proof by contradiction assume the derive an impossibility √2 = p/q forces p, q both
opposite even

Real numbers ℝ=ℚ∪ every point of the line the number line has no
irrationals gaps

Page 55 of 55