NCERT Solutions for Class 9th Maths Chapter 3 .6.3 Irrational Decimals: Chaos and Infinity — Exercise Set 3.5

Book page 613 Updated on2026-09-08

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.

Answer

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)

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?

Answer

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:

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.

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.

Answer

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

(vi) 21 digits after the point, so
= 23560185612239874790120/10²¹
divide top and bottom by 40
= 589004640305996869753/25000000000000000000

Why it happens: (ii) is irrational by the parity argument of Section 3.5.1 — √12 = 2√3, and √3 is irrational, so √12 cannot be a ratio of integers. (v) is the interesting one: the digits do follow a pattern, but the gaps of zeros grow 1, 2, 3, 4, … and never settle into a block of fixed length that repeats for ever. A rational number must repeat a block of constant length, so (v) is irrational. A visible pattern is not the same thing as a repeating block.

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

Q4.

The number 0.9 (which means 0.99999 …) is a rational number. Using algebra (let x = 0.9, multiply by 10, and subtract), explain why 0.9 is exactly equal to 1.

Answer

let x = 0.9 = 0.9999…
one digit repeats, so multiply by 10¹ = 10
10x = 9.9999… = 9.9
subtract: 10x − x = 9.9999… − 0.9999…
9x = 9
x = 1

So 0.9 and 1 are two decimal names for the same number.

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.

Answer

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 ✓

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.

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