NCERT Solutions for Class 9 Maths Chapter 3 Exercise 3.5 | 2026-27

Think and Reflect

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

Answer:

No, √2 cannot be written as a rational number in the form p/q, where p and q are integers and q ≠ 0.

Think and Reflect

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

Answer:

Assume that √3 is a rational number.

√3 = p/q

where p and q are coprime integers and q ≠ 0.

Squaring both sides:

3 = p²/q²

p² = 3q²

This means p² is divisible by 3, so p is also divisible by 3.

Let:

p = 3k

Substituting:

(3k)² = 3q²

9k² = 3q²

q² = 3k²

So, q² is also divisible by 3, which means q is divisible by 3.

Therefore, both p and q are divisible by 3, which contradicts the fact that p and q are coprime.

Hence, our assumption is wrong.

Therefore, √3 is an irrational number.

The same method works for √5, √7, and √10 because their prime factors create the same contradiction. Hence, √5, √7, and √10 are also irrational numbers.

Think and Reflect 

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

Solution:

Do it yourself.

Think and Reflect 

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. 

Solution:

For √3:
Start with the construction of √2. From the endpoint of √2, draw a perpendicular of 1 unit and join the new point to O.

By the Baudhayana-Pythagoras theorem:

(√2)² + 1² = 2 + 1 = 3

Therefore, the new hypotenuse is √3. Using this length as the radius, locate √3 on the number line.

For √5:
First, construct √4 = 2 on the number line. At point 2, draw a perpendicular of 1 unit and join the new point to O.

Then:

2² + 1² = 4 + 1 = 5

Therefore, the new hypotenuse is √5. Using this length as the radius, locate √5 on the number line.

Generalisation for √n:

Take the previously constructed length √(n − 1). Draw a perpendicular of 1 unit and join the new point to O.

Then:

(√(n − 1))² + 1² = (n − 1) + 1 = n

Thus, the new hypotenuse gives √n. Repeating this process, we can construct √2, √3, √4, √5, … and hence √n, where n is a positive integer.

Think and Reflect 

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? 

Solution:

Here,

10/3 = 3.3333... = 3.3̅

and

11/12 = 0.916666... = 0.916̅

We observe that in 10/3, the digit 3 repeats continuously after the decimal point. In 11/12, the digit 6 repeats continuously after the initial digits 91.

Hence, both decimal expansions are non-terminating but repeating.

Think and Reflect

The decimal expansion of p/q will be terminating precisely when the prime factors of q are only 2, only 5, or both 2 and 5. Can you explain why? 

Solution:

A rational number p/q has a terminating decimal expansion when the denominator q has no prime factors other than 2 and 5.

This is because our decimal system is based on 10, and

10 = 2 × 5.

For example,

1/2 = 0.5

1/5 = 0.2

1/20 = 0.05

All these decimals terminate because the denominators contain only the prime factors 2 and 5.

If q has any other prime factor, such as 3 or 7, the decimal expansion will be non-terminating and repeating.

Hence, the decimal expansion of p/q terminates precisely when the prime factors of q are only 2, only 5, or both 2 and 5.

Exercise Set 3.5 

1. 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.

Solution:

For 7/20:

20 = 2² × 5

Only 2 and 5 appear in the denominator, so the decimal terminates.

Now check by long division:

Therefore,

7/20 = 0.35

For 4/15:

15 = 3 × 5

Since 3 appears in the denominator, the decimal will be repeating.

Now check by long division:

Therefore,

4/15 = 0.266...

For 13/250:

250 = 2 × 5³

Only 2 and 5 appear in the denominator, so the decimal terminates.

Now check by long division:

Therefore,

13/250 = 0.052

Hence, 7/20 and 13/250 have terminating decimals, while 4/15 has a non-terminating repeating decimal.

2. 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? 

Solution:

Long division for 1/13:

So, the repeating block is: 0.07692 

repeating digits are the product of 2, 3, 4, 5, … to the repeating digit of 1/13.

3. 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.

Solution:

Classify the following numbers as rational or irrational:

(i) √81

√81 = 9

Since 9 is an integer, √81 is rational.

9 = 9/1

Therefore, √81 is rational.

(ii) √12

√12 = 2√3

Since √3 is irrational, √12 is irrational.

Therefore, √12 is irrational.

(iii) 0.33333...

The digit 3 repeats continuously.

Therefore, it is a non-terminating repeating decimal and is rational.

0.33333... = 1/3

Therefore, 0.33333... is rational.

(iv) 0.123451234512345...

The block 12345 repeats continuously.

Therefore, it is a non-terminating repeating decimal and is rational.

Let x = 0.123451234512345...

100000x = 12345.1234512345...

Subtracting,

100000x − x = 12345

99999x = 12345

x = 12345/99999

Therefore, 0.123451234512345... is rational.

(v) 1.01001000100001 ...

The digits do not repeat as a single fixed block. The number of zeros between the 1s keeps increasing.

Therefore, it is a non-terminating, non-repeating decimal.

Hence, 1.01001000100001... is irrational.

(vi) 23.560185612239874790120 

This is a terminating decimal, so it is rational.

As a fraction:

= 23560185612239874790120/10²¹

This may also be simplified to:

589004640305996869753/250000000000000000000

4. 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.

Solution:

Let x = 0.9999…

Multiply both sides by 10:

10x = 9.9999…

Now subtract the original equation from this:

10x – x = 9.9999…-0.9999…

9x = 9

Divide both sides by 9: x = 1

But we assumed x = 0.9999…,

So 0.9999… = 1

Hence, 0.9 (i.e., 0.9999…) is exactly equal to 1. \

5. 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.

Solution:

A cyclic number is one where its repeating digits rotate when multiplied by numbers.

We already know:

1/7 = 0.142857̅

The block 142857 is cyclic because multiplying it by 2, 3, 4, 5, or 6 gives rotations of the same digits.

Now, to find more such numbers, we look for values of n where:

The decimal expansion of 1/n is repeating.

The repeating block has special rotation properties.

These usually occur when:

n is a prime number.

The length of the repeating cycle is n − 1 (called a full reptend prime).

Examples:

1/17 = 0.0588235294117647̅

The repeating block has 16 digits and shows cyclic-like properties.

1/19 = 0.052631578947368421̅

A number n (more precisely, a prime p) has this cyclic property when the decimal expansion of 1/p repeats with the maximum possible length, which is p − 1. Such primes are called full reptend primes. These primes generate repeating decimals where the digit block cycles under multiplication, forming cyclic numbers.

Key Benefits of Vedantu’s NCERT Class 9 Maths Chapter 3 Exercise 3.5 Solutions

• Step-by-step solutions: Helps students understand how to solve each question systematically.

• Clear explanation of rational and irrational numbers: Makes concepts such as terminating and repeating decimals easier to understand.

• Easy understanding of decimal expansions: Helps students identify and classify different types of decimal representations.

• Supports construction-based questions: Explains the steps involved in constructing required line segments using a ruler and compass.

• Useful for exam preparation: Helps students revise important concepts and write answers with the correct steps.

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