Prove that 3 + 2√5 is irrational.

∴ 3 + 2√5 is irrational. ∎

Prove that 6 + √2 is irrational.

∴ 6 + √2 is irrational. ∎

NCERT Class 10 Maths Exercise 1.2 Solution – Real Numbers

Class 10MathsChapter 1: Real NumbersExercise 1.2CBSE

NCERT Class 10 Maths Chapter 1 Exercise 1.2 is all about proving that certain numbers — like sqrt(5), 3+2sqrt(5), 1/sqrt(2), 7sqrt(5), and 6+sqrt(2) — are irrational. Every question in this exercise uses the same powerful technique: the Contradiction Method (also called Proof by Contradiction), which you first learned when studying the irrationality of sqrt(2) in your textbook.

In this post, you will find clear, step-by-step solutions for all 3 questions (with sub-parts) of Exercise 1.2, written in a way that is easy to understand and reproduce in your CBSE board exam. By the end, you will know exactly how to set up a contradiction proof, where to use the Fundamental Theorem of Arithmetic, and how to write the final conclusion — all within the word limit expected in board answers.

Quick answers – Exercise 1.2

Question 1: $∴ \sqrt{5} is irrational. ■$
Question 2: $∴ 3 + 2 \sqrt{5} is irrational. ■$
Question 3 (i): $∴ \frac{1}{\sqrt{2}} is irrational. ■$
Question 3 (ii): $∴ 7 \sqrt{5} is irrational. ■$
Question 3 (iii): $∴ 6 + \sqrt{2} is irrational. ■$

Open the questions in the official book: NCERT Class 10 Maths Chapter 1 (PDF) or read it on the NCERT website.

Video Solution – Exercise 1.2

Watch the Exercise 1.2 solution on YouTube

Step-by-Step Solutions – Exercise 1.2

Tap a question to see its solution.

Question 1

Prove that

\sqrt{5}

is irrational.

Solution:

$Assume: \sqrt{5} is rational$
We use proof by contradiction. Let us assume that √5 is rational.
$\sqrt{5} = \frac{a}{b} (a, b are co-prime, b \neq 0)$
So √5 = a/b, where a and b are co-prime integers. This means they have no common factor other than 1.
$5 b^{2} = a^{2} (on squaring)$
Square both sides and multiply by b². This gives 5b² = a².
$5 ∣ a^{2} \Rightarrow 5 ∣ a$
So 5 divides a². By our theorem, 5 also divides a.
$a = 5 c \Rightarrow 5 b^{2} = 25 c^{2} \Rightarrow b^{2} = 5 c^{2}$
So we can write a = 5c. Squaring gives 25c². Cancelling 5 gives b² = 5c².
$5 ∣ b^{2} \Rightarrow 5 ∣ b$
Now 5 divides b². By the same theorem, 5 also divides b.
$5 is common to a and b! But they were co-prime \Rightarrow Contradiction$
See, 5 is now a common factor of both a and b. But we assumed they are co-prime. This is a contradiction.
$∴ \sqrt{5} is irrational. ■$
So our assumption was wrong. Therefore √5 is irrational. Proved!

Final answer

∴ \sqrt{5} is irrational. ■

Shortcut / Trick:

\begin{array}{l} Exam Tip: write it in 5 steps \\ Assume > a/b > square > theorem > contradiction \\ You get full marks \end{array}

Exam tip: write this in five steps: assume, a/b, square, theorem, and contradiction. This format gets full marks in the board exam.

Question 2

Prove that

3 + 2 \sqrt{5}

is irrational.

Solution:

$Think first! Do we have to square a/b again?$
Wait a moment. Do we need to write the whole proof again? Think about it first.
$Real life: legs 2, 4 \Rightarrow hypotenuse = 2 \sqrt{5}$
In a right triangle with sides 2 and 4, the hypotenuse is exactly 2√5. Let us see what happens when we add 3 to it.
$Assume 3 + 2 \sqrt{5} = \frac{a}{b} is rational$
The answer is no, you do not need to write the whole proof again. Just start like this. Assume that 3 + 2√5 is rational, which means it equals a/b.
$2 \sqrt{5} = \frac{a}{b} - 3 = \frac{a - 3 b}{b}$
Now move 3 to the other side. 2√5 = a/b – 3, which is (a – 3b)/b.
$\sqrt{5} = \frac{a - 3 b}{2 b}$
Divide by 2, so √5 = (a – 3b)/(2b).
$\frac{a - 3 b}{2 b} is rational \Rightarrow \sqrt{5} is rational$
Here a and b are integers, so the right side is a rational number. This means √5 is rational.
$But in Q1 we proved: \sqrt{5} irrational \Rightarrow Contradiction$
But we just proved in Q1 that √5 is irrational. This is a contradiction.
$∴ 3 + 2 \sqrt{5} is irrational. ■$
So 3 + 2√5 is irrational. Proved!

Final answer

∴ 3 + 2 \sqrt{5} is irrational. ■

Question 3

Question 3 (i)

Prove that

\frac{1}{\sqrt{2}}

is irrational.

Solution:

$\frac{1}{\sqrt{2}} = \sin 45^{\circ}$
This is the same value that you will see later in Trigonometry as sin 45°. That is why it is important to know that it is irrational.
$Assume \frac{1}{\sqrt{2}} = \frac{a}{b} is rational (a \neq 0)$
Assume that 1/√2 is rational, which means it equals a/b.
$\sqrt{2} = \frac{b}{a}$
Take the reciprocal of both sides. So √2 = b/a.
$\sqrt{2} rational? But \sqrt{2} is irrational \Rightarrow Contradiction$
This means √2 is rational. But we know that √2 is irrational. Contradiction!
$∴ \frac{1}{\sqrt{2}} is irrational. ■$
So 1/√2 is irrational.

Final answer

∴ \frac{1}{\sqrt{2}} is irrational. ■

Question 3 (ii)

Prove that

7 \sqrt{5}

is irrational.

Solution:

$Assume 7 \sqrt{5} = \frac{a}{b} is rational$
Assume that 7√5 is rational, which means it equals a/b.
$\sqrt{5} = \frac{a}{7 b}$
Divide by 7. So √5 = a/(7b).
$\frac{a}{7 b} rational \Rightarrow \sqrt{5} rational. But Q1 proved it irrational \Rightarrow Contradiction$
Here a and 7b are both integers, so the right side is rational. This means √5 is rational. But we proved in Q1 that it is irrational. Contradiction!
$∴ 7 \sqrt{5} is irrational. ■$
So 7√5 is irrational.

Final answer

∴ 7 \sqrt{5} is irrational. ■

Question 3 (iii)

Prove that

6 + \sqrt{2}

is irrational.

Solution:

$Assume 6 + \sqrt{2} = \frac{a}{b} is rational$
Assume that 6 + √2 is rational, which means it equals a/b.
$\sqrt{2} = \frac{a}{b} - 6 = \frac{a - 6 b}{b}$
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Move 6 to the other side. So √2 = (a – 6b)/b.
$Right side rational \Rightarrow \sqrt{2} rational. But \sqrt{2} irrational \Rightarrow Contradiction$
The right side is rational, so √2 is rational. But √2 is irrational. Contradiction!
$∴ 6 + \sqrt{2} is irrational. ■$
So 6 + √2 is irrational.

Final answer

∴ 6 + \sqrt{2} is irrational. ■

Shortcut / Trick:

\begin{array}{l} Super Trick (remember it in one line): \\ Rational + Irrational = Irrational \\ Rational x Irrational (non-zero) = Irrational \\ 1/Irrational = Irrational \end{array}

Here is the rule to remember. Rational + irrational is always irrational. A non-zero rational × irrational is also irrational. The reciprocal of an irrational number is also irrational. Use this rule to answer objective questions at once. In a proof, you only need to repeat it.

Key Takeaways – Exercise 1.2

Every question in Exercise 1.2 is solved using Proof by Contradiction — assume the number is rational, then show that leads to a contradiction.
The key step in each proof is using the fact that sqrt(5) (or sqrt(2)) is irrational — a result you must quote as a previously proved theorem.
If p/q (in lowest terms) satisfies an equation, then any common factor of p and q contradicts the assumption that p and q are coprime (HCF = 1).
Questions like Q2 and Q3 build on Q1 — first prove the simple surd is irrational, then use it to prove the composite expression is irrational.
This exercise carries direct 2–3 mark questions in CBSE board exams almost every year, so mastering the proof format is a must.

CBSE Board Exam Tips

Always start your proof with the standard opening line: ‘Let us assume, to the contrary, that [the number] is rational.’ — CBSE markers look for this exact setup.
When you write p/q, immediately state ‘where p and q are integers, q ≠ 0, and HCF(p, q) = 1’ — missing this line costs marks in board exams.
Quote the result ‘sqrt(5) is irrational’ (or ‘sqrt(2) is irrational’) as a known/proved fact; do not re-prove it inside Q2 or Q3 answers — it wastes time and space.
End every proof with a clear conclusion sentence such as ‘This contradicts our assumption. Therefore, [the number] is irrational.’ — the conclusion line is worth marks on its own.

Practice MCQs – Exercise 1.2

1. Which of the following is the correct first step when proving that sqrt(7) is irrational by contradiction?

AAssume sqrt(7) = p/q where p, q are integers, q != 0, and HCF(p, q) = 1
BAssume sqrt(7) is irrational and find its decimal expansion
CAssume 7 is a composite number and find its prime factors
DAssume sqrt(7) = p/q where p and q may have common factors

Show answer

Correct answer: A. Assume sqrt(7) = p/q where p, q are integers, q != 0, and HCF(p, q) = 1

The standard opening of a contradiction proof for irrationality is to assume the number equals p/q in lowest terms (HCF = 1). This is the assumption that is later contradicted to complete the proof.

2. In the proof that sqrt(3) is irrational, after writing sqrt(3) = p/q and squaring both sides, we get 3q^2 = p^2. What does this tell us about p?

A3 divides p
B3 divides q only
Cp divides 3
Dp and q are both odd

Show answer

Correct answer: A. 3 divides p

If 3q^2 = p^2, then 3 divides p^2. Since 3 is prime, by Euclid’s lemma, 3 must also divide p. This is the key step that eventually shows 3 divides both p and q, contradicting HCF(p,q) = 1.

3. If 5 + 3*sqrt(2) is assumed to be rational, which of the following expressions would also have to be rational?

Asqrt(2)
Bsqrt(5)
Csqrt(3)
Dsqrt(10)

Show answer

Correct answer: A. sqrt(2)

If 5 + 3*sqrt(2) = p/q (rational), then 3*sqrt(2) = p/q – 5, so sqrt(2) = (p – 5q)/(3q), which is a ratio of integers — making sqrt(2) rational. Since sqrt(2) is known to be irrational, this is a contradiction.

4. Which of the following numbers is irrational?

Asqrt(49)
Bsqrt(36)/sqrt(4)
C4 + sqrt(11)
Dsqrt(16) + sqrt(9)

Show answer

Correct answer: C. 4 + sqrt(11)

sqrt(49)=7, sqrt(36)/sqrt(4)=6/2=3, and sqrt(16)+sqrt(9)=4+3=7 are all rational. However, 4+sqrt(11) is irrational because sqrt(11) is irrational and adding a rational number to an irrational number always gives an irrational number.

5. In proving that 4*sqrt(3) is irrational, after assuming 4*sqrt(3) = p/q, we rearrange to get sqrt(3) = p/(4q). Why is this a contradiction?

ABecause p/(4q) is rational but sqrt(3) is irrational
BBecause p/(4q) is always greater than 1
CBecause 4q cannot be a denominator
DBecause p must equal 4 in this case

Show answer

Correct answer: A. Because p/(4q) is rational but sqrt(3) is irrational

p/(4q) is a ratio of integers (with non-zero denominator), so it is rational. But sqrt(3) is a known irrational number. A number cannot be both rational and irrational, so this is a contradiction.

6. Which of the following correctly describes an irrational number?

AA number whose decimal expansion is non-terminating and non-repeating
BA number that cannot be negative
CA number whose decimal expansion is non-terminating but repeating
DA number that is not a whole number

Show answer

Correct answer: A. A number whose decimal expansion is non-terminating and non-repeating

By definition, an irrational number has a decimal expansion that goes on forever without repeating any block of digits. Rational numbers, by contrast, have terminating or non-terminating repeating decimals.

7. If p^2 is divisible by 11, which of the following must be true?

Ap is divisible by 11
Bp is divisible by 121
Cp is divisible by 22
Dp is divisible by 11 only if p is even

Show answer

Correct answer: A. p is divisible by 11

Since 11 is a prime number, if 11 divides p^2 then by Euclid’s lemma (or the Fundamental Theorem of Arithmetic) 11 must also divide p. This property of primes is the backbone of all irrationality proofs in this exercise.

8. Which of the following is a rational number?

Asqrt(2) + sqrt(3)
Bsqrt(5) – sqrt(5)
C2*sqrt(7)
D1/sqrt(3)

Show answer

Correct answer: B. sqrt(5) – sqrt(5)

sqrt(5) – sqrt(5) = 0, which is a rational number (0 = 0/1). All the other options involve irrational surds that do not cancel, so they remain irrational.

9. To prove that 2 + 5*sqrt(3) is irrational, which known irrational result must you use as a supporting fact?

Asqrt(3) is irrational
Bsqrt(2) is irrational
Csqrt(5) is irrational
Dsqrt(6) is irrational

Show answer

Correct answer: A. sqrt(3) is irrational

After rearranging the assumption 2 + 5*sqrt(3) = p/q, you get sqrt(3) = (p – 2q)/(5q). To reach a contradiction, you need the fact that sqrt(3) is irrational. The irrationality of sqrt(2) or sqrt(5) is not directly relevant here.

10. How many questions (including all sub-parts) are there in NCERT Class 10 Maths Exercise 1.2?

Show answer

Correct answer: C. 5

Exercise 1.2 has Q1 (1 part), Q2 (1 part), and Q3 with sub-parts (i), (ii), and (iii) — making 5 proofs in total. Students should practise all 5 as any of them can appear in the board exam.

Frequently Asked Questions

How many questions are in NCERT Class 10 Maths Exercise 1.2?

Exercise 1.2 has 3 questions in total. Q1 and Q2 are single-part questions, while Q3 has three sub-parts — (i), (ii), and (iii) — making it 5 proofs to practise altogether. All questions are based on proving that given numbers are irrational.

Is Exercise 1.2 of Class 10 Maths important for the CBSE board exam?

Yes, Exercise 1.2 is very important for the CBSE Class 10 board exam. Proof-based questions from this exercise — especially proving that expressions like 3+2sqrt(5) or 7sqrt(5) are irrational — appear regularly as 2-mark or 3-mark questions. Practising the standard format of Proof by Contradiction is essential.

What method is used to prove numbers are irrational in Exercise 1.2?

All proofs in Exercise 1.2 use the Proof by Contradiction method. You assume the number is rational (can be written as p/q in lowest terms), perform algebraic steps, and arrive at a statement that contradicts a known fact — typically that sqrt(5) or sqrt(2) is irrational. This contradiction proves the original number must be irrational.

Why do we write HCF(p, q) = 1 at the beginning of every irrational proof?

Writing HCF(p, q) = 1 means p and q have no common factors other than 1, i.e., the fraction p/q is already in its simplest (lowest) form. This is the key assumption that gets contradicted later in the proof — when we show p and q actually share a common factor, it proves our initial assumption of rationality was wrong.

Can I use the proof of sqrt(5) being irrational inside the proof for Q2 (3 + 2sqrt(5))?

Yes, absolutely — and you should! In Q2, after algebraic rearrangement you get sqrt(5) = (p – 3q) / 2q, which would make sqrt(5) rational. Since sqrt(5) is already proved to be irrational (in Q1), this is a direct contradiction. You do not need to re-prove sqrt(5) is irrational; just cite it as a ‘previously proved result’ or ‘known fact’.

What is the difference between a rational number and an irrational number for Class 10?

A rational number can be expressed in the form p/q where p and q are integers and q ≠ 0; its decimal expansion is either terminating or non-terminating repeating. An irrational number cannot be expressed in this form — its decimal expansion is non-terminating and non-repeating. Examples of irrational numbers from Exercise 1.2 include sqrt(5), 3+2sqrt(5), and 7sqrt(5).

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NCERT Class 10 Maths Solutions – All Exercises

Chapter 1: Real Numbers – Exercise 1.1, Exercise 1.2

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Source: Questions from the NCERT Mathematics Textbook for Class 10, Chapter 1. Solutions written by IrfanEdu.

Video Solution – Exercise 1.2

Step-by-Step Solutions – Exercise 1.2

Question 1

Question 2

Question 3

Question 3 (i)

Question 3 (ii)

Question 3 (iii)

Related Posts

Key Takeaways – Exercise 1.2

CBSE Board Exam Tips

Practice MCQs – Exercise 1.2

Frequently Asked Questions

How many questions are in NCERT Class 10 Maths Exercise 1.2?

Is Exercise 1.2 of Class 10 Maths important for the CBSE board exam?

What method is used to prove numbers are irrational in Exercise 1.2?

Why do we write HCF(p, q) = 1 at the beginning of every irrational proof?

Can I use the proof of sqrt(5) being irrational inside the proof for Q2 (3 + 2sqrt(5))?

What is the difference between a rational number and an irrational number for Class 10?

NCERT Class 10 Maths Solutions – All Exercises

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