---
title: "NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.2"
url: https://www.swavid.com/maths/class/10/chapter/real-numbers/ncert-solutions/exercise-1-2
dateModified: 2026-10-07T14:41:11+00:00
---

# NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.2

Chapter 1: Real Numbers. Every question from Exercise 1.2, with full working and the final answer.

Free PDF (7 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-10/swavid-ncert-solutions-class-10-maths-chapter-1-real-numbers-a484766ea2.pdf

## EXERCISE 1.2

### Question 1

*3 marks · Proof*

Prove that $\sqrt{5}$ is irrational.

**Solution**

1. Let us assume, to the contrary, that $\sqrt{5}$ is rational.
2. That is, we can find integers $a$ and $b$ ($b \neq 0$) such that $\sqrt{5} = \frac{a}{b}$.
3. Suppose $a$ and $b$ have a common factor other than 1, then we divide by the common factor, and assume that $a$ and $b$ are coprime.
4. So, $b\sqrt{5} = a$.
5. Squaring on both sides, and rearranging, we get $5b^2 = a^2$.
6. Therefore, $a^2$ is divisible by 5, and by Theorem 1.2, it follows that $a$ is also divisible by 5.
7. So, we can write $a = 5c$ for some integer $c$.
8. Substituting for $a$, we get $5b^2 = 25c^2$, that is, $b^2 = 5c^2$.
9. This means that $b^2$ is divisible by 5, and so $b$ is also divisible by 5 (using Theorem 1.2).
10. Therefore, $a$ and $b$ have at least 5 as a common factor.
11. But this contradicts the fact that $a$ and $b$ are coprime.
12. This contradiction has arisen because of our incorrect assumption that $\sqrt{5}$ is rational.
13. So, we conclude that $\sqrt{5}$ is irrational. Hence proved.

**Answer:** $\sqrt{5}$ is irrational.

> Common mistake: Forgetting to state that $a$ and $b$ are coprime after cancelling common factors.

### Question 2

*3 marks · Proof*

Prove that $3 + 2\sqrt{5}$ is irrational.

**Solution**

1. Let us assume, to the contrary, that $3 + 2\sqrt{5}$ is rational.
2. That is, we can find coprime integers $a$ and $b$ ($b \neq 0$) such that $3 + 2\sqrt{5} = \frac{a}{b}$.
3. Rearranging this equation, we get $2\sqrt{5} = \frac{a}{b} - 3 = \frac{a - 3b}{b}$.
4. Therefore, $\sqrt{5} = \frac{a - 3b}{2b}$.
5. Since $3$, $a$, and $b$ are integers, $\frac{a - 3b}{2b}$ is rational, and so $\sqrt{5}$ is rational.
6. But this contradicts the fact that $\sqrt{5}$ is irrational.
7. This contradiction has arisen because of our incorrect assumption that $3 + 2\sqrt{5}$ is rational.
8. So, we conclude that $3 + 2\sqrt{5}$ is irrational. Hence proved.

**Answer:** $3 + 2\sqrt{5}$ is irrational.

> Common mistake: Attempting to prove $\sqrt{5}$ is irrational all over again instead of using the known result.

### Question 3

*3 marks · Proof*

Prove that the following are irrationals :
(i) $\frac{1}{\sqrt{2}}$
(ii) $7\sqrt{5}$
(iii) $6 + \sqrt{2}$

**Part (i)**

1. Let us assume, to the contrary, that $\frac{1}{\sqrt{2}}$ is rational.
2. That is, we can find coprime integers $a$ and $b$ ($b \neq 0$) such that $\frac{1}{\sqrt{2}} = \frac{a}{b}$.
3. Taking the reciprocal, we get $\sqrt{2} = \frac{b}{a}$.
4. Since $a$ and $b$ are integers, $\frac{b}{a}$ is rational, which implies that $\sqrt{2}$ is rational.
5. This contradicts the fact that $\sqrt{2}$ is irrational.
6. So, we conclude that $\frac{1}{\sqrt{2}}$ is irrational. Hence proved.

Answer (i): $\frac{1}{\sqrt{2}}$ is irrational.

**Part (ii)**

1. Let us assume, to the contrary, that $7\sqrt{5}$ is rational.
2. That is, we can find coprime integers $a$ and $b$ ($b \neq 0$) such that $7\sqrt{5} = \frac{a}{b}$.
3. Rearranging, we get $\sqrt{5} = \frac{a}{7b}$.
4. Since $7$, $a$, and $b$ are integers, $\frac{a}{7b}$ is rational, and so $\sqrt{5}$ is rational.
5. This contradicts the fact that $\sqrt{5}$ is irrational.
6. So, we conclude that $7\sqrt{5}$ is irrational. Hence proved.

Answer (ii): $7\sqrt{5}$ is irrational.

**Part (iii)**

1. Let us assume, to the contrary, that $6 + \sqrt{2}$ is rational.
2. That is, we can find coprime integers $a$ and $b$ ($b \neq 0$) such that $6 + \sqrt{2} = \frac{a}{b}$.
3. Rearranging, we get $\sqrt{2} = \frac{a}{b} - 6 = \frac{a - 6b}{b}$.
4. Since $6$, $a$, and $b$ are integers, $\frac{a - 6b}{b}$ is rational, and so $\sqrt{2}$ is rational.
5. This contradicts the fact that $\sqrt{2}$ is irrational.
6. So, we conclude that $6 + \sqrt{2}$ is irrational. Hence proved.

Answer (iii): $6 + \sqrt{2}$ is irrational.

**Answer:** All given numbers are irrational.

> Common mistake: Rationalising the denominator for part (i) incorrectly or missing the contradiction step.

## Related pages

- [All Chapter 1 NCERT solutions](https://www.swavid.com/maths/class/10/chapter/real-numbers/ncert-solutions)
- [Exercise 1.1](https://www.swavid.com/maths/class/10/chapter/real-numbers/ncert-solutions/exercise-1-1)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student