## Class 10 Math (NCERT) Chapter 1 Exercise 1.3 Solutions in (Hindi & English)

**Class 10 Maths Chapter 1 Real Numbers Exercise 1.3 Solutions** are given below in both Hindi and English medium. Exercise 1.2 has been solved by the method given by the book so that all of you can understand this solution easily and you get full marks in the board exam and all of you will be easy to prepare. Below in this page, first in English medium and then in Hindi medium, the exercises / questions are solved. Hope this solution will help you in your preparation and you will like it.

**Ex 1.3 Class 10 Math Solutions**

**Question 1. **Prove that √5 is irrational.

Solution: Let √5q = p => 5q²=p²

=> p²-Sq² ... (i)

Since 5 divides p², so it will divide p also.

Let p = 5r

Then p²-25r² [Squaring both sides]

=>5q² = 25r² [Form(i)]

=> q² = 5r²

Since 5 divides q², so it will divide q also. Thus, 5 is a common factor of both p and q.

This contradicts our assumption that √5 is rational.

Hence, √5 is irrational. Hence, proved.

**Question 2. Show that 3 + √5 is irrational.**

Solution: Let 3+2√5=p/q be a rational number, where p and q are co-prime and q≠0.

Then, 2√5=p/q-3=p-3q/q

=> √5=p-3q/2q

Since p-3q/2q is a rational number,

Therefore, √5 is a rational number. But, it is a contradiction.

Hence, 3+√5 is irrational. Hence, proved.

**Question 3. Prove that the following are irrational. (i) 1/√2 (ii) 7√5 (iii) 6+√2.**

Solution: (i) Let them assume that 1/√2 isrational.

∴ there exists co-prime integers a and b (b≠ 0) such that 1/√2=a/b => √2=b/a

Since a and b are integers, we get b/a is rational and so √2 is rational.

But this contradicts the fact that √2 is irrational.

This contradiction has arisen because of our incorrect assumption that 1/√2 is rational.

Hence, we conclude that 1/√2 is irrational.

(ii) Let us assume that 7√5 is rational.

∴ there exists co-prime integers a and b (b≠ 0) Such that 7√5=a/b => √5=a/7b

Since a and b are integers, we get a/7b is rational and so √5 is rational.

But this contradicts the fact that √5 is irrational.

This contradiction has arisen because of our incorrect assumption that 7√5 is rational.

Hence, we conclude that 7√5 is rational.

(iii) Let us assume that 6+√2 is rational.

∴ there exists co-prime integers a and b (b≠0) such that 6+√2=a/b => √2=a/b-6

Since a and b are integers, we get a/b-6 is rational and so √2 is rational.

But this contradicts the fact that √2 is irrational.

This contradicts has arisen because of our incorrect assumption that 6+√2 is rational.

Hence, we conclude that 6+√2 is irrational.