These NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.3 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.3

Question 1.

Prove that \(\sqrt{5}\) is irrational.

Solution:

Let us assume, to the contrary, that \(\sqrt{5}\) is irrational that is we can find integers a and b (b ≠ 0) such that \(\sqrt{5}\) = \(\frac { a }{ b }\). suppose a and b have a common factor other than 1 then we can divide by the common factor and assume that a and b are coprime.

So b\(\sqrt{5}\) = a

Squaring on both sides, and rearranging we get 5b² = a³.

Thus for a² is divisible by 5, it follows that a is also divisible by 5.

So, we can write a 5c for some integer c.

Substituting for a, we get

5b² = 25c²

b² = 5c²

This means that b2 is divisible by 5 and so b is also divisible by 5.

Therefore, a and b have at least 5 as common factor.

But this contradicts the fact that a and b are coprime.

This contradiction has arisen because of our incorrect assumption that \(\sqrt{5}\) is irrational.

So we conclude that \(\sqrt{5}\) is irrational.

Question 2.

Prove that 3 + 2A\(\sqrt{5}\) is irrational.

Solution:

Let us assume, to contrary, that 3 + 2\(\sqrt{5}\) is rational.

That is, we can find coprime a and b (b ≠ 0)

such that 3 + 2\(\sqrt{5}\) = \(\frac { a }{ b }\)

Therefore 3 – \(\frac { a }{ b }\) = – 2\(\sqrt{5}\)

Rearranging this equation we get 2\(\sqrt{5}\) = \(\frac { a }{ b }\) – 3 = \(\frac { a – 3b }{ b }\)

Since a and b are integers, \(\frac { a }{ b }\) – 3 we get is

rational and so 2\(\sqrt{5}\) is rational and so \(\sqrt{5}\) is rational.

But this contradicts the fact \(\sqrt{5}\) is irrational.

This contadiction has arise because of our incorrect assumption 3 + 2\(\sqrt{5}\) is rational.

Thus, we conclude that 3 + 2\(\sqrt{5}\) is irrational.

Question 3.

Prove that the following are irrationals:

(i) \(\frac{1}{\sqrt{2}}\)

(ii) 7\(\sqrt{5}\)

(iii) 6 + \(\sqrt{2}\)

Solution:

Let us assume to the contrary, that \(\frac{1}{\sqrt{2}}\) is rational that is, we can find coprime a and b(b ≠ 0) such that = \(\frac{1}{\sqrt{2}}\) = \(\frac { a }{ b }\)

Since a and b are integers so \(\frac { a }{ b }\) is rational and so \(\sqrt{2}\) is rational.

But this contradicts the fact that \(\sqrt{2}\) is irrational.

Thus, we conclude that \(\frac{1}{\sqrt{2}}\) is irrational

(ii) 7\(\sqrt{5}\)

Let us assume, to the contrary that 7\(\sqrt{5}\) is rational that is, we can find coprime a and b (≠ 0)

such that 7\(\sqrt{5}\) = \(\frac { a }{ b }\) Rearranging, we get \(\sqrt{5}\) = \(\frac { a }{ b }\)

Since 7, a and b are integers, \(\frac { a }{ 7b }\) is rational and so \(\sqrt{5}\) is rational

But this contradicts the fact that \(\sqrt{5}\) is irrational. So, we conclude that 7\(\sqrt{5}\) is irrational.

(iii) 6 + \(\sqrt{2}\)

Let us assume, to the contrary, that 6 + \(\sqrt{2}\) is irrational.

That is, we can find co prime a and b (* 0) such that 6 + \(\sqrt{2}\) = 7 b

Rearranging, we get \(\sqrt{2}\) = \(\frac { a-6b }{ b }\)

Since a, b and 6 are integers, so \(\frac { a-6b }{ b }\) is rational and so is rational.

But this contradicts the fact that \(\sqrt{2}\) is

irrational. So we conclude that 6 + \(\sqrt{2}\) is irrational.