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

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.

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