Well-designed AP Board Solutions Class 10 Maths Chapter 1 Real Numbers Exercise 1.2 offers step-by-step explanations to help students understand problem-solving strategies.

## Real Numbers Class 10 Exercise 1.2 Solutions – 10th Class Maths 1.2 Exercise Solutions

Question 1.

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

Solution:

Let us assume, to the contrary, that \(\sqrt{5}\) is a rational.

Then there exist positive integers a and b such that \(\sqrt{5}\) = \(\frac{a}{b}\)

Squaring on both sides,

\((\sqrt{5})^2\) = \(\left(\frac{a}{b}\right)^2\)

5 = \(\frac{\mathrm{a}^2}{\mathrm{~b}^2}\)

5b^{2} = a^{2} …..(1)

5 divides a^{2} i.e. 5/a^{2}

5 divides a i.e. 5/a

For some integer ‘c’

Let a = 5c

a^{2} = (5c)^{2}

a^{2} = 25c^{2}

5b^{2} = 25c^{2} [∵ a^{2} = 5b^{2}]

b^{2} = 5c^{2}

5 divides b^{2}

5 divides b i.e., 5/b ………. (2)

From (1) and (2), 5 is the common factor of a and b.

But this contradicts the fact that a and b have no common factor other than 1. This means that our supposition is wrong.

Hence, \(\sqrt{5}\) is an irrational number.

Question 2.

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

Solution:

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

That is, we can find coprimes a and b where b ≠ 0 such that

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

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

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

\(\sqrt{5}\) = \(\frac{a-3 b}{b}\)

Since, 2, 3, a and b are integers.

\(\frac{a-3 b}{2 b}\) is a rational and so \(\sqrt{5}\) is rational.

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

So, 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:

i) Let us assume to the contrary that \(\frac{1}{\sqrt{2}}\) is rational.

Then there exist positive integers a and b, such that \(\frac{1}{\sqrt{2}}\) = \(\frac{a}{b}\)

where b ≠ 0 and a, b are coprimes.

\(\frac{\sqrt{2}}{1}\) = \(\frac{\mathrm{b}}{\mathrm{a}}\)

i.e., \(\sqrt{2}\) = \(\frac{\mathrm{b}}{\mathrm{a}}\)

\(\sqrt{2}\) is a rational number and it is a contradiction.

This contradiction arised due to our supposition that \(\frac{1}{\sqrt{2}}\) is a rational number.

Hence \(\frac{1}{\sqrt{2}}\) is an irrational number.

ii) Let us assume to the contrary, that 7\(\sqrt{5}\) is rational.

That is, we can find coprimes a and b where b ≠ 0 such that

7\(\sqrt{5}\) = \(\frac{\mathrm{b}}{\mathrm{a}}\)

We get \(\sqrt{5}\) = \(\frac{\mathrm{a}}{7 \mathrm{~b}}\)

Since, a, b and 7 are integers.

\(\frac{\mathrm{a}}{7 \mathrm{~b}}\) 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) Let us suppose that 6 + \(\sqrt{2}\) is a rational number.

That is, we can find coprimes a and b where b ≠ 0 such that

6 + \(\sqrt{2}\) = \(\frac{a}{b}\)

∴ \(\sqrt{2}\) = \(\frac{a}{b}\) – 6 ⇒ \(\sqrt{2}\) = \(\frac{a-6 b}{b}\)

Since, a, b and 6 are integers.

\(\frac{a-6 b}{b}\) is rational and so, \(\sqrt{2}\) is rational.

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

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