# AP 10th Class Maths 1st Chapter Real Numbers Exercise 1.2 Solutions

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 = $$\frac{\mathrm{a}^2}{\mathrm{~b}^2}$$
5b2 = a2 …..(1)
5 divides a2 i.e. 5/a2
5 divides a i.e. 5/a
For some integer ‘c’
Let a = 5c
a2 = (5c)2
a2 = 25c2
5b2 = 25c2 [∵ a2 = 5b2]
b2 = 5c2
5 divides b2
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.