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.