MP Board Class 10th Maths Solutions Chapter 1 Real Numbers Ex 1.3

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 1 Real Numbers Ex 1.3 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 1 Real Numbers Ex 1.3

Question 1.
Prove that \(\sqrt{5}\) is irrational.
Solution:
Let us assume, to the contrary, that \(\sqrt{5}\) is rational.
∴ \(\sqrt{5}=\frac{a}{b}\)
∴ b × \(\sqrt{5}\) = a
By Squaring on both sides,
5b2 = a2 …………. (i)
∴ 5 divides a2.
5 divides a.
∴ We can write a = 5c.
Substituting the value of ‘a’ in eqn. (i),
5b2 = (5c)2 = 25c2
b2 = 5c2
It means 5 divides b2.
∴ 5 divides b.
∴ ‘a’ and ‘b’ have at least 5 as a common factor.
But this contradicts the fact that a’ and ‘b’ are prime numbers.
∴ \(\sqrt{5}\) is an irrational number.

MP Board Class 10th Maths Solutions Chapter 1 Real Numbers Ex 1.3

Question 2.
Prove that 3 +2\(\sqrt{5}\) is irrational.
Solution:
Let 3 + 2\(\sqrt{5}\) is rational.
MP Board Class 10th Maths Solutions Chapter 1 Real Numbers Ex 1.3 2
⇒ From (1), \(\sqrt{5}\) is rational
But this contradicts the fact that \(\sqrt{5}\) is irrational.
∴ Our supposition is wrong.
Hence, 3 + 2\(\sqrt{5}\) is irrational.

MP Board Class 10th Maths Solutions Chapter 1 Real Numbers Ex 1.3

Question 3.
Prove that the following are irrationals.
(i) \(\frac{1}{\sqrt{2}}\)
(ii) 7\(\sqrt{5}\)
(iii) 6 + \(\sqrt{2}\)
Solution:
(i) We have
MP Board Class 10th Maths Solutions Chapter 1 Real Numbers Ex 1.3 4
From (1), \(\sqrt{2}\) is rational number which contradicts the fact that \(\sqrt{2}\) is irrational.
∴ Our assumption is wrong.
Thus, \(\frac{1}{\sqrt{2}}\) is irrational.

(ii) Let \(7 \sqrt{5}\) is rational.
∴ We can find two co-prime integers a and b such that \(7 \sqrt{5}=\frac{a}{b}\), where b ≠ 0
MP Board Class 10th Maths Solutions Chapter 1 Real Numbers Ex 1.3 5
This contradicts the fact that \(\sqrt{5}\) is irrational.
∴ Out assumption is wrong.
Thus, we conclude that 7\(\sqrt{5}\) is irrational.

(iii) Let 6 + \(\sqrt{2}\) is rational.
∴ We can find two co-prime integers a and b
MP Board Class 10th Maths Solutions Chapter 1 Real Numbers Ex 1.3 6
= Rational [ ∵ a and b are integers]
From (1), \(\sqrt{2}\) is a rational number,
which contradicts the fact that \(\sqrt{2}\) is an irrational number.
∴ Our supposition is wrong.
⇒ 6 + \(\sqrt{2}\) is an irrational number.