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

NCERT Solutions for Class 10 Maths 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 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.

NCERT Solutions for Class 10 Maths 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:
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.

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

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