Math is tough but solving them is impossible said no one ever!! Make use of these NCERT Solutions For Class 10 Maths for finding solutions and also note that NCERT Solutions for Class 10 Maths PDF are available online for extra reference. This chapter includes NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.3. Practice more so you score more!!
NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 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.
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.
We hope you found these NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.3 helped you with finding the solutions easily.
Make sure to practice the remaining parts of Class 10 Maths NCERT Solutions English Medium Chapter 1.