NCERT solution class 10 chapter 1 Real numbers exercise 1.3

Exercise 1.3

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Q - 1, Ex 1.3 - Real Numbers - Chapter 1 - Maths Class 10th - NCERT

Question 1
Prove that $\sqrt{5}$ is irrational .
Sol :
Let $\sqrt{5}$ is a rational number.
Therefore, we can find two integers a , b $(b\not=0)$ such that $\sqrt{5}=\dfrac{a}{b}$
Let a and b have a common factor other than 1 . Then we can divide them by the common factor and assume that a and b are co-prime .
$a=\sqrt{5}b$
$a^2=5b^2$ .....eq-(1)
Therefore, a² is divisible by 5 and it can be said that a is divisible by 5.
Let a = 5k, where k is an integer and putting this value in eq-(1) , we get
$\begin{align}(5k)^2&=5b^2\\5k^2&=b^2\end{align}$
This means that b² is divisible by 5 and hence, b is divisible by 5.
And also this implies that a and b have 5 as a common factor .
And this is a contradiction to the fact that a and b are co-prime .
Hence, $\sqrt{5}$ cannot be expressed as $\dfrac{p}{q}$ or it can be said that $\sqrt{5}$ is irrational .

Q - 2, Ex 1.3 - Real Numbers - Chapter 1 - Maths Class 10th - NCERT

Question 2
Prove that $3+2\sqrt{5}$ is irrational
Sol :
Let $3+2\sqrt{5}$ is rational .
Therefore, we can find two integers a , b $(b\not=0)$ such that $3+2\sqrt{5}=\dfrac{a}{b}$
$2\sqrt{5}=\dfrac{a}{b}-3$
$2\sqrt{5}=\dfrac{a}{b}-3$
$\sqrt{5}=\dfrac{1}{2}\bigg(\dfrac{a}{b}-3\bigg)$
Since a and b are integers,$\dfrac{1}{2}\bigg(\dfrac{a}{b}-3\bigg)$ will also be rational and therefore $\sqrt{5}$ is rational.
This contradicts the fact that $\sqrt{5}$ is irrational. Hence, our assumption that $3+2\sqrt{5}$ is rational is false . Therefore , $3+2\sqrt{5}$ is irrational

Q - 3, Ex 1.3 - Real Numbers - Chapter 1 - Maths Class 10th - NCERT

Question 3
Prove that the following are irrationals:
(i) $\dfrac{1}{\sqrt{2}}$
Sol :
Let $\dfrac{1}{\sqrt{2}}$ is rational .
Therefore, we can find two integers a , b $(b\not=0)$ such that $\dfrac{1}{\sqrt{2}}=\dfrac{a}{b}$
$\sqrt{2}=\dfrac{b}{a}$
$\dfrac{b}{a}$ is rational as a and b are integers.
$\sqrt{2}$ will also be rational and this contradicts the fact that $\sqrt{2}$ is irrational.
Hence, our assumption that $\dfrac{1}{\sqrt{2}}$ is rational is false . Therefore , $\dfrac{1}{\sqrt{2}}$ is irrational

(ii) $7\sqrt{5}$
Sol :
Let $7\sqrt{5}$ is rational .
Therefore, we can find two integers a , b $(b\not=0)$ such that $7\sqrt{5}=\dfrac{a}{b}$
$\sqrt{5}=\dfrac{a}{7b}$
$\dfrac{a}{7b}$ is rational as a and b are integers.
$\sqrt{5}$ will also be rational and this contradicts the fact that $\sqrt{5}$ is irrational.
Hence, our assumption that $7\sqrt{5}$ is rational is false . Therefore , $7\sqrt{5}$ is irrational

(iii) $6+\sqrt{2}$
Sol :
Let $6+\sqrt{2}$ is rational .
Therefore, we can find two integers a , b $(b\not=0)$ such that $6+\sqrt{2}$
$\sqrt{2}=\dfrac{a}{b}-6$
$\dfrac{a}{b}-6$ is rational as a and b are integers.
$\sqrt{2}$ will also be rational and this contradicts the fact that $\sqrt{2}$ is irrational.
Hence, our assumption that $6+\sqrt{2}$ is rational is false . Therefore , $6+\sqrt{2}$ is irrational