Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4

Exercise 1.3

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

(1) $\begin{align*}(5k)^2&=5b^2\\5k^2&=b^2\end{align*}$

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 .

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 $3+2\sqrt{5}=\dfrac{a}{b}$

$2\sqrt{5}=\dfrac{a}{b}-3$

$\sqrt{5}=\dfrac{1}{2}\bigg(\dfrac{a}{b}-3\bigg)$

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

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

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NCERT solution class 10 chapter 1 Real numbers

Exercise 1.3

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