NCERT solution class 10 chapter 1 Real numbers

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 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

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

1 thought on “NCERT solution class 10 chapter 1 Real numbers

Leave a Reply

Your email address will not be published. Required fields are marked *