NCERT Solutions For Class 10 Maths – Chapter 1 Exercise 1.3 Real Number

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•  Exercise 1.2

CHAPTER 1 – REAL NUMBER – EXERCISE 1.3

Q1. Prove that Square Root  (5)  is irrational .

Solution:-   Let  root( 5) is a rational number, 

So we can write it as –

=> root(5) = a/b, where a and b are co prime numbers )

On Square on both sides ,

square (Root(5) ) = a2/b2

${\left(b\sqrt{5}\right)}^2$ $a^2$

=5

$b^2$ $b^2$

=

$\frac{a^2}{5}$

where 5 divide =

$a^2$

a both ……….Eq 1

$\frac{a}{5}$

=c (where c is co-prime number)

a=5c

putting the value of a in eq 1

$b^2$

=

$\frac{{\left(5c\right)}^2}{5}$

=

$\frac{25c^2}{5}=5c^2$ $c^2=\frac{b^2}{5}$

where 5 divided by

$b^2$

and b both……….eq2

from the equation 1 and 2 we say that 5 is the common factor of a and b but a and b are co -prime number

so, a and b are not have common factor .

Our assumption

$\sqrt{5}$

is rational number is wrong

$\sqrt{5}$

is irrational number

Q2. Prove that 3+2

$\sqrt{5}$

is irrational.

Solution:- 

Let  3+2

$\sqrt{5}$

is rational

Therefore, we can find two co-prime integers a,b(b

$\not\equiv$

0) such that

$$\begin{gathered} 3+2\sqrt{5} =\frac{a}{b} \\ 2\sqrt{5 } =\frac{a}{b}–3 \\ \sqrt{5 }=\frac{1}{2}(\frac{a}{b}–3) \end{gathered}$$

Since a and b are integers,

$\frac{1}{2}(\frac{a}{b}–3)$

will also rational and therefore ,

$\sqrt{5}$

is rational .Our assumption is wrong therefor 3+2

$\sqrt{5}$

is irrational number.

Q3. Prove that the following are irrationals :

i)

$\frac{1}{\sqrt{2}}$

solution:-

Let

$\frac{1}{\sqrt{2}}$

is rational

Therefore, we can find two co-prime integers a,b(b

$\not\equiv$

0) such that

$\frac{1}{\sqrt{2}}=\frac{a}{b}$ $\sqrt{2} =\frac{a}{b}$ $\frac{b}{a}$

is rational as a a and b are integers.

Therefore ,

$\sqrt{2 }$

is rational which contradicts to the fact that

$\sqrt{2}$

is rational .

Hence,our assumption is wrong and

$\frac{1}{\sqrt{2}}$

is rational

ii)

$7\sqrt{5}$

Solutions :-

Let

$7\sqrt{5}$

is rational

Therefore, we can find two co-prime integers a,b(b

$\not\equiv$

0) such that

$$\begin{gathered} 7\sqrt{5}=\frac{a}{b} \\ \sqrt{5}=\frac{a}{7b} \end{gathered}$$ $\frac{a}{7b}$

is rational as a a and b are integers.

Therefore,

$\sqrt{5}$

should be rational .

This contradicts to the fact that

$\sqrt{5}$

is irrational .

Hence,our assumption is wrong and

$7\sqrt{5}$

is rational .

iii) 6+

$\sqrt{2}$

Solution:-

Let 6+

$\sqrt{2}$

is rational

Therefore, we can find two co-prime integers a,b(b

$\not\equiv$

0) such that

$$\begin{gathered} 6+\sqrt{2}=\frac{a}{b} \\ \sqrt{2}=\frac{a}{b}–6 \end{gathered}$$

Since a and b are integer ,

$\frac{a}{b}–6$

is also rational and hence ,

$\sqrt{2}$

should be rational .This contradicts the fact that

$\sqrt{2}$

is irrational.Therefore ,our assumption is false and hence,

$6+\sqrt{2}$

is irrational.

You can see the solution for complete chapter here –

• Complete 10th Class Maths Solution 

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