CBSE 10th Maths - Real Numbers - Exercise 1.3

Solution:

Let $\sqrt{5}$ is a rational number.

Therefore, we can find two integers a, b (b ≠ 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=$\sqrt{5b}$

$a^{2}$=$5b^{2}$

Therefore, $a^{2}$ is divisible by 5 and it can be said that a is divisible by 5.

Let a = 5k, where k is an integer

(5k)$^{2}$=5b$^{2}$

This means that $b^{2}$ is divisible by 5 and hence, b is divisible by 5.

b$^{2}$=5k$^{2}$

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 is irrational.

Solution:

Let 3+2$\sqrt{5}$ is rational.

Therefore, we can find two integers a, b (b ≠ 0) such that

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

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

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

Since a and b are integers,$\dfrac{1}{2}$($\dfrac{a}{b}$-3) 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.

Solution:

i.$\dfrac{1}{\sqrt{2}}$

Let $\dfrac{1}{\sqrt{2}}$ is rational.

Therefore, we can find two integers a, b (b ≠ 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.

Therefore,$\sqrt{2}$ is rational which contradicts to the fact that $\sqrt{2}$ is irrational.

Hence, our assumption is false and $\dfrac{1}{\sqrt{2}}$ is irrational.

ii.$7\sqrt{5}$

Let $7\sqrt{5}$ is rational.

Therefore, we can find two integers a, b (b ≠ 0) such that

$7\sqrt{5}$ = $\dfrac{a}{b}$ for some integers a and b

$\therefore \sqrt{5} $=$\dfrac{a}{7b}$

$\dfrac{a}{7b}$ is rational as a and b are integers.

Therefore, $\sqrt{5}$ should be rational.

This contradicts the fact that $\sqrt{5}$ is irrational. Therefore, our assumption that $7\sqrt{5}$ is rational is false. Hence, $7\sqrt{5}$ is irrational.

iii.$6+\sqrt{2}$

Let $6+\sqrt{2}$ be rational.

Therefore, we can find two integers a, b (b ≠ 0) such that

$6+\sqrt{2}$=$\dfrac{a}{b}$

$\sqrt{2}$ =$\dfrac{a}{b}$ -6

Since a and b are integers, $\dfrac{a}{b}$ -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.