(i) Irrationality of \( \dfrac{1}{\sqrt{2}} \)

Assume that \( \dfrac{1}{\sqrt{2}} \) is rational. Then there exists a rational number \(k\) such that

\[ \dfrac{1}{\sqrt{2}} = k. \]

Taking reciprocals on both sides (which is allowed because \(k \neq 0\)), we get

\[ \sqrt{2} = \dfrac{1}{k}. \]

The right-hand side is the reciprocal of a rational number and hence is rational. Thus \( \sqrt{2} \) would be rational, which contradicts the known fact that \( \sqrt{2} \) is irrational. Therefore, \( \dfrac{1}{\sqrt{2}} \) must be irrational.

(ii) Irrationality of \( 7\sqrt{5} \)

Assume that \( 7\sqrt{5} \) is rational. Then there exists a rational number \(m\) such that

\[ 7\sqrt{5} = m. \]

Dividing both sides by 7 (a non-zero rational number), we get

\[ \sqrt{5} = \dfrac{m}{7}. \]

The right-hand side is a quotient of two rational numbers, so it is rational. Hence \( \sqrt{5} \) would be rational, which contradicts the fact that \( \sqrt{5} \) is irrational. Therefore, \( 7\sqrt{5} \) is irrational.

(iii) Irrationality of \( 6 + \sqrt{2} \)

Assume that \( 6 + \sqrt{2} \) is rational. Then there exists a rational number \(n\) such that

\[ 6 + \sqrt{2} = n. \]

Rearranging, we get

\[ \sqrt{2} = n - 6. \]

Since \(n\) and 6 are rational, their difference \(n - 6\) is rational. Thus, \( \sqrt{2} \) would be rational, which contradicts the fact that \( \sqrt{2} \) is irrational.

Hence the assumption is false, and \( 6 + \sqrt{2} \) is irrational.

Therefore, each of \( \dfrac{1}{\sqrt{2}} \), \( 7\sqrt{5} \) and \( 6 + \sqrt{2} \) is irrational.

NCERT SolutionsClass 10 - Mathematics - Chapter 1: REAL NUMBERSEXERCISE 1.2

Question. 1

Question. 2

Question. 3

NCERT Solutions

NCERT Solutions
Class 10 - Mathematics - Chapter 1: REAL NUMBERS
EXERCISE 1.2