Prove that the following are irrationals :

(i) $\dfrac{1}{\sqrt{2}}$

(ii) $7\sqrt{5}$

(iii) 6 + $\sqrt{2}$

(i) Let us assume, to the contrary, that $\dfrac{1}{\sqrt{2}}$ is a rational number.

Then, $\dfrac{1}{\sqrt{2}} = \dfrac{a}{b}$, where a and b have no common factors other than 1.

Solving above equation,

$\Rightarrow b = \sqrt{2}a \\[1em] \Rightarrow \sqrt{2} = \dfrac{b}{a}$

Since b and a are integers, $\dfrac{b}{a}$ is a rational number and so, $\sqrt{2}$ is rational.

We know that $\sqrt{2}$ is irrational. So, our assumption was wrong.

Hence, proved that $\dfrac{1}{\sqrt{2}}$ is an irrational number.

(ii) Let us assume, to the contrary, that $7\sqrt{5}$ is a rational number.

Then, $7\sqrt{5} = \dfrac{a}{b}$, where a and b have no common factors other than 1.

$\Rightarrow 7\sqrt{5}b = a \\[1em] \Rightarrow \sqrt{5} = \dfrac{a}{7b}$

Since, a, 7, and b are integers, so, $\dfrac{a}{7b}$ is a rational number. This means $\sqrt{5}$ is rational but this contradicts the fact that $\sqrt{5}$ is irrational. So, our assumption was wrong.

Hence, proved that $7\sqrt{5}$ is an irrational number.

(iii) Let us assume, to the contrary, that $6 + \sqrt{2}$ is rational.

Then, $6 + \sqrt{2} = \dfrac{a}{b}$, where a and b have no common factors other than 1.

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

Since, a, b, and 6 are integers, so, $\dfrac{a}{b} - 6$ is a rational number. This means $\sqrt{2}$ is also a rational number.

This contradicts the fact that $\sqrt{2}$ is irrational. So, our assumption was wrong.

Hence, proved that $6 + \sqrt{2}$ is an irrational number.