Show that the following numbers are irrational.

Solution:

(i) Let us assume that $\frac{1}{\sqrt{2}}$ is rational .Then, there exist positive co primes $a$ and $b$ such that

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

$\frac{1}{\sqrt{2}}=\left(\frac{a}{b}\right)^{2}$

$\Rightarrow \frac{1}{2}=\frac{a^{2}}{b^{2}}$

$\Rightarrow b^{2}=2 a^{2}$

$\Rightarrow 2 \mid b^{2}\left(\because 2 \mid 2 a^{2}\right)$

$\Rightarrow 2 \mid b$

$\Rightarrow b=2 c$ for some positive integer $c$

$\Rightarrow 2 a^{2}=b^{2}$

$\Rightarrow 2 a^{2}=4 c^{2}(\because a=p c)$

$\Rightarrow a^{2}=2 c^{2}$

$\Rightarrow 2 \mid a^{2}\left(\because 2 \mid 2 c^{2}\right)$

$\Rightarrow 2 \mid a$

$\Rightarrow 2 \mid a$ and $2 \mid b$

This contradicts the fact that $a$ and $b$ are co primes.

Hence $\frac{1}{\sqrt{2}}$ is irrational

(ii) Let us assume that $7 \sqrt{5}$ is rational .Then, there exist positive co primes $a$ and $b$ such that

$7 \sqrt{5}=\frac{a}{b}$

$\sqrt{5}=\frac{a}{7 b}$

We know that $\sqrt{5}$ is an irrational number

Here we see that $\sqrt{5}$ is a rational number which is a contradiction

Hence $7 \sqrt{5}$ is irrational

(iii) Let us assume that $6+\sqrt{2}$ is rational. Then, there exist positive co primes $a$ and $b$ such that

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

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

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

Here we see that $\sqrt{2}$ is a rational number which is a contradiction as we know that $\sqrt{2}$ is an irrational number

Hence $6+\sqrt{2}$ is irrational

(iv) Let us assume that $3-\sqrt{5}$ is rational .Then, there exist positive co primes $a$ and $b$ such that

$3-\sqrt{5}=\frac{a}{b}$

$\sqrt{5}=3-\frac{a}{b}$

$\sqrt{5}=\frac{3 b-a}{b}$

Here we see that $\sqrt{5}$ is a rational number which is a contradiction as we know that $\sqrt{5}$ is an irrational number

Hence $3-\sqrt{5}$ is irrational