Prove that following numbers are irrationals:

Solution:

(i) Let us assume that $\frac{2}{\sqrt{7}}$ is rational . Then, there exist positive co primes a and b such that

$\frac{2}{\sqrt{7}}=\frac{a}{b}$

$\sqrt{7}=\frac{2 b}{a}$

$\sqrt{7}$ is rational number which is a contradication.

Hence $\frac{2}{\sqrt{7}}$ is irrational

(ii) Let us assume that $\frac{3}{2 \sqrt{5}}$ is rational .Then, there exist positive co primes a and $b$ such that

$\frac{3}{2 \sqrt{5}}=\frac{a}{b}$

$\sqrt{5}=\frac{3 b}{2 a}$

$\sqrt{5}$ is a rational number which is a contradication.

Hence $\frac{3}{2 \sqrt{5}}$ is irrational

(iii) Let us assume that $4+\sqrt{2}$ is rational .Then, there exist positive co primes $a$ and $b$ such that

$4+\sqrt{2}=\frac{a}{b}$

$\sqrt{2}=\frac{a}{b}-4$

$\sqrt{2}=\frac{a-4 b}{b}$

$\sqrt{2}$ is a rational number which is a contradication.

Hence $4+\sqrt{2}$ is irrational

(iv) Let us assume that $5 \sqrt{2}$ is rational .Then, there exist positive co primes a and $b$ such that

$5 \sqrt{2}=\frac{a}{b}$

$\sqrt{2}=\frac{a}{b}-5$

$\sqrt{2}=\frac{a-5 b}{b}$

\sqrt{2} \text { is a rational number which is a contradication. }

Hence $5 \sqrt{2}$ is irrational