Prove that the following numbers are irrational:

(i) $3 + \sqrt{5}$

(ii) $15 - 2\sqrt{7}$

(iii) $\dfrac{1}{3 - \sqrt5}$

$\text{(i) } 3 + \sqrt{5}$

Let us assume that $3 + \sqrt{5}$ is a rational number, say r.

Then,

$3 + \sqrt{5} = r \\[1.5em] \Rightarrow \sqrt{5} = r - 3$

As r is rational, r - 3 is rational

$\Rightarrow \sqrt{5}$ is rational

But this contradicts the fact that $\sqrt{5}$ is irrational.

Hence, our assumption is wrong.

∴ $\bold{3 + \sqrt{5}}$ is an irrational number.

$\text{(ii) } 15 - 2\sqrt{7}$

Let us assume that $15 - 2\sqrt{7}$ is a rational number, say r.

Then,

$15 - 2\sqrt{7} = r \\[1.5em] \Rightarrow 2\sqrt{7} = 15 - r \\[1.5em] \Rightarrow \sqrt{7} = \dfrac{15 - r}{2} \\[1.5em]$

As r is rational, 15 - r is rational

$\Rightarrow \dfrac{15 - r}{2}$ is rational

$\Rightarrow \sqrt{7}$ is rational

But this contradicts the fact that $\sqrt{7}$ is irrational.

Hence, our assumption is wrong.

∴ $\bold{15 - 2\sqrt{7}}$ is an irrational number.

$\text{(iii) }\dfrac{1}{3 - \sqrt5}$

Let us rationalise the denominator

$\dfrac{1}{3 - \sqrt5} = \dfrac{1}{3 - \sqrt5} × \dfrac{3 + \sqrt5}{3 + \sqrt5} \\[1.5em] = \dfrac{3 + \sqrt5}{(3)^2 - (\sqrt5)^2} \\[1.5em] = \dfrac{3 + \sqrt5}{9 - 5} \\[1.5em] = \dfrac{3 + \sqrt5}{4} \\[1.5em]$

Let us assume that $\dfrac{3 + \sqrt5}{4}$ is a rational number, say r.

Then,

$\dfrac{3 + \sqrt5}{4} = r \\[1.5em] \Rightarrow 3 +\sqrt{5} = 4r \\[1.5em] \Rightarrow \sqrt{5} = 4r - 3 \\[1.5em]$

As r is rational, 4r is rational

$\Rightarrow$ 4r - 3 is also rational

$\Rightarrow \sqrt{5}$ is rational

But this contradicts the fact that $\sqrt{5}$ is irrational.

Hence, our assumption is wrong.

$\Rightarrow \dfrac{3 + \sqrt5}{4}$ is an irrational number.

∴ $\bold{\dfrac{1}{3 - \sqrt5}}$ is an irrational number.