Prove that the following numbers are irrational:

$\begin{matrix} \text{(i)} & 5 + \sqrt{2} \\[0.5em] \text{(ii)} & 3 - 5\sqrt{3} \\[0.5em] \text{(iii)} & 2\sqrt{3} - 7 \\[0.5em] \text{(iv)} & \sqrt{2} - \sqrt{5} \end{matrix}$

$\text{(i) } 5 + \sqrt{2}$

Let us assume that $5 + \sqrt{2}$ is a rational number, say r.

Then,

$5 + \sqrt{2} = r \\[0.5em] \Rightarrow \sqrt{2} = r - 5$

As r is rational, r - 5 is rational

$\Rightarrow \sqrt{2}$ is rational

But this contradicts the fact that $\sqrt{2}$ is irrational.

Hence, our assumption is wrong.

∴ $5 + \sqrt{2}$ is an irrational number.

$\text{(ii) } 3 - 5\sqrt{3}$

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

Then,

$3 - 5\sqrt{3} = r \\[0.5em] \Rightarrow 5\sqrt{3} = 3 - r \\[0.5em] \Rightarrow \sqrt{3} = \dfrac{3 - r}{5} \\[0.5em]$

As r is rational, 3 - r is rational

$\Rightarrow \dfrac{3 - r}{5}$ is rational

$\Rightarrow \sqrt{3}$ is rational

But this contradicts the fact that $\sqrt{3}$ is irrational.

Hence, our assumption is wrong.

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

$\text{(iii) } 2\sqrt{3} - 7$

Let us assume that $2\sqrt{3} - 7$ is a rational number, say r.

Then,

$2\sqrt{3} - 7 = r \\[0.5em] \Rightarrow 2\sqrt{3} = r + 7 \\[0.5em] \Rightarrow \sqrt{3} = \dfrac{r + 7}{2} \\[0.5em]$

As r is rational, r + 7 is rational

$\Rightarrow \dfrac{r + 7}{2}$ is rational

$\Rightarrow \sqrt{3}$ is rational

But this contradicts the fact that $\sqrt{3}$ is irrational.

Hence, our assumption is wrong.

∴ $2\sqrt{3} - 7$ is an irrational number.

$\text{(iv) } \sqrt{2} + \sqrt{5}$

Let us assume that $\sqrt{2} + \sqrt{5}$ is a rational number, say r.

Then,

$\sqrt{2} + \sqrt{5} = r \\[0.5em] \Rightarrow \sqrt{5} = r - \sqrt{2} \\[0.5em] \Rightarrow (\sqrt{5})^2 = (r - \sqrt{2})^2 \\[0.5em] \Rightarrow 5 = r^2 + 2 - 2\sqrt{2}r \\[0.5em] \Rightarrow 2\sqrt{2}r = r^2 + 2 - 5 \\[0.5em] \Rightarrow 2\sqrt{2}r = r^2 - 3 \\[0.5em] \Rightarrow \sqrt{2} = \dfrac{r^2 - 3}{2r} \\[0.5em]$

As r is rational,

$\Rightarrow \dfrac{r^2 - 3}{2r}$ is rational

$\Rightarrow \sqrt{2}$ is rational

But this contradicts the fact that $\sqrt{2}$ is irrational.

Hence, our assumption is wrong.

∴ $\sqrt{2} + \sqrt{5}$ is an irrational number.