Question 2

We will prove this by contradiction.

Step 1 — Assume it is rational

Let's assume 3 + 2$\sqrt{5}$ is a rational number.

So, we can write it as $\frac{a}{b}$.

Here, $a$ and $b$ are integers.

$b$ is not equal to zero.

Also, $a$ and $b$ are coprime.

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

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

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

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

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

Step 2 — Contradiction

We know $a$ and $b$ are integers.

So, $a - 3b$ is an integer.

Also, $2b$ is an integer.

Since $b \neq 0$, $2b \neq 0$.

Thus, $\frac{a - 3b}{2b}$ is a rational number.

This means $\sqrt{5}$ is rational.

But we know $\sqrt{5}$ is irrational.

This creates a contradiction.

Our initial assumption was wrong.

Answer

(i) Therefore, $3 + 2\sqrt{5}$ is irrational.