Proof:

Assume, to the contrary, that $3 + 2\sqrt{5}$ is rational.

Then we can write $3 + 2\sqrt{5} = \frac{a}{b}$, where $a$ and $b$ are coprime integers and $b \neq 0$.

Rearranging:

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

Analysis:

Since $a$ and $b$ are integers, $(a - 3b)$ and $(2b)$ are also integers.

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

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

Contradiction:

But we know that $\sqrt{5}$ is irrational (proved in Exercise 1.2, Question 1).

This is a contradiction.

This contradiction has arisen because of our incorrect assumption that $3 + 2\sqrt{5}$ is rational.

Conclusion: $3 + 2\sqrt{5}$ is irrational.