Prove that $3+2 \sqrt{5}$ is irrational.

Solution:

Let us assume, to the contrary, that $3+2 \sqrt{5}$ is rational. That is, we can find coprime integers a and $b(b \neq 0)$ such that $3+2 \sqrt{5}=\frac{\mathbf{a}}{\mathbf{b}}$

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

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

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

Since a and b are integers, we get $\frac{\mathbf{a}}{\mathbf{2 b}}-\frac{\mathbf{3}}{\mathbf{2}}$ is rational, and so $\frac{\mathbf{a}-\mathbf{3} \mathbf{b}}{\mathbf{2 b}}=\sqrt{\mathbf{5}}$ is rational.

But this contradicts the fact that $\sqrt{5}$ is irrational. This contradiction has arisen because of our

incorrect assumption that $3+2 \sqrt{5}$ is rational.

So, we conclude that $3+2 \sqrt{5}$ is irrational.