Prove that \( 3 + 2\sqrt{5} \) is irrational.
Assume that \( 3 + 2\sqrt{5} \) is rational. Then there exists a rational number \(r\) such that
\[ 3 + 2\sqrt{5} = r. \]
Rewriting, we get
\[ 2\sqrt{5} = r - 3. \]
Since \(r\) and 3 are rational numbers, their difference \(r - 3\) is also rational. Thus, the right-hand side \(r - 3\) is rational. Dividing both sides by 2, we obtain
\[ \sqrt{5} = \dfrac{r - 3}{2}. \]
The right-hand side is a quotient of rational numbers, so it is rational. Hence \( \sqrt{5} \) would be rational.
But from the previous result, \( \sqrt{5} \) is irrational. This is a contradiction.
Therefore, our assumption that \( 3 + 2\sqrt{5} \) is rational is false. Hence \( 3 + 2\sqrt{5} \) is irrational.