Question

Given that \(\sqrt{5}\) is an irrational number, prove that \(2 + 3\sqrt{5}\) is an irrational number.

Hint:

If adding or multiplying an irrational number with a rational gives a rational, then the irrational number must become rational, which leads to contradiction.

Answer 1

Given:
\(\sqrt{5}\) is irrational.
We need to prove \(2 + 3\sqrt{5}\) is irrational.

Step 1: Assume the contrary
Suppose \(2 + 3\sqrt{5}\) is rational.
Let \(2 + 3\sqrt{5} = r\), where \(r\) is a rational number.

Step 2: Rearrange to isolate \(\sqrt{5}\)
\[ 3\sqrt{5} = r - 2 \] \[ \sqrt{5} = \frac{r - 2}{3} \]

Step 3: Analyze the right side
Since \(r\) and 2 are rational, \(\frac{r - 2}{3}\) is rational.
This implies \(\sqrt{5}\) is rational.

Step 4: Contradiction
This contradicts the given fact that \(\sqrt{5}\) is irrational.

Conclusion:
Our assumption is wrong.
Hence, \(2 + 3\sqrt{5}\) is irrational.

Final statement:
\[ \boxed{2 + 3\sqrt{5} \text{ is irrational}} \]