Question

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

Hint:

In contradiction proofs, always aim to isolate the known irrational part (like \(\sqrt{5}\)) on one side of the equation.

Answer 1

Step 1: Understanding the Concept:
We use the Method of Contradiction. We assume the opposite of what we want to prove (that the number is rational) and show that this leads to a logical impossibility (a contradiction).
Step 2: Key Formula or Approach:
A rational number can be expressed as \(\frac{a}{b}\), where \(a\) and \(b\) are integers and \(b \neq 0\).
Step 3: Detailed Explanation:
1. Let us assume, to the contrary, that \(3 + 2\sqrt{5}\) is rational.
2. Therefore, we can find co-prime integers \(a\) and \(b\) (\(b \neq 0\)) such that:
\[ 3 + 2\sqrt{5} = \frac{a}{b} \] 3. Rearrange the equation to isolate \(\sqrt{5}\):
\[ 2\sqrt{5} = \frac{a}{b} - 3 \]
\[ 2\sqrt{5} = \frac{a - 3b}{b} \]
\[ \sqrt{5} = \frac{a - 3b}{2b} \]
4. Since \(a\) and \(b\) are integers, \(a - 3b\) and \(2b\) are also integers. Thus, \(\frac{a - 3b}{2b}\) is a rational number.
5. This implies that \(\sqrt{5}\) is rational.
6. But this contradicts the given fact that \(\sqrt{5}\) is irrational.
7. The contradiction has arisen because of our incorrect assumption.
Step 4: Final Answer:
Hence, \(3 + 2\sqrt{5}\) is an irrational number.