Question

Prove that \(\frac{2 + 3\sqrt{5}}{7}\) is an irrational number, given that \(\sqrt{5}\) is an irrational number.

Hint:

In board exams, clearly state that the combination of integers under standard arithmetic operations (multiplication, subtraction, division) always yields a rational number.

Answer 1

Step 1: Understanding the Question:
We need to prove that the given expression is irrational. We are allowed to use the fact that \(\sqrt{5}\) is known to be irrational.

Step 2: Key Formula or Approach:
We will use proof by contradiction.
We assume that the number is rational, and show that this leads to a contradiction of the given fact that \(\sqrt{5}\) is irrational.

Step 3: Detailed Explanation:

• Assume on the contrary that \(\frac{2 + 3\sqrt{5}}{7}\) is a rational number.
Therefore, it can be written in the form \(\frac{a}{b}\), where \(a\) and \(b\) are integers, \(b \neq 0\), and \(a, b\) are co-prime:
\[ \frac{2 + 3\sqrt{5}}{7} = \frac{a}{b} \]

• Rearrange the equation to isolate the radical term \(\sqrt{5}\):
\[ 2 + 3\sqrt{5} = \frac{7a}{b} \]
\[ 3\sqrt{5} = \frac{7a}{b} - 2 \]
\[ 3\sqrt{5} = \frac{7a - 2b}{b} \]
\[ \sqrt{5} = \frac{7a - 2b}{3b} \]

• Analyze both sides of the equation:
- Since \(a\) and \(b\) are integers, \(7a - 2b\) and \(3b\) are also integers (with \(3b \neq 0\)).
- This means the RHS \(\frac{7a - 2b}{3b}\) is a rational number.
- Consequently, the LHS \(\sqrt{5}\) must also be a rational number.

• Identify the contradiction:
This contradicts our given hypothesis that \(\sqrt{5}\) is an irrational number.
Our assumption that \(\frac{2 + 3\sqrt{5}}{7}\) is rational must be false.


Step 4: Final Answer:
Therefore, \(\frac{2 + 3\sqrt{5}}{7}\) is an irrational number.