Question

Prove that \(6 - 4\sqrt{5}\) is an irrational number, given that \(\sqrt{5}\) is an irrational number.

Answer 1

Step 1: Understanding the given problem:

We are asked to prove that \( 6 - 4\sqrt{5} \) is an irrational number, given that \( \sqrt{5} \) is an irrational number.

Step 2: Assume the contrary:

We begin by assuming the opposite, that \( 6 - 4\sqrt{5} \) is a rational number. If it were rational, we could express it as a fraction of two integers, say:
\[ 6 - 4\sqrt{5} = \frac{p}{q} \] where \( p \) and \( q \) are integers and \( q \neq 0 \).

Step 3: Solving for \( \sqrt{5} \):

Rearranging the equation above to isolate \( \sqrt{5} \):
\[ 4\sqrt{5} = 6 - \frac{p}{q} \] \[ \sqrt{5} = \frac{6 - \frac{p}{q}}{4} \] \[ \sqrt{5} = \frac{6q - p}{4q} \] Thus, \( \sqrt{5} \) is expressed as the ratio of two integers \( \frac{6q - p}{4q} \), which means that \( \sqrt{5} \) is a rational number.

Step 4: Contradiction:

However, this contradicts the given fact that \( \sqrt{5} \) is an irrational number. Therefore, our assumption that \( 6 - 4\sqrt{5} \) is rational must be false.

Step 5: Conclusion:

Since assuming \( 6 - 4\sqrt{5} \) is rational leads to a contradiction, we conclude that \( 6 - 4\sqrt{5} \) must be an irrational number.