Question

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

Answer 1

Step 1: Problem Statement:
We aim to demonstrate that \( 6 - 4\sqrt{5} \) is irrational, given that \( \sqrt{5} \) is irrational.
Our proof will utilize the principle that the sum or difference of a rational and an irrational number is always irrational.

Step 2: Component Analysis:
The number \( 6 - 4\sqrt{5} \) can be decomposed into a rational component and an irrational component.
- The number \( 6 \) is rational.
- The term \( 4\sqrt{5} \) is the product of the rational number \( 4 \) and the irrational number \( \sqrt{5} \), making \( 4\sqrt{5} \) irrational.

Step 3: Irrationality Proof:
The difference between a rational number and an irrational number is inherently irrational. In this case, \( 6 \) is rational and \( 4\sqrt{5} \) is irrational. Consequently, \( 6 - 4\sqrt{5} \) must be irrational.

Step 4: Conclusion:
As \( 6 - 4\sqrt{5} \) is the difference between a rational and an irrational number, it is classified as irrational. This concludes the proof that \( 6 - 4\sqrt{5} \) is an irrational number.