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 problem:
We are asked to prove that \( 6 - 4\sqrt{5} \) is an irrational number, given that \( \sqrt{5} \) is an irrational number.
To prove this, we will use the property that the sum or difference of a rational number and an irrational number is always irrational.

Step 2: Expressing the number in terms of rational and irrational components:
The number \( 6 - 4\sqrt{5} \) can be written as the sum of two terms: one rational (6) and one involving \( \sqrt{5} \) (which is irrational). Specifically: - \( 6 \) is a rational number. - \( 4\sqrt{5} \) is the product of 4 (a rational number) and \( \sqrt{5} \) (an irrational number), which is irrational.

Step 3: Proving the irrationality:
We know that the difference between a rational number and an irrational number is always irrational. Here, \( 6 \) is rational and \( 4\sqrt{5} \) is irrational. Therefore, the expression \( 6 - 4\sqrt{5} \) must be irrational.

Step 4: Conclusion:
Since \( 6 - 4\sqrt{5} \) is the difference of a rational number and an irrational number, it is irrational. Thus, we have proven that \( 6 - 4\sqrt{5} \) is an irrational number.