Question

Prove that \(5 - \sqrt{3}\) is an irrational number.

Hint:

The key to this type of proof is to isolate the known irrational root (like \(\sqrt{2}\), \(\sqrt{3}\), etc.) on one side of the equation. The other side will be an expression of rational numbers, which is itself rational, leading to the necessary contradiction.

Answer 1

Step 1: Understanding the Concept:
This is a proof by contradiction. We start by assuming the opposite of what we want to prove. If this assumption leads to a logical contradiction, then our original statement must be true. The proof relies on the known fact that \(\sqrt{3}\) is an irrational number.

Step 2: Detailed Explanation:
Assumption: Let us assume, to the contrary, that \(5 - \sqrt{3}\) is a rational number.
By the definition of a rational number, it can be expressed in the form \(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q \neq 0\). \[ 5 - \sqrt{3} = \frac{p}{q} \] Now, we rearrange the equation to isolate the irrational part (\(\sqrt{3}\)): \[ 5 - \frac{p}{q} = \sqrt{3} \] \[ \sqrt{3} = \frac{5q - p}{q} \] Contradiction:
Since \(p\) and \(q\) are integers, \(5q\) is an integer and \(5q - p\) is also an integer. Also, \(q \neq 0\).
Therefore, \(\frac{5q - p}{q}\) is a rational number.
This implies that \(\sqrt{3}\) is a rational number.
However, this contradicts the well-established fact that \(\sqrt{3}\) is an irrational number.
This contradiction arises because our initial assumption that \(5 - \sqrt{3}\) is rational was incorrect.

Step 3: Final Answer:
Therefore, we must conclude that \(5 - \sqrt{3}\) is an irrational number. Hence proved.