Question

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

Hint:

This proof relies on the property: "If a prime number \(p\) divides \(a^2\), then \(p\) divides \(a\)." It is a standard derivation frequently asked in board exams.

Answer 1

Step 1: Understanding the Concept:
We use the method of contradiction. We assume \(\sqrt{3}\) is rational and show that this leads to a logical impossibility regarding its co-prime factors.
Step 3: Detailed Explanation:
1. Assume \(\sqrt{3}\) is rational. Then \(\sqrt{3} = \frac{p}{q}\) where \(p\) and \(q\) are co-prime integers (\(q \neq 0\)).
2. Squaring both sides: \(3 = \frac{p^2}{q^2} \implies p^2 = 3q^2\).
3. This implies that \(p^2\) is divisible by 3. By theorem, if \(p^2\) is divisible by 3, then \(p\) is also divisible by 3.
4. Let \(p = 3k\) for some integer \(k\). Substitute this in \(p^2 = 3q^2\):
\[ (3k)^2 = 3q^2 \implies 9k^2 = 3q^2 \implies 3k^2 = q^2 \]
5. This implies \(q^2\) is divisible by 3, so \(q\) is also divisible by 3.
6. This means \(p\) and \(q\) have at least 3 as a common factor.
7. This contradicts our assumption that \(p\) and \(q\) are co-prime.
Step 4: Final Answer:
Since the assumption is wrong, \(\sqrt{3}\) is an irrational number.