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

We are asked to prove that \(\sqrt{3}\) is an irrational number.
Step 1: Assume the contrary
- Suppose \(\sqrt{3}\) is rational.
- Then it can be expressed as a fraction of two integers in lowest terms:
\[ \sqrt{3} = \frac{p}{q} \] where \(p\) and \(q\) are integers with no common factors (other than 1), and \(q \neq 0\).
Step 2: Square both sides
\[ 3 = \frac{p^2}{q^2} \implies p^2 = 3 q^2 \]
Step 3: Analyze divisibility
- From \(p^2 = 3 q^2\), we see that \(p^2\) is divisible by 3.
- Therefore, \(p\) must also be divisible by 3 (since 3 is prime). Let \(p = 3r\) for some integer \(r\).
Substitute \(p = 3r\) into \(p^2 = 3 q^2\):
\[ (3r)^2 = 3 q^2 \implies 9 r^2 = 3 q^2 \implies q^2 = 3 r^2 \]
- This shows that \(q^2\) is also divisible by 3, so \(q\) is divisible by 3.
Step 4: Contradiction
- Now both \(p\) and \(q\) are divisible by 3, which contradicts our assumption that \(p/q\) is in lowest terms.
Step 5: Conclusion
- Since assuming \(\sqrt{3}\) is rational leads to a contradiction, our assumption is wrong.
- Therefore, \(\sqrt{3}\) is irrational.
Answer:
\(\sqrt{3}\) is an irrational number.