Question

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

Hint:

To prove square roots of primes are irrational, assume a rational form and show both numerator and denominator become divisible by the same prime — creating a contradiction.

Answer 1

Concept: A number is irrational if it cannot be written in the form \( \frac{p}{q} \), where \( p, q \) are integers and \( q \neq 0 \). To prove a number is irrational, we commonly use the proof by contradiction method:

• Assume the number is rational.
• Write it in lowest terms.
• Show a contradiction (both numerator and denominator divisible by same number).

Step 1: Assume \( \sqrt{3} \) is rational.
Suppose \( \sqrt{3} = \frac{p}{q} \), where \( p \) and \( q \) are coprime integers (no common factor).
Step 2: Square both sides.
\[ 3 = \frac{p^2}{q^2} \quad \Rightarrow \quad p^2 = 3q^2. \] This means \( p^2 \) is divisible by 3, so \( p \) must also be divisible by 3.
Step 3: Substitute \( p = 3k \).
Let \( p = 3k \). Then: \[ (3k)^2 = 3q^2 \quad \Rightarrow \quad 9k^2 = 3q^2 \quad \Rightarrow \quad q^2 = 3k^2. \] Thus, \( q^2 \) is divisible by 3, so \( q \) is also divisible by 3.
Step 4: Contradiction.
Both \( p \) and \( q \) are divisible by 3, contradicting the assumption that they are coprime. Hence, our assumption is false. Conclusion: Therefore, \( \sqrt{3} \) is irrational. Similarly for \( \sqrt{5} \):
Assume \( \sqrt{5} = \frac{p}{q} \) in lowest terms. \[ 5 = \frac{p^2}{q^2} \Rightarrow p^2 = 5q^2. \] This implies \( p \) is divisible by 5. Let \( p = 5k \). Then: \[ 25k^2 = 5q^2 \Rightarrow q^2 = 5k^2. \] So \( q \) is also divisible by 5 — again a contradiction. Hence, \( \sqrt{5} \) is irrational.