Question

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

Answer 1

Step 1: Problem Definition:
The objective is to prove that \( \sqrt{3} \) is irrational. This means it cannot be represented as a fraction \( \frac{p}{q} \) where \( p \) and \( q \) are integers, \( q eq 0 \), and \( \gcd(p, q) = 1 \).

Step 2: Hypothesis for Contradiction:
Assume \( \sqrt{3} \) is rational. Therefore, \( \sqrt{3} = \frac{p}{q} \) for coprime integers \( p \) and \( q \) with \( q eq 0 \).
Squaring both sides yields \( 3 = \frac{p^2}{q^2} \).
Multiplying by \( q^2 \) gives \( 3q^2 = p^2 \).
This implies \( p^2 \) is divisible by 3.

Step 3: Divisibility Property Application:
If \( p^2 \) is divisible by 3, then \( p \) must also be divisible by 3 (as 3 is prime). Thus, \( p = 3k \) for some integer \( k \).

Step 4: Substitution and Simplification:
Substitute \( p = 3k \) into \( 3q^2 = p^2 \):
\[ 3q^2 = (3k)^2 = 9k^2 \]
Dividing by 3, we get \( q^2 = 3k^2 \).
This indicates \( q^2 \) is divisible by 3, and consequently, \( q \) must also be divisible by 3.

Step 5: Identification of Contradiction:
We have established that both \( p \) and \( q \) are divisible by 3. This directly contradicts the initial assumption that \( p \) and \( q \) are coprime (\( \gcd(p, q) = 1 \)), as their common divisor is at least 3.

Step 6: Final Conclusion:
The assumption that \( \sqrt{3} \) is rational leads to a contradiction. Therefore, \( \sqrt{3} \) must be irrational.