Question

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

Hint:

To prove that a number is irrational, assume that it is rational and show that this leads to a contradiction.

Answer 1

We will prove that \( \sqrt{3} \) is irrational by contradiction. Assume that \( \sqrt{3} \) is rational. Then it can be expressed as the ratio of two integers \( \frac{p}{q} \), where \( p \) and \( q \) are coprime (i.e., their greatest common divisor is 1). Thus, we assume that: \[ \sqrt{3} = \frac{p}{q}. \] Squaring both sides: \[ 3 = \frac{p^2}{q^2} \quad \implies \quad 3q^2 = p^2. \] This shows that \( p^2 \) is divisible by 3, which implies that \( p \) must also be divisible by 3 (since if a square of a number is divisible by a prime, the number itself must be divisible by that prime). Let \( p = 3k \), where \( k \) is an integer. Substituting \( p = 3k \) into the equation: \[ 3q^2 = (3k)^2 = 9k^2. \] Dividing both sides by 3: \[ q^2 = 3k^2. \] This shows that \( q^2 \) is divisible by 3, which implies that \( q \) must also be divisible by 3. Thus, both \( p \) and \( q \) are divisible by 3, which contradicts our assumption that \( p \) and \( q \) are coprime. Therefore, \( \sqrt{3} \) cannot be rational.
Conclusion:
Since assuming \( \sqrt{3} \) is rational leads to a contradiction, we conclude that \( \sqrt{3} \) is irrational.