Question

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

Answer 1

Step 1: Understand the problem:
We are asked to prove that \( \sqrt{3} \) is an irrational number, which means that it cannot be expressed as a ratio of two integers, i.e., it cannot be written as \( \frac{p}{q} \), where \( p \) and \( q \) are integers and \( q \neq 0 \), and \( \gcd(p, q) = 1 \).

Step 2: Assume the opposite (contradiction approach):
Assume, for the sake of contradiction, that \( \sqrt{3} \) is a rational number. Then, it can be written as a fraction \( \frac{p}{q} \), where \( p \) and \( q \) are coprime integers (i.e., \( \gcd(p, q) = 1 \)) and \( q \neq 0 \). Thus, we assume:
\[ \sqrt{3} = \frac{p}{q} \] Squaring both sides, we get:
\[ 3 = \frac{p^2}{q^2} \] Multiplying both sides by \( q^2 \), we get:
\[ 3q^2 = p^2 \] This means that \( p^2 \) is divisible by 3.

Step 3: Use the property of divisibility:
Since \( p^2 \) is divisible by 3, we can conclude that \( p \) must also be divisible by 3. This is because if a prime number divides the square of an integer, it must divide the integer itself. Therefore, we can write \( p = 3k \), where \( k \) is some integer.

Step 4: Substitute \( p = 3k \) into the equation:
Substitute \( p = 3k \) into the equation \( 3q^2 = p^2 \):
\[ 3q^2 = (3k)^2 = 9k^2 \] Simplifying this equation, we get:
\[ 3q^2 = 9k^2 \] Dividing both sides by 3, we get:
\[ q^2 = 3k^2 \] This means that \( q^2 \) is also divisible by 3, which implies that \( q \) must also be divisible by 3.

Step 5: Arrive at a contradiction:
We have now shown that both \( p \) and \( q \) are divisible by 3. However, this contradicts our original assumption that \( \gcd(p, q) = 1 \), because if both \( p \) and \( q \) are divisible by 3, their greatest common divisor is at least 3, not 1.

Step 6: Conclusion:
Since assuming that \( \sqrt{3} \) is rational leads to a contradiction, we conclude that \( \sqrt{3} \) cannot be rational. Therefore, \( \sqrt{3} \) is an irrational number.