Question

Prove that \( \sqrt{2} \) is an irrational number.

Hint:

To prove a number is irrational, assume it is rational, derive a contradiction, and conclude that the assumption must be false.

Answer 1

To prove that \( \sqrt{2} \) is an irrational number, we will use proof by contradiction.
Step 1: Assume, for the sake of contradiction, that \( \sqrt{2} \) is a rational number. Then, we can express it as a fraction in the form: \[ \sqrt{2} = \frac{p}{q}, \] where \( p \) and \( q \) are coprime integers (i.e., the greatest common divisor of \( p \) and \( q \) is 1), and \( q \neq 0 \).
Step 2: Square both sides: \[ 2 = \frac{p^2}{q^2} \quad \Rightarrow \quad 2q^2 = p^2. \]
Step 3: This implies that \( p^2 \) is an even number because it is divisible by 2. Since \( p^2 \) is even, \( p \) must also be even (because the square of an odd number is odd).
Step 4: Let \( p = 2k \), where \( k \) is some integer. Substituting this into the equation \( 2q^2 = p^2 \), we get: \[ 2q^2 = (2k)^2 \quad \Rightarrow \quad 2q^2 = 4k^2 \quad \Rightarrow \quad q^2 = 2k^2. \]
Step 5: This implies that \( q^2 \) is also even, and therefore \( q \) must be even as well.
Step 6: We have now shown that both \( p \) and \( q \) are even, which contradicts our assumption that \( p \) and \( q \) are coprime (since both are divisible by 2).
Step 7: Therefore, our assumption that \( \sqrt{2} \) is rational must be false, and thus \( \sqrt{2} \) is an irrational number.
Conclusion: Hence, \( \sqrt{2} \) is an irrational number.