Question

Prove that \( \sqrt{2} \) is not a rational number.

Hint:

To prove that a number is irrational, use proof by contradiction, assuming it is rational and showing that this leads to an inconsistency.

Answer 1

We will prove that \( \sqrt{2} \) is not a rational number by contradiction.
Assume that \( \sqrt{2} \) is a rational number. Then it can be expressed as a fraction: \[ \sqrt{2} = \frac{p}{q}, \] where \( p \) and \( q \) are integers and \( \gcd(p, q) = 1 \) (i.e., the fraction is in its simplest form). Now, square both sides: \[ 2 = \frac{p^2}{q^2}. \] Multiply both sides by \( q^2 \): \[ 2q^2 = p^2. \] This shows that \( p^2 \) is even, because it is two times some integer \( q^2 \). If \( p^2 \) is even, then \( p \) must also be even (since the square of an odd number is odd). Let \( p = 2k \), where \( k \) is an integer. Substitute \( p = 2k \) into the equation \( 2q^2 = p^2 \): \[ 2q^2 = (2k)^2 = 4k^2. \] Simplify: \[ q^2 = 2k^2. \] This shows that \( q^2 \) is also even, which means \( q \) must also be even. Thus, both \( p \) and \( q \) are even, which contradicts our assumption that \( \frac{p}{q} \) is in its simplest form. Therefore, our assumption that \( \sqrt{2} \) is rational is incorrect. Hence, \( \sqrt{2} \) is irrational.
Conclusion:
Since assuming \( \sqrt{2} \) is rational leads to a contradiction, we conclude that \( \sqrt{2} \) is irrational.