Question

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

Hint:

To prove irrationality, assume the opposite and reach a contradiction using properties of even and odd numbers.

Answer 1

Step 1: Assume the contrary.
Suppose \(\sqrt{2}\) is rational. Then it can be written as \(\sqrt{2} = \frac{p}{q}\), where \(p\) and \(q\) are co-prime integers and \(q \neq 0\).
Step 2: Square both sides.
\[ 2 = \frac{p^2}{q^2} \implies p^2 = 2q^2 \] This shows that \(p^2\) is even, hence \(p\) is even.
Step 3: Let \(p = 2k\).
\[ p^2 = (2k)^2 = 4k^2 \implies 4k^2 = 2q^2 \implies q^2 = 2k^2 \] Thus, \(q\) is also even.
Step 4: Contradiction.
Since both \(p\) and \(q\) are even, they have a common factor 2, contradicting the assumption that \(p\) and \(q\) are co-prime.
Step 5: Conclusion.
Hence, \(\sqrt{2}\) is irrational.