The irrationality of \( \sqrt(2) \)

Consider a rational number fully reduced, \( \frac{p}{q}\) such that \(\frac{p^2}{q^2} = 2\). Investigate whether \(p\) or \(q\) are odd or even numbers leads to a contradiction which shows that \(2\) cannot be rational.

• \(p\) and \(q\) can't both be even, as otherwise they could be simplified.
• \(p\) and \(q\) must both be odd, as an odd \(\times\) odd is odd, so if \(p\) is odd, then \(q\) must be odd also, as \(p^2\) (odd) will divide by \(q^2\) to produce 2 only if \(q^2\) is also odd. Similarly, if \(q\) is odd, so too must \(p\) be.
• \(p\) is even as \(p^2\) is even. This can be seen by rearranging the original expression to get \(p^2 = 2q^2\).

These points lead to a contradiction.