Just to demonstrate \(\sqrt{2}\) cannot be written as a fraction p/q, where p and q are integers:

If we can write \(\sqrt{2}\) as a fraction, we can write it as a completely reduced fraction. So suppose \(\sqrt{2} = \frac{p}{q}\) where \(p\) and \(q\) are integers, and \(\frac{p}{q}\) is a completely reduced fraction. That is, the GCD of p and q is 1. In particular, if we can write \(\sqrt{2}\) as a fraction p/q of two integers, we must be able to write it as a fraction where

p and q are not both even .

\(\sqrt{2} = \frac{p}{q} \\

2 = \frac{p^2}{q^2} \\

2q^2 = p^2\)

Since the left side is even, the right side must be even, so \(p^2\) must be even, and \(p\) must be even. But if \(p\) is even, \(p^2\) must be divisible by \(2^2 = 4\). If the right side of the equation is divisible by 4, so must be the left side of the equation: \(2q^2\) must be divisible by 4, which means \(q\) must be divisible by 2. But we've just shown that \(p\) and \(q\)

must both be even, which means \(\frac{p}{q}\) was not a completely reduced fraction, a contradiction. So it cannot be possible to write \(\sqrt{2}\) as a fraction of two integers, and \(\sqrt{2}\) must be an irrational number.

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