\(\sqrt{2}\) is not rational and rather an irrational number.


Proof:

Suppose it is a rational number and let \(\sqrt{2} = \frac{p}{q}\) where \(p\) and \(q\) are integers and \(q\) is not zero. Furthermore, assume that the fraction is in its simplist form. If it’s not then simplify it until you can’t simplify further. Now we can square each side:

$$ \begin{align*} (\sqrt{2})^2 &= \big(\frac{p}{q}\big)^2 \\ 2 &= \frac{p^2}{q^2} \\ p^2 &= 2q^2. \\ \end{align*} $$

From this we can conclude that \(p^2\) is even since it’s a multiple of 2. Since \(p^2\) is even, then \(p\) must be even as well. (Why? see proof here). Since \(p\) is even, we can write \(p\) as \(2r\) where \(r\) is an integer. Let’s plug in \(2r\) back in \(p^2 = 2q^2\) and simplify,

$$ \begin{align*} p^2 &= 2q^2 \\ (2r)^2 &= 2q^2 \\ 4r^2 &= 2q^2 \\ 2r^2 &= q^2. \end{align*} $$

But now \(q^2\) is even since it’s a multiple of 2! Since \(p^2\) is even and \(q^2\) is also even then they must have at least 2 as a common factor. But we said earlier that \(p^2/q^2\) is in its simplest form. Therefore, this is a contradiction and \(\sqrt{2}\) can not be a rational number!

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