Classic proof is that we assume $\sqrt{2}$ is rational, then so on.
But can we prove it directly, not using proof by contradiction?
If yes, how? If not, then why?
Thank you!
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You can have a look at : Question on a constructive proof of irrationality of $\sqrt 2$ – Watson Nov 08 '16 at 07:46
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The problem with trying to prove such a thing directly is in the definition of an irrational number. We define an irrational number as some number that can't be written in the form $\frac{p}{q}$, where $p$ and $q$ are both integers. So to prove that $\sqrt{2}$ is not irrational, we have to prove that it cannot be written in such a form, which means we have to do a contradiction somewhere.
Ashwin Trisal
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By Eisenstein's criterion, $x^2-2$ is irreducible. Thus $\sqrt{2}$ is not rational. You can prove that $\sqrt{2}$ is real using the mean value theorem.
pancini
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