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It is well known that the greeks proved the irrationality of $\sqrt{2} $ but they do so in a informal manner, i.e. they lack of formal definition of irrational number. Today we have several axiomatic constructions of the reals, and formal definitions of real numbers and irrational numbers, take the Dedekind cuts for example, a irrational number is now defined to be a Dedekind cut that is not a rational cut.

My question is:

"Is the classical proof of the irrationality of $\sqrt{2}$ correct/valid by modern standards?"

If it is not, can we use it's method (with some minor alterations of course) to prove that $\{x \in \mathbb Q : x^2 < 2 \text { or } x < 0 \}$ is not a rational cut ?

Thanks in advance.

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    $\begingroup$ Yes, it's valid. Irrationality is hardly a difficult concept: it simply means not rational. The proof shows that it can't be the quotient of two integers, which doesn't depend at all on any axiomatic construction of the reals. $\endgroup$ – Matt Samuel Mar 19 '18 at 4:10
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    $\begingroup$ You should define what you mean by the classical proof of the irrationality of $\sqrt 2$, at least the main idea. I would guess that the proof as presented does not stand up, but that the idea can be used to make a good proof. $\endgroup$ – Ross Millikan Mar 19 '18 at 4:14
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    $\begingroup$ This is one of the first proofs you see in any sort of 'intro to modern mathematics' course, so I'd certainly hope it is valid by modern standards.. $\endgroup$ – Grant M Mar 19 '18 at 4:16
  • $\begingroup$ Assume that the square root of two is the quotient of two integers that shares no common factor and deduce that they're both even contradicting the fact that that the original fraction was irreducible. $\endgroup$ – Vinicius L. Deloi Mar 19 '18 at 4:16
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    $\begingroup$ Yes it's valid. The ancients showed that if $m,n\in \Bbb N$ then $m^2\ne 2n^2,$ and therefore there cannot exist $x\in \Bbb Q$ such that $x^2=2.$ $\endgroup$ – DanielWainfleet Mar 19 '18 at 11:22

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