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Statement: $\exists a,b \notin \Bbb Q : a^b \in \Bbb Q$

Proof: Set $\lambda=\sqrt 2^\sqrt2$

if $\lambda\in\Bbb Q$, then $\sqrt 2^\sqrt2 \in\Bbb Q,\space a=b=\sqrt 2 \notin \Bbb Q$.

if $\lambda\notin\Bbb Q$, then $\lambda^\sqrt2=\left(\sqrt2^{\sqrt2}\right)^{\sqrt2}=\sqrt 2^{\sqrt2\sqrt2}=\sqrt2^2 =2,$, so take $a= \lambda,\space b=\sqrt 2$

I watched a video saying that this is not a valid proof, because it doesn't consider the case that $\lambda \in\Bbb Q$ is a meanigless statement or something like that as there is no way to check if a number is rational or not. But... why would that spoil the proof?

PinkyWay
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Tony
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  • This proof is absolutely fine. Worrying whether $\sqrt{2}^{\sqrt{2}}$ is an entirely different issue. – pisco Jul 22 '17 at 14:54
  • The proof is alright and I believe it's a quite known one. $\lambda \in Q$ isn't a meaningless statement considering that it has either a true or false value. – Stefan4024 Jul 22 '17 at 14:54
  • @Stefan4024 Here is the exact statement that was made: https://youtu.be/Z7ynOygOj5c?t=3m50s – Tony Jul 22 '17 at 15:03
  • The proof is correct if you assume the Law of excluded middle (https://en.wikipedia.org/wiki/Law_of_excluded_middle) like most people do. Some people, the so called intuitionists, do not allow this kind of proof but require to give explicitly a and b. But the above proof doesn't. – tiefi Jul 22 '17 at 15:06
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    Before looking at the proof, what can the sentence $$[\exists c :a\notin \mathbb Q, b \notin \mathbb Q, c=a^b \in \mathbb Q]$$ even mean? – Did Jul 22 '17 at 15:30
  • @Did In the original it was: $\exists a,b \notin \mathbb Q : a^b \in \mathbb Q $ but I think it's clear enough – Tony Jul 22 '17 at 15:59
  • It is not. By far. – Did Jul 22 '17 at 16:00
  • @did I don't understand your confusion. The statement is "there exist two irrationals, a and b, so that $a^b$ is rational". Which is perfectly clear. The proof seems valid (assuming irrational to an irrational power is well defined, and $(a^b)^c = a^{bc}$ is proven) . But I'll have to watch the video to see precisely what the objection was. If it is what the OP says it isn't valid, but it may (or may not) be that the objection is not being accurately described. – fleablood Jul 22 '17 at 16:10
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    http://math.andrej.com/2009/12/28/constructive-gem-irrational-to-the-power-of-irrational-that-is-rational/ – Patrick Stevens Jul 22 '17 at 16:11
  • @fleablood You are referring to the corrected version of the title, please read the previous version if you want to "understand" the point in my comment. Or do you think that $$[\exists c:a\notin\mathbb Q,b\notin\mathbb Q,c=a^b\in\mathbb Q]$$ is legit? – Did Jul 22 '17 at 16:17
  • Okay, two seconds in and it's clear this person is an intuitionist as described by tiefi. At 5:22 we get the the gyst of his argment. "The law of excluded middle is legitimate if we are talking about finite sets" and and 3:50 "$\sqrt{2}^{\sqrt{2}}$ is meaningless because there are an infinite number of integers and you can't test them all". If you aren't an intuitionist you may believe in the excluded middle and you may believe in the untested. – fleablood Jul 22 '17 at 16:25
  • I think think the statement as you wrote it is ungrammatical. It's meaning is clear and it's as legit as any other badly stated statement but it the equivalent of an english language dangling participal or misplaced modifier. ("Walking along the passage to the town, a memorial archway blocked our way."). But no, I did not see the original title. It's worth pointing out but it's a bit heavy handed to say it can't mean anything or is "far from clear". – fleablood Jul 22 '17 at 16:34
  • So..his arguments are 1)that the law of excluded middle is only legitimate for finite statements and 2)Statements that "x is this" or "x is that" are only meaningful if they can be verified directly. Thing is, if you accept the law of excluded middle for infinite cases or reject the intutitionist premise that only the directly verifiable are meaningful, those objections are not valid. Which I guess is his point. Whether you accept or reject those premises is a matter of "religious dogma". Maybe. Not sure I agree. But a valid viewpoint. – fleablood Jul 22 '17 at 16:42

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