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How cani demonstrate that $Q(\pi)\neq \mathbb{R}$, that is the field of rational functions of $Q(x)$ evaluated in $\pi$ is a strict subset of $\mathbb{R}$?

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    $\begingroup$ A counting argument would work ($\Bbb Q(\pi)$ is countable, although it requires a bit of work to show, while $\Bbb R$ is not, which also requires some work to show). Is that allowed? $\endgroup$ – Arthur Oct 11 '17 at 13:39
  • $\begingroup$ Do you know about the distinction between countable and uncountable sets? $\endgroup$ – Noah Schweber Oct 11 '17 at 13:39
  • $\begingroup$ It does not contain $\sqrt{2}$. $\endgroup$ – Orest Bucicovschi Oct 11 '17 at 13:42
  • $\begingroup$ @orangeskid How do you know? ;) $\endgroup$ – M. Winter Oct 11 '17 at 13:46
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    $\begingroup$ It's common knowledge that $\pi$ is transcendental, but hard to prove. One can avoid this fundamental fact by reasoning in the alternative case too: even if $\pi$ were algebraic, we couldn't have $\Bbb Q(\pi)=\Bbb R$, either by cardinality argument, or the fact $\Bbb R$ contains algebraic elements of arbitrarily high degree. $\endgroup$ – anon Oct 11 '17 at 13:57
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If $p\in\Bbb Q(x)$ and $p(\pi)=\sqrt 2$ then $p(\pi)^2=2$, which is impossible because $\pi$ is not algebraic.

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  • $\begingroup$ Of course, this assumes that the OP knows that $\pi$ is not algebraic. $\endgroup$ – Noah Schweber Oct 11 '17 at 13:53
  • $\begingroup$ Against the general advice of this forum, the OP asked a homework-type question, but gave us no hints what he knows. So he has no basis for objecting to anything used in an answer. $\endgroup$ – GEdgar Oct 11 '17 at 13:55
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    $\begingroup$ @GEdgar I wasn't objecting to the answer (I upvoted it in fact) - I was just pointing out that it relies on a nontrivial fact. $\endgroup$ – Noah Schweber Oct 11 '17 at 13:55
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    $\begingroup$ @NoahSchweber ... OK. I also upvoted this. My first reaction to the question was that $\sqrt{\pi} \not\in \mathbb Q(\pi)$. $\endgroup$ – GEdgar Oct 11 '17 at 14:00

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