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 $\mathbb{R}$?

  • 6
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

If $p\in\Bbb Q(x)$ and $p(\pi)=\sqrt 2$ then $p(\pi)^2=2$, which is impossible because $\pi$ is not algebraic.

  • $\begingroup$ 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
  • 2
    $\begingroup$ @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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.