# $Q(\pi)\neq \mathbb{R}$

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 $\mathbb{R}$?

• 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? – Arthur Oct 11 '17 at 13:39
• Do you know about the distinction between countable and uncountable sets? – Noah Schweber Oct 11 '17 at 13:39
• It does not contain $\sqrt{2}$. – Orest Bucicovschi Oct 11 '17 at 13:42
• @orangeskid How do you know? ;) – M. Winter Oct 11 '17 at 13:46
• 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. – 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.
• Of course, this assumes that the OP knows that $\pi$ is not algebraic. – Noah Schweber Oct 11 '17 at 13:53
• @NoahSchweber ... OK. I also upvoted this. My first reaction to the question was that $\sqrt{\pi} \not\in \mathbb Q(\pi)$. – GEdgar Oct 11 '17 at 14:00