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Since $\Bbb Q[\pi]$ consists of expressions of the form $$a_0 + a_1\pi + \ldots + a_n\pi^n \quad\quad a_i \in \Bbb Q$$ for $n\in \Bbb N$, the following isomorphism of sets is immediate: $$\Bbb Q[\pi] \cong \bigsqcup_{n\ge 1} \Bbb Q^n$$ It follows that $\Bbb Q[\pi]$ is countable, as it is a countable union of countable sets. Therefore, $\Bbb Q[\pi]$ is a proper subset of $\Bbb R$, an uncountable set. Certainly, all numbers in $\mathbb R \setminus \mathbb Q[\pi]$ are irrational, but I haven't been able to come up with a concrete example of an element in $\mathbb R \setminus \mathbb Q[\pi]$. Could someone throw some light on this problem? Thanks!


Note: This question arises purely from curiosity, so I am not sure how easy or difficult it is to answer in terms of the required mathematical machinery.

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  • $\begingroup$ It might not be easy to prove, but elements such that $\sqrt{2},e$ or $\log(2)$ cannot be in $\mathbb{Q}[\pi]$. $\endgroup$
    – Marcos
    Jan 31, 2023 at 12:23
  • $\begingroup$ How do you know for sure? A proof shall quench my curiosity. @Marcos $\endgroup$ Jan 31, 2023 at 12:24
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    $\begingroup$ @Marcos, it is an open question whether $e + \pi$ is rational, which is even weaker than what OP is asking for. $\endgroup$ Jan 31, 2023 at 12:27
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    $\begingroup$ @esoteric-elliptic, this question is not quite a duplicate, but covers what you are asking for. More generally you could look up "algebraically independent numbers" (which are also stronger than what you are asking for). $\endgroup$ Jan 31, 2023 at 12:30
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    $\begingroup$ For $\sqrt{2}$ is easy indeed. If $\sqrt{2}=a_0+a_1\pi+\dots+a_n\pi^n$ this implies by squaring both sides that $2=a_0^2$ over $\mathbb{Q}$, which has no solutions. In general @MeesdeVries is right, it is a really difficult problem, which is not solved. $\endgroup$
    – Marcos
    Jan 31, 2023 at 12:32

3 Answers 3

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Let $\overline {\mathbb Q}$ denote the algebraic closure of $\mathbb Q$. Then $\overline{\mathbb Q} \cap \mathbb Q[\pi] = \mathbb Q$:

Suppose $p(\pi) \in \overline {\mathbb Q}$ for some polynomial with rational coefficients, then, as $\overline{\mathbb Q}$ is algebraically closed, either $\pi \in \overline{\mathbb Q}$ or $p(x)$ is a constant. We know that $\pi \notin \overline {\mathbb Q}$, hence $p(x) = q \in \mathbb Q$ is constant, which proves the statement above.

This shows that $(\overline {\mathbb Q}\cap \mathbb R) \setminus \mathbb Q \subseteq \mathbb R \setminus\mathbb Q[\pi]$, so we obtain many examples, including $\sqrt 2, \sqrt 3, \varphi$ etc. As the comments have already pointed out, it is very difficult to show that some non-algebraic numbers are not in $ {\mathbb Q}[\pi]$.

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Consider $t=\sqrt\pi$. If $t$ had the form $P(\pi)$ for some $P\in\Bbb{Q}[X]$, then by squaring the equality $\sqrt\pi=P(\pi)$ and rearranging the terms one would obtain a nonzero polynomial $R\in\Bbb{Q}[X]$ such that $R(\pi)=0$. On the other hand it is known (through Lindemann's theorem) that $\pi$ is transcendant, and such $R$ cannot exist.

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Enumerate all possible Turing machines $TM_1, TM_2, ...$, and define $x$ as the number with the infinite decimal expansion $0.d_1d_2d_3...$, where $d_i = \begin{cases} 5 \text{ if $TM_i$ halts on a blank input}\\ 6 \text{ otherwise} \end{cases}$

Then, $x$ is not computable, but everything in $\mathbb{Q}[\pi]$ is computable, so $x$ is an example of such a number.

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    $\begingroup$ You have to choose a computable enumeration of Turing Machines. Clearly there exists some enumeration of Turing Machines such that $TM_{2k}$ halts while $TM_{2k+1}$ doesn't. If you choose that enumeration, then $x = 0.65656565... \in \mathbb Q$. But a great answer of course! $\endgroup$ Jan 31, 2023 at 13:24

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