# What's an example of an element in $\mathbb R \setminus \mathbb Q[\pi]$?

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.

• It might not be easy to prove, but elements such that $\sqrt{2},e$ or $\log(2)$ cannot be in $\mathbb{Q}[\pi]$. Jan 31, 2023 at 12:23
• How do you know for sure? A proof shall quench my curiosity. @Marcos Jan 31, 2023 at 12:24
• @Marcos, it is an open question whether $e + \pi$ is rational, which is even weaker than what OP is asking for. Jan 31, 2023 at 12:27
• @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). Jan 31, 2023 at 12:30
• 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. Jan 31, 2023 at 12:32

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 $${\mathbb Q}[\pi]$$.

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.

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.

• 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! Jan 31, 2023 at 13:24