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

Question

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!

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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.

Answer 1

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]$.

Answer 2

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.

Answer 3

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.