Decidability of transcendental numbers
Kristoffer's solution can be used to show that, assuming reals are represented so that we can compute limits of sequences of reals which are computably Cauchy. Recall that a sequence $(a_n)_n$ is computably Cauchy if there is a computable map $f$ such that, given any $k$ we have $|a_{m} - a_{n}| &lt; 2^{-k}$ for all $m, n \geq f(k)$. The standard representations of reals are like that, for example the one where a real is represented by a machine that computes an arbitrarily good rational approximation. (We can also speak in terms of computing digits, but then we have to allow negative digits. This is a well known issue in computability theory of the reals.)


  Theorem: Suppose $S \subseteq \mathbb{R}$ is a subset such that there exists a computable sequence $(a_n)_n$ which is computably Cauchy and its limit $x = \lim_n a_n$ is outside $S$. Then question "is a real number $x$ an element of $S$" is undecidable.


Proof. 
Suppose $S$ were decidable. Given any Turing machine $T$, consider the sequence $b_n$ defined as
$$b_n = \begin{cases}
a_n &amp; \text{if $T$ has not halted in the first $n$ steps,}\\
a_m &amp; \text{if $T$ has halted in step $m$ and $m \leq n$.}
\end{cases}$$
It is easy to check that $b_n$ is computably Cauchy, therefore we can compute its limit $y = \lim_n b_n$. Now we have $y \in S$ iff $T$ halts, so we can solve the Halting Problem. QED.

There is a dual theorem in which we assume the sequence is outside $S$ but its limit is in $S$.

Examples of sets $S$ satisfying these conditions are: an open interval, a closed interval, the negative numbers, the singleton $\lbrace 0 \rbrace$, rational numbers, irrational numbers, transcedental numbers, algebraic numbers, etc.

A set which does not satisfy the conditions of the theorem is the set $S = \lbrace q + \alpha \mid q \in \mathbb{Q}\rbrace$ of rational numbers translated by a non-computable number $\alpha$. Exercise: is $S$ decidable?