Of course in regular logic, the answer is there aren't any. But in intuitionistic logic, there might be, as seen by this answer: https://math.stackexchange.com/a/1437130/49592.
My question is, as per that answer, what is a specific number that is neither rational nor irrational (note that the link above uses a different definition of irrational than the normal one).
Also per the question, you would have to cite what model you are using.
Edit:
Quoting from Mario's Answer:
On the other hand, there should be a model of the reals with
constructive logic + ¬LEM, such that there is a non-rational
non-irrational number, and I invite any constructive analysts to
supply such examples in the comments.
Best Answer
I've thought about this question for a while without an answer. The key is to consider the structure of the constructible real numbers. I was actually a bit cavalier with my original definition, "$x$ is irrational if $|x-p/q|<q^{-2}$ has infinitely many coprime solutions". The problem lies in what is meant by "infinitely many" here. If it means that there is an injection from $\Bbb N$, then we get the same set as described by the other definition using infinite continued fractions. In particular, if $x$ is irrational, then there is a function $f:\Bbb N\to\Bbb Q$ that converges to $x$, and under the "Russian constructivist" camp, we can assume that $f$ is a computable function, so $x$ is computable. And obviously the rational numbers are computable.
Thus Chaitin's constant $\Omega$ is neither rational nor irrational.
To be clear, in constructivist logic we don't necessarily know that $\Omega$ even exists; in fact, that's the whole point. But it does mean that we can take the Markovian model of constructible reals $\Bbb R_M$ and add $\Omega$ to get a model $\Bbb R_M[\Omega]=:\Bbb R_{\Omega}$ which can be viewed as a subset of $\Bbb R$ (in the ambient universe where LEM is true). Then $\Bbb Q=\Bbb Q_M=\Bbb Q_{\Omega}$, and $\Bbb I_M=\Bbb I_{\Omega}\subsetneq\Bbb I$, with $\Omega\in\Bbb I\setminus \Bbb I_{\Omega}$, yet at the same time $\Omega\in \Bbb R_{\Omega}$ by definition, so we have $\Bbb Q_{\Omega}\cup\Bbb I_{\Omega}\subsetneq\Bbb R_{\Omega}$. (We can even say $\Bbb Q_{\Omega}\cup\Bbb I_{\Omega}=\Bbb R_M$ but this is only valid as a proof in the full universe, using LEM.)
I mentioned that "infinitely many" had two interpretations above. The other one is that it is not finite, and this yields the poorer definition "$x$ is irrational if $x$ is not rational". In this case it is clear that no number can be neither rational nor irrational, because if it is not rational then it must be irrational by definition.