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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.

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    $\begingroup$ As mentioned in the link, the things intuitionist logic can prove is a subset of the things classical logic can prove. And, since classical logic proves there is no rational and irrational number, we can't prove the existence of one in intuitionist logic. Since you seem aware of this, to clarify: you're asking us to find a model of intuitionist mathematics (which will fail to be a model of classical mathematics, in that classically false statements are true in it), and a number in that model which is both rational and irrational? $\endgroup$ – Mike Haskel Sep 17 '15 at 1:13
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    $\begingroup$ So what is your question? $\endgroup$ – Mike Haskel Sep 17 '15 at 1:16
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    $\begingroup$ Okay. I will try to build something. Be aware though that asking for "a number that, in some model ..." isn't quite right. The model will determine what the numbers are (and they won't be numbers in the classical sense). So, we're looking for "a model, containing some number that ..." $\endgroup$ – Mike Haskel Sep 17 '15 at 1:19
  • $\begingroup$ @MikeHaskel Yeah, I just typing lazy (and not quite used to Model Theory.) Feel free to edit to make it make more sense. $\endgroup$ – PyRulez Sep 17 '15 at 1:20
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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.

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  • $\begingroup$ Never expected that, good answer. Quick question(s): 1) What is the markovian model? 2) How do you add a number to a model? $\endgroup$ – PyRulez Jan 15 '16 at 0:47
  • $\begingroup$ @PyRulez I am referring to the "Russian school of constructivism" here, where all functions are computable. In the context of this answer, you could build $\Bbb R_M$ as the set of computable numbers (i.e. the set of real numbers whose digit strings can be calculated by a Turing machine), and $\Bbb R_\Omega$ as the set of real numbers calculable by a Turing machine with an oracle for $\Omega$. (In other words, any numbers you can get with field operations and limits of computable sequences etc., but with $\Omega$ also thrown in the mix.) $\endgroup$ – Mario Carneiro Jan 17 '16 at 8:07

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