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Most of the examples of non-computable real numbers use some kind of a diagonalization construction over some turing computable model of computation. See Are there any examples of non-computable real numbers?.

I want to know if there are "natural" real numbers that are not computable. I'm having difficulty in formalizing what I mean by "natural". Here is a necessary condition for naturality: The description of that number should not mention any turing computable model of computation.

Ideally, this number should have existed in the literature even before Turing invented Turing machines. Somehow this is analogous to the way Solomon Feferman says:

Finally, we must take note of the fact that up to now, no previously (w.r.t. the day Gödel announced his incompleteness theorems) formulated open problem from number theory or finite combinatorics, such as the Goldbach conjecture or the Riemann Hypothesis or the twin prime conjecture or the P=NP problem, is known to be independent of the kinds of formal systems we have been talking about,not even of PA.

in http://math.stanford.edu/~feferman/papers/newaxioms.pdf. The parts in parenthesis have been added by me to put his quote in proper context. My question was partly motivated by this quote.

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Abhishek Anand
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    So you want a number which is interesting or useful for some reason other than because it's non-computable? – Jack M May 29 '14 at 22:02
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    Also, this section on Wikipedia seems relevant. If the answer to the titular question is "yes", it would seem there's no answer to this question. – Jack M May 29 '14 at 22:07
  • @JackM : yes, I want a number which is interesting or useful for some reason other than because it's non-computable – Abhishek Anand May 29 '14 at 22:10
  • Does the halting set ${ (e,n) : \text{program $e$ halts on input $n$}}$ qualify? The definition involves Turing machines, of course, but not models or diagonalization. The fact that it is interesting is related to the fact that it is not computable, but I wouldn't say that it's only interesting because it's not computable. – Trevor Wilson May 29 '14 at 22:15
  • Your definition (or at least sufficient condition) of natural is possibly the most unnatural thing I've ever heard of in mathematics :-) I've never heard of anyone defining a mathematical notion in terms of when it was first published. – Dan Piponi May 29 '14 at 22:17
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    Second question: when you mention the idea "all the reals that are needed in practice are computable," what do you mean by "in practice"? The practice of what? – Trevor Wilson May 29 '14 at 22:19
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    @TrevorWilson : It does not qualify since it's description uses Turing machines, which is a Turing complete model of computation. Sorry, I can't think of a good way to define "practice". – Abhishek Anand May 29 '14 at 22:30
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    How about $0^\sharp$ then? – Trevor Wilson May 29 '14 at 22:36
  • @Trevor: How about $0^{\dagger}$? – Asaf Karagila May 29 '14 at 22:36
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    @Asaf Heh, I would be very surprised if the OP said $0^\dagger$ qualified and $0^\sharp$ didn't :-) – Trevor Wilson May 29 '14 at 22:37
  • @Trevor: Fine. How about Random reals? Or even Jensen reals? – Asaf Karagila May 29 '14 at 22:40
  • @Asaf One could even ask about random reals in the non-set-theoretic definition of "random". I don't know whether it's possible to develop probability theory in a framework where every real is computable. – Trevor Wilson May 29 '14 at 22:44
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    $0^{\sharp}$ has the problem, of course, that it might not exist... :-) More seriously, the fact that we can meaningfully talk about 'worlds' where the computable reals are the only ones that exist means that any examples are bound to seem at least somewhat artificial. – Steven Stadnicki May 29 '14 at 22:58
  • @Trevor Wilson: probability theory requires noncomputable reals to prove simple things like "every closed subset of $2^{\omega}$ of positive measure is nonempty". – Carl Mummert May 30 '14 at 00:30
  • @CarlMummert Doesn't the fact that the empty set has measure zero follow directly from the definition of Lebesgue measure? – Trevor Wilson May 30 '14 at 00:32
  • @Trevor Wilson: yes, it does in ZFC, but not in the weaker frameworks where we would study computable measure theory. – Carl Mummert May 30 '14 at 00:37
  • I seem to recall that things like the probability that a random Turing machine eventually halts are examples of this sort of thing. – Michael Hardy May 30 '14 at 00:41
  • @Michael Hardy: see my comments in this thread: http://math.stackexchange.com/questions/462790/are-there-any-examples-of-non-computable-real-numbers#comment996656_462795 – Carl Mummert May 30 '14 at 00:50

2 Answers2

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The correct intuition is that there are no examples of particular natural noncomputable real numbers.

One significant obstacle to finding an example is that computability is more directly about sets of naturals, not about real numbers. Most examples of noncomputable real numbers are constructed by coding a noncomputable set of naturals into a real number. The coding method is always somewhat arbitrary, which goes against the uniqueness of the real number being constructed.

Examples of coding methods

Here are two examples of different coding methods. The trouble is that the best way of encoding information into a real number seems to be to use the decimal (or binary, or base 813) expansion in some way, but there is no obvious "canonical" or "best" way to do this. Suppose I have an infinite set $A$ of natural numbers. I can make a real number $r$ that "computes" $A$ in several ways:

  • I can make it so that $r \in [0,1]$ and the decimal expansion of $r$ is the characteristic function of $A$. I could also do this so that $r \in [0,1]$ and the binary expansion of $r$ is the characteristic function of $A$. Neither of these seems more canonical than the other.

  • I can make it so that the "gaps" between the nonzero digits in the binary expansion of $r$ tell me about $A$. So if $A = \{1, 3, 5, \ldots\}$ then I take $$ r = 0.1010001000001\ldots$$

Carl Mummert
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    Exactly!${}{}{}{}$ – dtldarek May 30 '14 at 00:48
  • Is there a "natural" non-recursive set of positive integers then? – Conifold May 30 '14 at 01:39
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    @Conifold: There are r.e. sets that are not recursive. The most "natural" one is probably the set of codes of proofs from PA. – UserB1234 May 31 '14 at 18:14
  • Thanks @Carl Mummert . I want to ask 2 questions to confirm that I understand your answer:

    1. Did you emphasize the word "particular" to rule out answers such as "a randomly generated real"? Indeed, I wanted examples of particular reals in this sense.

    2. I'm not sure I understand your last clause "which goes against the uniqueness of the real number being constructed". How do arbitrary coding methods go against the uniqueness of the real being constructed?

     – Abhishek Anand Jun 02 '14 at 15:26
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    @Abhishek: (1) Yes, I emphasize "particular" to rule out constructions that simultaneously produce many reals each of which is noncomputable. The cardinality argument showing not all reals are computable is the epitome of this. For (2) I will add two examples to the answer. – Carl Mummert Jun 02 '14 at 15:40
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    @CarlMummert I would include also continued fractions (my favorite). – dtldarek Jun 04 '14 at 20:03
  • @UserB1234 You probably meant the set of codes of theorems of PA. The set of codes of proofs from PA is computable (assuming, of ourse, reasonable coding scheme). – Andreas Blass Jan 01 '18 at 00:18
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I'm not aware of any previously known number which is not computable, but one can easily construct one that does not involve any Turing machines or any other similar ways of computation (e.g. $\lambda$-terms, grammars and so on).

To give an example, take a look at Wikipedia's list of undecidable problems and pick one you like.

For example, we could take the matrix mortality problem and make the inputs into a sequence, e.g. pick your favourite bijection $\mathbb{N} \to \mathbb{Z}^{2\cdot 15^2}$ and create a sequence $(a_n)_{n \in \mathbb{N}}$ of pairs of integer $15 \times 15$ matrices. Then define $f : M_{15\times 15} \times M_{15\times 15} \to \{0,1\}$ as $f(A,B) = 1$ if there exists $k \in \mathbb{N}$ and function $h : \{0,..,k\} \to \{A,B\}$ such that $\prod_{i = 0}^k h(i) = \mathbf{0}$ and $f(A,B) = 0$ otherwise.

We can now define our number as $$\sum_{k \in \mathbb{N}}\frac{f(a_k)}{2^k}$$

and because matrix mortality problem is undecidable, then the above number is uncomputable.

I know this is not exactly what you were looking for, but perhaps it might help $\ddot\smile$

dtldarek
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  • I was going to give this answer with Goodstein's theorem :) – Ryan Reich May 30 '14 at 01:20
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    @RyanReich And how your sequence would look like? As far as I know, the Goodstein's sequence is computable. – dtldarek May 30 '14 at 08:56
  • Of course you meant to write $f$ rather than $g$. – r.e.s. Jun 01 '14 at 22:48
  • @r.e.s. Yes, thank you for noticing. – dtldarek Jun 01 '14 at 23:31
  • Thanks @dtldarek . Your parametrized construction of an uncomputable real is perhaps the most general one I've seen so far. Also, you pointed out holes in my definition of "natural". – Abhishek Anand Jun 04 '14 at 19:31
  • do you really need a bijection? I think a function with a computable right inverse will suffice. – Abhishek Anand Nov 11 '14 at 16:34
  • Also, there is a (little ?) more to the proof. Assuming (for the sake of contradiction) that there exists a machine that can compute an arbitrarily close rational approximation to the proposed number ($\sum_{k \in \mathbb{N}}\frac{f(a_k)}{2^k}$), one has to decide an arbitrary instance of the undecidable problem. – Abhishek Anand Nov 11 '14 at 18:56
  • @Abhishek The only thing you need is to reduce an undecidable problem to that number (i.e. if we had such number, we could solve an undecidable problem). You are correct that you don't need the bijection, only a way to calculate the index to look at given some input (hence, also that any input is represented by some index). I prefer to use bijections, because they are easier to reason about. Finally, my answer was just a sketch, not a full proof (although it should be easy enough to make it into a full proof) - I'm sorry if you found it unhelpful or, worse, misleading in any way. – dtldarek Nov 11 '14 at 20:27
  • @dtldarek Your answer was very helpful. I was just wondering if you already had an easy reduction to that number (when represented as a function that computes arbitrarily close rational approximations to it). – Abhishek Anand Nov 11 '14 at 21:47