This question came from the answer I've given to the question An easy example of a non-constructive proof without an obvious "fix"?.

Rereading my answer I had some doubt about the existence of an uncomputable integer number, as that proposed in my answer. For convenience I report here the definition:

Let $\gamma$ be a real number defined as, say: $\gamma= > 12.23873560031\cdots$ and define: $c=3*10^2+3*10^5+3*10^{10}+\cdots$ where we have chosen the digit $3$, but it could be any digit, and the exponents of $10$ are the positions of this digit in the decimal expansion. Then $c$ may be a number or not, depending on whether this expansion has a finite number of $3$s in its expansion. If we chose $ > \gamma= \zeta(5)$ (the Riemann zeta function), we don't know today if it has a finite number of digits $3$ (so far as I know), so we don't know if $c$ is a number or not. If someone proves that $\zeta(5)$ actually has a finite number of $3$ in its expansion then we will have two possibilities: The proof is not be constructive, i.e. we cannot find the position of the last $3$, and $c$ is really a number but it is a non-computable integer. Or the proof is constructive and we know the position of the last $3$ and the number $c$ is computable.

Searching on the web for "uncomputable integers" I've not found pertinent pages (while there are a lot of pages for "uncomputable numbers", but related to real numbers).

So, there exist uncomputable integer numbers or my answer was completely wrong?

  • $\begingroup$ Even all algebraic irrational numbers are computable. "Computable" means that there is a (finite ) algorithm that can (in principle) calculate ALL the digits of the number. In the case of an integer $N$ , the program "print($N$)" already does the job, since the input of $N$ is finite. So, $N$ clearly must be computable. $\endgroup$ – Peter Jul 24 at 18:11

No, there is not. If you consider an integer with the usual meaning, it is finite information. All finite information can be computed.

For example, you can build a machine $M$ and the proposition "$M$ halts on input $0$" can be impossible to prove in any known usual theory. So the integer $i$ defined by $i=0$ if $M$ halts on input $0$ else $i=1$ is not "findable". But anyway it is computable.

Remember that computable means : A program exists that can compute that information. But the existence can sometimes be proved with non constructive steps...

  • $\begingroup$ So, if I well understand, the number $c$ in my definition is computable but not "findable"? And what "findable" means exactly? $\endgroup$ – Emilio Novati Jan 26 '15 at 22:42
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    $\begingroup$ If $c$ is an integer, it is computable. But the proposition "this program P computes c" can be non provable for all P even if it's true for some P. $\endgroup$ – Xoff Jan 26 '15 at 22:45

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