The following is a quote from Denis' answer in another thread where the context centers on creating a noncomputable real number. It's old and I'm new here so I can't comment on the answer itself, thus I'm asking the question here (note, you shouldn't have to read the other thread for this):

Are there any examples of non-computable real numbers?

"Any language can be turned into a number, by setting the $i^{th}$ decimal to 1 if the $i^{th}$ word is in the language, and to 0 otherwise. So we can build for instance the number $H$, which describes the halting problem and is therefore uncomputable."

I'm trying to grasp this example. I assume the language of course has an alphabet, like {a, b, c, ..., x, y, z}. We then have an enumeration of all combinations of letters (e.g., s1 = a, s2 = b, s3 = c, ... s27 = aa, s28 = ab, s29 = ac, ...). From there, we say each $s_i$ is either a word or it isn't, where then i = 1 or i = 0, respectively.

My question is, assuming the English language is used where the largest word will contain only a 'small' number of letters, wouldn't we just produce a number $H$ that has a trailing string of 0's? I assume $H$ must be in (0, 1) so as to not be an 'infinitely large' integer...

If $H$ is guaranteed to have a trailing string of 0's, then it will be a rational number, yes? And, aren't all rationals computable by definition?

Ok, so I'm clearly missing something because I assume the answer provided by Denis is correct. A clearer explanation is appreciated.


You're right that if the English language (or any other finite language) is used you get a computable number. Denis's example is not the English language, though: it's the language that encodes the halting problem. That is, this "language" consists of all words that are encodings of Turing machines that halt.

  • $\begingroup$ Ok, so he meant 'formal language' where a word represents a program that can be run by a Turing machine. That makes more sense. (slap face)... :) Thank you. $\endgroup$ – AplanisTophet Jun 14 '17 at 2:48

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