Consider any (random) irrational number $N$ such that there is an algorithm that defines this number. Now we can be interested in the exact answer to the following question: “Does the decimal expansion of $N$ contain all finite sequences of decimal digits?”. As far as I understand, there are only two possibilities: Yes and No. Note that if the answer is No (for some particular irrational number), there exists some smallest integer $X$ such that it will never appear in the decimal expansion of $N$.
I want to ask the following question: is the problem of finding the exact Yes/No answer for any $N$ computable? For example, consider Pi. If the answer to my question is “Yes”, then there exists some algorithm that will terminate with the output containing only one bit (0 or 1) that answers to this (or this) question. Is it possible at all? In other words, we don’t know if the answer is Yes for Pi, but do we know if it is possible to prove this?

lyrically wicked
  • 740
  • 4
  • 10
  • Are you interested in provability or computability? Lots of things are provable but not computable – Stella Biderman Mar 10 '17 at 06:02
  • @StellaBiderman: this question is only about computability, but yes, I am also interested in provability (assuming that the proof contains a finite number of symbols) of Yes/No answer for any (random) irrational number. The problem is: what formal language would one use for such proofs? Set theory language? Which version of the language? – lyrically wicked Mar 10 '17 at 06:20
  • 2
    Let $c$ and $d$ be two irrational numbers whose decimal expansions do and do not have this property respectively. Suppose there is a Turing machine $M$ which you want to know whether it ever halts. Let $x$ be the irrational number whose $n$th digit is defined as follows: run $M$ for up to $n$ steps; if it has not halted, take the $n$th digit of $c$; if it did halt, take the $n$th digit of $d$. Now if you can compute whether $x$ has the desired property, you can decide whether $M$ ever halts. –  Mar 10 '17 at 07:21
  • @Rahul: please, see the edited question. – lyrically wicked Mar 10 '17 at 07:51
  • I have just seen and downvoted it. –  Mar 10 '17 at 07:51

1 Answers1


First: It's perhaps a nitpick, but the set of all numbers which can be 'defined by some algorithm' is going to be at most countable; so you must be clearer about exactly what you mean by 'random' in this context when you say "Consider any (random) irrational number such that...(etc.)". The naive statement "Choose a natural number at random, and then..." is not rigorous enough to be evaluated to start with.

Second: You say, "As far as I understand, there are only two possibilities: Yes and No."

Well, that is a common and reasonable philosophical stance. But a constructivist would not agree with this formulation in the abstract.

Thirdly: Let $C$ be Champernowne's constant; and $C_n$ be the $n$th digit in the decimal expansion of $C$. Let $T$ be some arbitrary Turing machine; and let $T_n$ be $0$ if $T$ halts at or before step $n$, and $1$ otherwise. Let $H(T)$ be the number whose $n$th decimal digit is $C_n$ if $T_n=1$; and $0$ otherwise.

Then the question "Does the decimal expansion of $H(T)$ contain all finite sequences of decimal digits?" is equivalent to the question "does $T$ not halt?". But this is not always provable for arbitrary $T$ (for whatever particular formal system one chooses); it's the Halting Problem. So in general one would say that one cannot demonstrate that, for all arbitrary $T$, either a proof exists that $H(T)$ meets your constraints, or a proof exists that $H(T)$ does not meet your constraints.

Thus, there are numbers $N$ such that there is an algorithm that defines this number; for which we cannot say we have an "exact answer" to your question, as you describe it.

As far as how all this applies to $\pi$... well, that's rather too specific, so who knows? :)

Chas Brown
  • 1,620
  • 1
  • 10
  • 13