Let's assume a have an arbitrarily long number, take π for example. Since we know π is infinite, there will at some point be a group of numbers like "2015201620172018...", correct? My question is, for any arbitrarily long number, is there a mathematical way of determining the position of a specified set of numbers in said long number? The only conceivable way I can think of is similar to how we find the coefficient of x using the binomial theorem, but that leaves the problem of determining the number in the format of a binomial expansion.

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    $\pi$ is a finite irrational number, so with an non-recurring decimal expression. Not all irrationals have every finite sequence of digits in their decimal expression. Even so, for most of those which do, there is no shortcut for finding the first location where a given sequence appears (if you think there are an uncountable number of such numbers and a countable number of shortcuts) – Henry May 09 '22 at 15:00
  • "*...similar to how we find the coefficient of $x$ using the binomial theorem...*" That problem and this are completely different and have absolutely nothing to do with one another. – JMoravitz May 09 '22 at 23:10
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    Does this answer your question? [Does $\pi$ contain all possible number combinations?](https://math.stackexchange.com/questions/216343/does-pi-contain-all-possible-number-combinations) – JMoravitz May 09 '22 at 23:14

1 Answers1


A string being infinite does not necessarily mean any finite string is included in it. Take for example the decimal expansion


Clearly there are an infinite number of decimals, but that does not mean that you can find any finite string of numbers you want in there. Furthermore, the answer to your question is that no, there is no way to do so for any arbitrary number which satisfies the conditions you need. Indeed even proving that a number has such a property is an extremely difficult thing to do to begin with.

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  • I should have explained better, I meant to include that I am not talking about rational numbers(because of the problem you have pointed out above) but for any irrational number, there should be sequences that occur due to their infinite nature, sort of like how the library of babel works(every possible combination of every possible character) – Déjan Venter May 09 '22 at 15:03
  • @DéjanVenter That's not necessarily true either. See the answer to [this question](https://math.stackexchange.com/questions/216343/does-pi-contain-all-possible-number-combinations). – Lorago May 09 '22 at 15:05
  • @DéjanVenter $0.10110111011110111110111111\dots$ where we continually append ever-increasing lengths of $1$'s between individual $0$'s is an irrational number and yet it does not have any digits other than $0$ or $1$ appearing... and notably doesn't have even the substring "2" appearing in it, much less "2015201620172018". Go back and revisit the *actual* definition of irrational numbers and do not conflate them with other unrelated concepts like "normal" numbers. – JMoravitz May 09 '22 at 23:14