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This question is directly inspired by "Does Pi contain all possible number combinations?". I would like to state firstly for the record that I have no serious number theory education. I think I probably could solve all exercises from Elements of Number Theory by Vinogradov given enough time but that is pretty much my limit.

Non the less I have given some thought to similar questions. They pup up in my subject (Dynamical Systems) on several places. I learned about weakly and strongly irrational numbers from trying to understand proofs (yes Arnold and Moser proofs are different and they do not prove the identical statements) of KAM theorem. All possible combinations argument appears when one uses symbolic dynamics to show existence of transitive orbit in let say logistical map.

$\pi$ is just an example of irrational number. As every student of Calculus in U.S. learned during the fist semester each real number has a decimal representation. If the number is irrational that decimal representations contains infinite number of digits without repeating patterns. So the natural questions are:

  1. Does a decimal representation of a weakly irrational number contain all possible number combinations?
  2. Does a decimal representation of a strongly irrational number contain all possible number combinations?
  3. Does a decimal representation of an algebraic irrational number contain all possible number combinations?
  4. Does a decimal representation of a transcendental irrational number contain all possible number combinations ($\pi$ is just a special case)?
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Predrag Punosevac
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    Some of this has been discussed here many times before. Liouville's number $\sum 10^{-i!} = 0.110001000000000000000001000\ldots$ is an example of a transcendental number that does *not* contain every possible sequence of digits. – MJD Jan 27 '15 at 02:10
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    You say "$\pi$ is just a special case." Do you think $\pi$ does or doesn't contain all possible finite sequences? If you think it does, then might I recommend you read again the post you linked to. If you think it doesn't, then why would you think $\pi$ would behave differently than others? – JMoravitz Jan 27 '15 at 02:33
  • What is a "strongly irrational" number? Is it the same as a strong Liouville number? – TonyK Jan 27 '15 at 10:08
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    Despite it is usually very hard to prove or disprove any specific number is normal, we do know the set of non-normal numbers has measure zero. If you pick a real number randomly, you will be almost sure it contains all possible finite sequences. – achille hui Jan 27 '15 at 10:21

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The current state is that determing if an irratioal number contains every finite sequence is completely out of reach.

Every algebraic irrational number and numbers like $\pi$ and $e$ are widely believed to be even NORMAL ( which is a stronger property ), but there is no irrational number which was not specially constructed for which it could be proven or disproven, that it contains every finite sequence.

Peter
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