We do not know whether $\pi$ is normal or it is not and many other weaker statements, e.g. (*) $\pi$ contains infinitely many $0$s.

Inspired by the Godel's incompleteness theorem that there are some true statements which cannot be proved, do we currently know that

1) if $\pi$ is normal then there is a proof of it,

2) if $\pi$ is not normal then there is a proof of it?

I suppose that we do not know much in this direction, however is there at least a proof for a much weaker property of $\pi$ of a similar kind to (*)?


Does Pi contain all possible number combinations?


Dávid Natingga
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  • Your title seems to be a bit misleading. Are you asking about whether a proof _does_ exist either way, or are you asking whether we know if a proof _could_ exist? – Semiclassical Aug 02 '14 at 21:02
  • You may be inspired by things like: If Goldbach is false then there exists a proof of its falseness (i.e. a counterexample that can be tested in finite time). Normality (or not normality) is different (it seems) in that either proof would "use" all infninitely many digits. – Hagen von Eitzen Aug 02 '14 at 21:12
  • That title is more helpful, thanks. (As for the answers? I suspect the answer is "no" to all of them.) – Semiclassical Aug 02 '14 at 21:14
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    Intuitively, that's true about Goldbach - that is, if it is undecidable, it is intuitively true. But I don't think that means that you can find a counter-example if it is false. If it is undecidable, then there is a model where it is true and a model where it is false; it's just that the model where it is true is the axiom we'd choose to model our intuition. Maybe I'm wrong, @HagenvonEitzen – Thomas Andrews Aug 02 '14 at 21:17
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    No.${}{}{}{}{}{}{} $ – Andrés E. Caicedo Aug 02 '14 at 22:28
  • @thomas If the goldbach conjecture is FALSE, then this can be proven in principle, although it may be unfeasible to find a counterexample. If it is UNDECIDABLE in PA, this fact cannot be proven in PA itself, otherwise, PA would have proven that no counterexample exists, hence that goldbachs conjecture is true. Maybe, ZFC can prove that goldbachs conjecture is undecidable in PA, then it must be true. Finally, if Goldbach's conjeture is UNDECIDABLE in ZFC, then we are lost because this cannot be proven in ZFC. – Peter Jan 18 '15 at 12:16
  • If goldbachs conjecture is provably true, then it is true in every model. – Peter Jan 18 '15 at 12:17

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