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Let $\Pi \in \Sigma^\infty$ be the decimal representation of $\pi$ where $\Sigma=\{0..9\}$. It is not known whether $\Pi$ contains each natural number. On the other hand, no number is known not to occur in $\Pi$. (Hope I've got that right)

First of all: is this property unique to $\pi$ or do all irrational numbers share it? UPDATE: As pointed out by Brian and MJD, irrationality is completely unrelated: 1. there are irrational numbers which are known not to contain all numbers in their representation (Brian) 2. there are irrational numbers whose representation contains all numbers (MJD).

At any event, I think it should be much easier to show that almost all numbers are present, i.e. $ \exists N\in\mathbb{N}\forall k>N\exists p\in\Sigma^\star\exists s\in\Sigma^\infty: \Pi=pR(k)s$ where $R:\mathbb{N} \to \Sigma^\star$ maps each number to its decimal representation. So "allmost all" here means all except finitely many.

Has something like this been proven or disproven?

In general I don't think it's possible to construct a non-repetitive stream of symbols $\sigma \in \Sigma^\infty $ that does not contain almost all $w\in\Sigma^\star$ ("almost all" with respect to some enumeration of $\Sigma^\star$). (EDIT: This is possible as Brian points out)

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Not all irrational numbers are normal (that's what the property is called), but those known not to be normal are rather "unnatural" (Lioville numbers ans such) –  Hagen von Eitzen May 15 '13 at 18:32
    
Note that $\pi$ is more than simply irrational. It is transcendental. –  Fly by Night May 15 '13 at 18:33
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@user17762 I don't think this is quite the same question. Although it has some of the same misconceptions (every non-repeating decimal should have almost all numbers somewhere in the expansion), it's not asking whether or not $\pi$ is normal, nor is it asking if all number combinations show up in it (which is weaker than normality). It's asking if almost all natural numbers (I assume artistoex means except a set of measure 0) can be found somewhere in the decimal expansion of $\pi$, and requires nothing about the density of finding those numbers. –  Stahl May 15 '13 at 18:38
    
@stahl Exactly. That point remains unanswered although it sounds much harder now. –  artistoex May 15 '13 at 18:57
    
Also voting to reopen (see my previous comment for why this is not a duplicate of the question "does Pi contain all possible number combinations?") –  Stahl May 15 '13 at 19:46

3 Answers 3

Every decimal that is not eventually repeating represents an irrational number, so it’s very easy to construct irrational numbers that do not have the property, e.g., $0.01001000100001\dots~$.

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Champernowne's constant is irrational and obviously contains every integer in its digits.

Not every irrational number contains every integer in its digits. For example, consider $0.1010010001000010000001\ldots$ or $0.101011010101011010101\ldots$ where the $n$th block is $1$ if $n$ is a square, $01$ otherwise. Obviously both of these omit nearly all sequences of digits; they even omit nearly all sequences of zeroes and ones.

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that's a very interesting number, it's even transcendental. –  artistoex May 16 '13 at 8:58

There are many irrational numbers known not to contain some particular number. There are uncountably many irrational numbers which do not contain the decimal digit 7, for example -- and therefore don't contain 17, 73, or any numbers with a 7. Since almost all numbers contain a given decimal digit, this shows that there are uncountably many irrational numbers which fail to contain most integers as substrings.

As far as I know it is compatible with current knowledge that all digits of $\pi$ after some point are either 3 or 7 (no significance to these choices, obviously). So we can't even prove that almost all numbers appear.

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This is correct. –  user77335 May 15 '13 at 19:41

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