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Is an irrational number, such as $\pi$ or $\sqrt2$, guaranteed to contain every possible digit sequence somewhere within it? Is there no proof for this? Is there any clue as to whether this is so? It seems logical to me, seeing that irrational numbers continue infinitely and are essentially patternless.

If it is true that every possible digit sequence can be found in any irrational number, that would imply that one could find any set of data (such as an encoded version of the Human Genome Project or something like that) within an irrational number, which would be quite intriguing in a philosophical context.

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relevant: en.wikipedia.org/wiki/Normal_number –  vadim123 May 7 '13 at 13:01
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Consider $.1010010001\cdots$. –  David Mitra May 7 '13 at 13:02
    
possible duplicate of Prove there are no hidden messages in Pi –  Ross Millikan May 7 '13 at 13:02
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It would be quite uninteresting in a philosophical context, since you could also find encoded faulty versions of the Human Genome Project, and no reliable way to tell the correct version from the faulty ones. –  Gerry Myerson May 7 '13 at 13:07
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"seeing that irrational numbers continue infinitely and are essentially patternless" is a misunderstanding. The decimal expansion of an irrational number can never keep repeating indefinitely -- but that's just one kind of pattern, and every other kind of pattern in the decimals will produce an irrational. –  Henning Makholm May 7 '13 at 13:13

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up vote 4 down vote accepted

An irrational number is not guaranteed to contain every possible digit sequence. For example, the irrational number $\sum_{i=1}^\infty 10^{-i!}$ contains only very specific subsequences of 0's and 1's.

As far numbers having these properties, see the link to the Wikipedia article on normal numbers in the comments.

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On the other hand, $\sum_{i=1}^\infty i\cdot 10^{-i^2}$ is a number that does contain every possible finte digit sequence. –  Hagen von Eitzen May 7 '13 at 13:20

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