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If pi is an irrational number that goes on infinitely forever, does it mean that I can get any sequence of numbers of any length, and somewhere in the decimals of Pi, this sequence will exist.

Eg. say if my sequence was 26.

this would appear in pi here 3.14*26*...

Would any random number (such as 654376546579687598635254235342564397564), I pick as the sequence appear somewhere along the sequence of the decimals of pi and all other irrational numbers?

Is this possible to prove?

stickman
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2 Answers2

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No; in fact the opposite can quickly be proved, that many irrational numbers do not have this property. For instance, the number $0.101001000100001\ldots$ (one more $0$ each time) is irrational (since its decimal expansion doesn't repeat) but clearly doesn't have this property.

joriki
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  • @joriki: are there examples of numbers that do have this property? – confused Jun 27 '11 at 13:55
  • @joriki : mmm ok fair enough. What about for pi? Does pi follow any sort of a pattern that would rule out certain random number sequences? – stickman Jun 27 '11 at 13:58
  • @confused: Yes, for instance the Champernowne number: http://en.wikipedia.org/wiki/Champernowne_constant – joriki Jun 27 '11 at 13:58
  • @joriki: of course, thanks. @stickman: pi is believed to be normal, but it hasn't been proven. – confused Jun 27 '11 at 13:59
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    @stickman: Not only is $\pi$ not known to be normal; according to Wikipedia (http://en.wikipedia.org/wiki/Normal_number#Properties_and_examples), "It is not even known whether all digits occur infinitely often in the decimal expansions of those constants [$\sqrt{2}$, $\pi$ and $\ln 2$]." I think the situation can be summarized by saying that very little indeed is known about these sorts of properties of decimal expansions. As far as I'm aware there are also no known patterns, and no patterns are expected to be found in the decimal expansions of these numbers. – joriki Jun 27 '11 at 14:02
  • @joriki : thank you. I found your answer helpful. – stickman Jun 27 '11 at 14:07
  • @stickman: You're welcome. – joriki Jun 27 '11 at 14:18
  • @joriki, perhaps you could add your comments to the answer. – lhf Jun 27 '11 at 16:34
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No. There are irrational numbers whose decimal expansion has only two different digits (say 0 and 1) so that no sequence that contains any of the other digits appears in it. What you are asking is part of the definition of normal numbers.

Julián Aguirre
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    And no one knows whether $\pi$ is normal. – lhf Jun 27 '11 at 14:00
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    And even irrationals that have all ten digits may not have all 100 2-digit numbers. And even irrationals that have all 100 2-digit numbers may not have all 1000 3-digit numbers. And so on. – Gerry Myerson Jun 28 '11 at 04:43