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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?

2 Answers 2

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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.

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    @joriki, perhaps you could add your comments to the answer.2011-06-27
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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.

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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.2011-06-28