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As a teenager - the concept of irrational numbers fascinated me. The idea that all possible numbers existed in PI.

From that I reasoned that any piece of data you have now also existed in PI somewhere. For a moment I thought that this could lead to a brilliant compression algorithm, where you could simply point to the index and range in PI where your particular piece of data existed. When I got older I realised that the index was likely to be larger than the piece of data you were storing, making it a bad compression trade-off.

Now I'm sure this line of thinking must fit into a branch of Mathematics somewhere - but I'm not sure where to look.

My question is: What is the term for 'PI-indexing'?

EDIT: A related example - here is an example of a filesystem that stores files as locations in PI.

hawkeye
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    I don't think all possible numbers exist within $\pi$. Or at least it hasn't been proven that they do. (Also "all possible numbers" is quite vague.) It is very easy to construct irrational numbers which do not contain all possible numbers. As an exmaple consider the irrational number: $0.10110111011110111110111111\cdots$ – Ian Miller Jul 09 '16 at 00:07
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    You should realize that although it may be true that every finite sequence of decimal digits occurs in the decimal expansion of $\pi$, it is certainly not necessary for that to happen in order that a number be irrational. For example, the number $$ 0.1,0,1,00,1,000,1,\underbrace{0000}\text{ four 0s} ,1, \underbrace{00000}\text{five 0s}, 1, \underbrace{000000}\text{six 0s}, 1, \underbrace{0000000}\text{seven 0s},1,\ldots $$ is irrational, and its decimal expansion certainly does not contain every finite sequence of digits, nor even every finite sequence of 0s and 1s. – Michael Hardy Jul 09 '16 at 00:08
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    Furthermore, although a variety of proofs that $\pi$ is irrational are known, it has never yet been proved that every finite sequence of digits occurs in its decimal expansion. That is so far only a conjecture. $\qquad$ – Michael Hardy Jul 09 '16 at 00:09
  • @pjs36: Actually, that's not quite true. "Normal" is a stronger condition that has to do with how often all the numbers appear as substrings of the decimal expansion. There's a term for the weaker condition you mentioned but I don't remember what it is. – Daniel McLaury Jul 09 '16 at 00:16

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First of all, it's only suspected that all numbers occur as substrings of the decimal expansion of $\pi$. Almost every number has this property, but it's actually incredibly difficult to show that the property holds for any one specific number, with the exception of certain numbers constructed for the sole purpose of having this property.

The general area of math here would be "irrational number theory."

Daniel McLaury
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  • One should add (in view of the way the question was phrased) that it has been proved that $\pi$ is irrational. That was first proved in the 18th century. In the 20th century Mary Cartwright and Ivan Niven discovered proofs of the irrationality of $\pi$ that can be understood by those who know nothing beyond first-year calculus. $\qquad$ – Michael Hardy Jul 09 '16 at 00:12
  • Understood using nothing beyond first-year calculus, perhaps. Understood by those who know nothing beyond first-year calculus? I'm not so sure. – Daniel McLaury Jul 09 '16 at 00:18
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    Certainly it cannot be understood by most people who, in typical state universities in the USA, got grades of $\text{A}+$ in calculus, but reasonable students who know nothing beyond first year calculus can understand them if they do a lot of work. $\qquad$ – Michael Hardy Jul 09 '16 at 00:19
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    Maybe I should add that I mean such a person can follow the steps and verify independently that the proof is valid, but there is much beyond that to understand about those proofs, and that does require more than first-year calculus. $\qquad$ – Michael Hardy Jul 09 '16 at 00:22
  • @DanielMcLaury "it's only suspected that all numbers occur as substrings of the decimal expansion of π." Who suspects this? A link to further info would be helpful. – spinkus Mar 10 '18 at 11:58
  • @spinkus the rationale for this suspicion is given in the answer. – Daniel McLaury Mar 10 '18 at 14:11
  • I don't see it. – spinkus Mar 12 '18 at 00:53