For any finite sequence of numbers, that sequence can be found in the decimal extension of pi. True or false.
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Anonymous2010-01-09 20:01
wouldn't surprise me if it was true
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Anonymous2010-01-09 20:21
Something about how PI goes on forever without repeating, i.e. an infinite sequence of numbers that never repeat. It therefore must contain infinite arrangements finite sequences. I'm not sure if you can then leap to the conclusion that an infinite arrangement of finite sequences must then contain every possible finite sequence. If so, I guess you could say the same thing about irrational roots.
Ultimately, I don't think there is a true way to prove this. There's no NP-Complete algorithm that will safely identify if any potential member x is in set Pi, when Pi itself never seems to end or wrap-around itself. Other irrational numbers may offer clues to the secret of Pi inclusiveness.
A number 0.99... is irrational, but will only ever contains all sequences of the number 9 repeated x times. In fact, it contains ALL combinations of any number of 9s repeating (an uncountable number of times).
An number like 0.7272... is irrational, but only contains sequences of 27, 72, and all numbers that are 27* or 72*. It will never contain the sequence 277, despite that also being made of the digits 2 and 7. A sequence that can be produced from the members of a member of a set does not have to be in the set.
The former indicates that an irrational number can contain every combination of the members of its set; and, the latter indicates that it will not necessarily contain a combination of the members of any single member in that set. 0.99... suggests that all the members of a set can be combined and represented in the irrational number, and we are not restricted by larger members such as in 0.7272... which creates roadblocks. For the moment, all we can say is that Pi must necessarily contain as its alphabet all the natural numbers. The "non-repeating" part of the definition is deceptive to us.
>>1
Likely untrue, but I don't think it's easily provable.
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Anonymous2010-01-10 8:07
I would say that OPs statement is true by definition
However, its "truthness" is of no particular use, like most maths, yet it does offer an interesting concept to ponder about... like most maths
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Anonymous2010-01-10 9:19
>>11
Agreed, I was most impressed my >>6-san's thoroughness and professionalism
>>13
It is not known if it is true. Please read >>10 then do some learnin'.
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Anonymous2010-01-10 10:54
But you don't know that at the nth digit, the decimal expansion becomes something like ...999.... And so there will only be 9's (or another number) after a certain decimal place. since you don't know whether this will happen or not, you can't know whether any given sequence will be found in the decimal expansion of pi.
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Anonymous2010-01-10 11:00
>>14
Even though I was completely wrong because I mistook one classification of numbers for another (and can not guarantee the same rules to apply for both).
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Anonymous2010-01-10 12:43
lol computer scientists attempt to solve a math problem
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Anonymous2010-01-10 14:43
>>16
Actually, we know that this won't happen, otherwise pi would be rational and that's just plain wrong.
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Anonymous2010-01-10 14:46
>>19
In fact, any student of freshman calculus should be able to write a rigorous proof of the irrationality of π. And that's how I can tell you're all still in high school.
>>10 >>15
It has been conjectured that every irrational algebraic number is normal; while no counterexamples are known, there also exists no algebraic number that has been proven to be normal in any base.
It's like proving the absence of god - formally impossible, yet no sound counterexamples exist to prove its existence either.
It's true we don't know for sure of the distribution of the 1284762184681681-21672174572147219216745217436745177 numbers after the coma in Pi, but as the starting sequence would imply there is absolutely no pattern to exclude it from being normal in base 10...
I want to believe!
I want to believe that within Pi this whole thread is encoded in lisp.
I want to believe that within Pi there is a 32bit bitmap of me hacking your anus.
I want to believe...
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Anonymous2010-01-11 7:06
I conjecture that, on average, to store the position of any data within π, you need a number that is as big as the data was to begin with.