Is Pi Truly Infinite and Nonrepetitive?

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If pi is infinite and nonrepetive and every number combination is in pi, somewhere, does that mean pi itself is in pi somewhere.. ? (that would make it periodic)
 
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Edi said:
If pi is infinite and nonrepetive and every number combination is in pi, somewhere, does that mean pi itself is in pi somewhere.. ? (that would make it periodic)

If pi is "normal in base ten" then that would mean that every _finite_ sequence of decimal digits occurs somewhere in the decimal expansion of pi.

There can only be countably many non-terminating decimal expansions found in consecutive digits in the decimal expansion of pi -- the one starting at the first digit, the one starting at the second digit, the one starting at the third digit, etc.

Since there are uncountably many non-terminating decimal expansions, it is certain that not all of them appear.
 
If pi is infinite and nonrepetive and every number combination is in pi, somewhere, does that mean pi itself is in pi somewhere.. ? (that would make it periodic)
Is there any proof that every (finite) number combination is in pi?
 
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[itex]\pi[/itex] is not periodic. If it were it would be in [itex]\mathbb{Q}[/itex] which it isn't.
 
telecomguy said:
The string 31415926 occurs at position 50,366,472 counting from the first digit after the decimal point.

http://www.angio.net/pi/bigpi.cgi
Interesting but does not answer the question which was about the entire countable string. On the other hand, Edi seems to be under the impression that we could have entire string, then additional digits which is not possible.

Edi, it is NOT known whether "every number combination is in pi" is true or not.
 
You are looking for a number n such that 10^n*pi-pi is an integer. Call this integer q.
Then (10^n-1)*pi = q
pi = q/(10^n-1)
To find such an n, pi would have to be a rational number, which is isn't.
So, no, pi cannot repeat or contain itself.
 
Khashishi said:
You are looking for a number n such that 10^n*pi-pi is an integer. Call this integer q.

As has been pointed out, n=0, q=0 satisfies this criterion. Pi contains itself -- in a trivial sense.

Then (10^n-1)*pi = q

Yes
pi = q/(10^n-1)

pi = 0/0 ?