Is Pi Truly Infinite and Nonrepetitive?

[Edi]

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)

 

[CompuChip]

Yes, it is, starting at the first digit :)

 

[jbriggs444]

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.

 

[Dead Boss]

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?

 

[telecomguy]

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

 

[jbriggs444]

Is there any proof that every (finite) number combination is in pi?

It is not known whether pi is normal in base 10.

 

[0xDEADBEEF]

\pi is not periodic. If it were it would be in \mathbb{Q} which it isn't.

 

[HallsofIvy]

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.

 

[Khashishi]

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.

 

[jbriggs444]

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 ?

 

[Khashishi]

you know what I meant.

 

[CompuChip]

You are looking for a number n > 0 such that 10^n*pi-pi is an integer.

Happy?

 

[jbriggs444]

Well, I could be even more picky and take n = 0.1200175... and q = 1.

 

[micromass]

Well, I could be even more picky and take n = 0.1200175... and q = 1.

Please stop hijacking this thread. We all know what Khashishi meant.