Proof by contradiction that π does not contain π
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--AugustO (talk) 19:14, August 4, 2021 (EDT)
- Your analysis above seems to assume a periodicity, which in turn assumes some kind of end point to the string of digits. But there is neither.--Andy Schlafly (talk) 20:13, August 4, 2021 (EDT)
Reddit Commentary
So, here are excerpts from the top three comments:
- *My guess is: No, it doesn't, because then it would be a repeating decimal.
- I think you're right about all of this, but it's currently unknown whether Pi contains all finite sequences of digits.
- Your proof of pi not containing itself is actually remarkably rigorous and proves that pi must not contain itself.
Andy, how do those make your point? --AugustO (talk) 19:19, August 4, 2021 (EDT)
- I'm being fair to both sides, and did not think the Reddit postings made my point.--Andy Schlafly (talk) 20:00, August 4, 2021 (EDT)
At last Andrew Schlafly says: π does not contain π
"Stated another way, no one disputes that pi includes every number that has a finite representation. Thus pi includes itself in all of its increasingly precise representations, without limit, and therefore pi contains pi itself within a vanishingly small margin of error."
So, π does not contain &pi, but π contains every finite approximation of π? That is nothing special: Emile Borel proofed a century ago, that almost all irrational numbers are normal, and therefore, almost all irrational numbers contain every finite approximation of π, e, √2, etc.
The difficulty is to prove for a given number that it is normal: no one has established this for π yet.
But I thought that your claim was that π contains π as a "consecutive sequence of digits in π" or as an "infinite and unbroken sequence of digits in π". How on earth could I get such a wrong impression? Oh, that's why:
Could you explain what you mean by "pi contains pi" exactly? Usually, when looking for birthdays, you would look for a consecutive string of six numbers. And mathematicians are certain that you could find the text of "Hamlet" if you take π mod 26 as a unbroken string. That seems to be different from your idea to find pi... --AugustO (talk) 15:25, July 29, 2021 (EDT) |
Not different at all. "Pi contains pi" means what it says: the full pi within pi. Infinity denial appears to be an obstacle here.--Andy Schlafly (talk) 16:38, July 29, 2021 (EDT) |
For clarification: π containing π means that there is a consecutive sequence of digits in π (not starting from the first one) which equals π? --AugustO (talk) 17:07, July 29, 2021 (EDT) |
Right, as I proved above.--Andy Schlafly (talk) 17:13, July 29, 2021 (EDT) |
Sorry for being a little bit thick (or nitpicking) - perhaps I have difficulties because English isn't my first language... OTOH: mathematics is universal!
You are saying that there is a infinite and unbroken sequence of digits in the representation of π (not starting with the first significant digit) equal to π ? |
Yes. Repeatedly asking for clarification of something that is clear could be the result of denial, in this case infinity denial. I don't think the disconnect has anything to do with communication in English!--Andy Schlafly (talk) 22:59, July 29, 2021 (EDT) |
Infinity denial indeed! --AugustO (talk) 03:48, August 5, 2021 (EDT)