Difference between revisions of "Talk:Essay:pi contains pi"

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(→‎Andy, you are nearly there!: For someone criticizing everybody else for their understanding of infinity, you write some...)
(→‎No one disputes that pi includes every number that has a finite representation: being a "normal number base 10" is vital to your "proof"!)
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It is a collective hunch that π is normal, but there is no proof (yet)! --[[User:AugustO|AugustO]] ([[User talk:AugustO|talk]]) 14:37, August 5, 2021 (EDT)
 
It is a collective hunch that π is normal, but there is no proof (yet)! --[[User:AugustO|AugustO]] ([[User talk:AugustO|talk]]) 14:37, August 5, 2021 (EDT)
 
:The issue of whether pi is normal is interesting but seems unrelated to the issue here of whether pi contains pi.--[[User:Aschlafly|Andy Schlafly]] ([[User talk:Aschlafly|talk]]) 11:05, August 6, 2021 (EDT)
 
:The issue of whether pi is normal is interesting but seems unrelated to the issue here of whether pi contains pi.--[[User:Aschlafly|Andy Schlafly]] ([[User talk:Aschlafly|talk]]) 11:05, August 6, 2021 (EDT)
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::Sorry to disagree - the issue is central: π has at least to be a ''normal number base 10'' to assure that all finite sequences of digits appear in its representation. Otherwise, there could be an index when π starts to look like ..010010001000010000010000001.... and so forth, with no chance to detect arbitrarily long approximations of π in it. --[[User:AugustO|AugustO]] ([[User talk:AugustO|talk]]) 17:37, August 6, 2021 (EDT)
  
 
== Andy, you are nearly there! ==
 
== Andy, you are nearly there! ==

Revision as of 21:37, August 6, 2021

Proof by contradiction that π does not contain π

  1. π is given as a digital expansion x0.x1x2x3x4x5...
  2. "finding π in π" means that there is an index n such that π starts all over again at xn
  3. If π starts again at n, then x0 = xn = 3, x1 = xn+1 = 1, x2 = xn+2 = 4, x3 = xn+3 = 1 and so on. "And so on" means that xk = xk+n for each natural number k, i.e.,. k ∈ N0.
  4. What happens if k equals n? What is the value of xn + n? According to the step above, xn + n =xn. But xn=x0=3 ! So, xn + n = xn=x0=3
  5. That's true for all multiples of n: if w ∈ N0, then xw n = x(w-1) n = .... = x2 n = xn = x0 = 3.
  6. What about xw n +k ( 0 ≤ k < n )? Again, xw n +k = x(w-1) n +k = x(w-2) n +k = x2 n +k = xn +k = xk
  7. So, for every natural number w and every number k with 0 ≤ k < n we have: xw n +k = xk. That makes π periodic, the length of the period is n.
  8. Therefore, π can be written as x0x1x2x3...xn-1 / (10n-1 -1) . But this is a rational number, and we know that π is irrational.
  9. We have a classical contradiction: As our result is wrong, our assumption that π is contained in π has to be wrong.

--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)
I don't assume periodicity, periodicity is the result of assuming that "π contains π" ... --AugustO (talk) 03:29, August 5, 2021 (EDT)

Reddit Commentary

So, here are excerpts from the top three comments:

  1. *My guess is: No, it doesn't, because then it would be a repeating decimal.
  2. I think you're right about all of this, but it's currently unknown whether Pi contains all finite sequences of digits.
  3. 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 proved 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 ?

--AugustO (talk) 17:34, July 29, 2021 (EDT)

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)

No one disputes that pi includes every number that has a finite representation

No one knows:

  • discovermaths "Is π normal?"
  • WolframMathWorld: "It is also known that pi is not a Liouville number (Mahler 1953), but it is not known if pi is normal to any base (Stoneham 1970)."

It is a collective hunch that π is normal, but there is no proof (yet)! --AugustO (talk) 14:37, August 5, 2021 (EDT)

The issue of whether pi is normal is interesting but seems unrelated to the issue here of whether pi contains pi.--Andy Schlafly (talk) 11:05, August 6, 2021 (EDT)
Sorry to disagree - the issue is central: π has at least to be a normal number base 10 to assure that all finite sequences of digits appear in its representation. Otherwise, there could be an index when π starts to look like ..010010001000010000010000001.... and so forth, with no chance to detect arbitrarily long approximations of π in it. --AugustO (talk) 17:37, August 6, 2021 (EDT)

Andy, you are nearly there!

  • "Pi containing pi means that there will be a string of numbers that repeat themselves: from the beginning of pi to when it first begins to repeat itself, which then would repeat itself again and again." Indeed, that's what I have shown above in #Proof_by_contradiction_that_π_does_not_contain_π
  • "However, it is generally assumed that there are no repeating patterns in pi." That's not a general assumption, that's a proven fact: it was proved for the first time in the 1760s by a certain Johann Heinrich Lambert, and since then, many more accessible proofs have been found.

So, what happens when an assumption leads to a contradiction to a proven fact? We get a proof by contradiction! --AugustO (talk) 03:38, August 6, 2021 (EDT)

The repetition likely won't happen until after an infinite number of digits, so this is not a problem. I've updated the essay on this point.--Andy Schlafly (talk) 11:01, August 6, 2021 (EDT)
Pray, what is the index of the first occurrence of this repetition? Is it x∞ + 12 ? Or x∞ + 4125 ? Or even x2∞? How is the weather on the day after the infinite reign of God?
For someone criticizing everybody else for their understanding of infinity, you write some weird stuff! --AugustO (talk) 11:32, August 6, 2021 (EDT)
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