View source for Talk:Essay:pi contains pi

==Proof by contradiction that &pi; does not contain &pi;==
{|
|style="background-color:lightblue"|
#&pi; is given as a digital expansion x<sub>0</sub>.x<sub>1</sub>x<sub>2</sub>x<sub>3</sub>x<sub>4</sub>x<sub>5</sub>...
#"finding &pi; in &pi;" means that there is an index <i>n</i> such that &pi; starts all over again at x<sub>n</sub>
#If &pi; starts again at <i>n</i>, then x<sub>0</sub> = x<sub>n</sub> = 3, x<sub>1</sub> = x<sub>n+1</sub> = 1, x<sub>2</sub> = x<sub>n+2</sub> = 4, x<sub>3</sub> = x<sub>n+3</sub> = 1 and so on. "And so on" means that x<sub>k</sub> = x<sub>k+n</sub> for each natural number <i>k</i>, i.e.,. <i>k &isin; <b>N</b><sub>0</sub></i>. 
#What happens if <i>k</i> equals <i>n</i>? What is the value of x<sub>n + n</sub>? According to the step above, x<sub>n + n</sub> =x<sub>n</sub>. But x<sub>n</sub>=x<sub>0</sub>=3 ! So, x<sub>n + n</sub> = x<sub>n</sub>=x<sub>0</sub>=3
#That's true for all multiples of <i>n</i>: if <i>w ∈ <b>N</b><sub>0</sub></i>, then  x<sub>w n</sub> = x<sub>(w-1) n</sub> = .... =  x<sub>2 n</sub> = x<sub>n</sub> =  x<sub>0</sub> = 3.
#What about  x<sub>w n +k</sub> (<i> 0 ≤ k &lt; n </i>)? Again, x<sub>w n +k</sub> = x<sub>(w-1) n +k</sub> = x<sub>(w-2) n +k</sub> = x<sub>2 n +k</sub> = x<sub>n +k</sub> = x<sub>k</sub>
#So, for every natural number <i>w</i> and every number <i>k</i> with <i>0 ≤ k &lt; n</i> we have: x<sub>w n +k</sub> = x<sub>k</sub>. That makes &pi; periodic, the length of the period is <i>n</i>.
#Therefore, &pi; can be written as x<sub>0</sub>x<sub>1</sub>x<sub>2</sub>x<sub>3</sub>...x<sub>n-1</sub> / (10<sup>n-1</sup> -1) . But this is a rational number, and we know that &pi; is irrational. 
#We have a classical contradiction: As our result is wrong, our assumption that &pi; is contained in &pi; has to be wrong.
|}

--[[User:AugustO|AugustO]] ([[User talk: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.--[[User:Aschlafly|Andy Schlafly]] ([[User talk:Aschlafly|talk]]) 20:13, August 4, 2021 (EDT)

::I don't assume periodicity, periodicity is the result of assuming that "&pi; contains &pi;" ... --[[User:AugustO|AugustO]] ([[User talk:AugustO|talk]]) 03:29, August 5, 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? --[[User:AugustO|AugustO]] ([[User talk: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.--[[User:Aschlafly|Andy Schlafly]] ([[User talk:Aschlafly|talk]]) 20:00, August 4, 2021 (EDT)

== At last Andrew Schlafly says: &pi; does not contain &pi; ==

<i>"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."</i>

So, '''&pi; does not contain &pi''', but &pi; contains every finite approximation of &pi;? That is nothing special: [[Emile Borel]] proved a century ago, that almost all irrational numbers are [[normal number|normal]], and therefore, almost all irrational numbers contain every finite approximation of &pi;, ''e'', &radic;2, etc.

The difficulty is to prove for a given number that it is normal: no one has established this for &pi; yet.

But I thought that your claim was that &pi; contains &pi; as a "consecutive sequence of digits in &pi;" or as an "infinite and unbroken sequence of digits in &pi;". How on earth could I get such a wrong impression? Oh, that's why:
{|
|style="background-color:lightblue"|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 &pi; mod 26 as a unbroken string. That seems to be different from your idea to find pi... --[[User:AugustO|AugustO]] ([[User talk:AugustO|talk]]) 15:25, July 29, 2021 (EDT)
|-
|style="background-color:pink"|Not different at all.  "Pi contains pi" means what it says:  the full pi within pi.  [[Infinity denial]] appears to be an obstacle here.--[[User:Aschlafly|Andy Schlafly]] ([[User talk:Aschlafly|talk]]) 16:38, July 29, 2021 (EDT)
|-
|style="background-color:lightblue"|For clarification: &pi; containing  &pi; means that there is a consecutive sequence of digits in &pi; (not starting from the first one) which equals &pi;? --[[User:AugustO|AugustO]] ([[User talk:AugustO|talk]]) 17:07, July 29, 2021 (EDT)
|-
|style="background-color:pink"|Right, as I proved above.--[[User:Aschlafly|Andy Schlafly]] ([[User talk:Aschlafly|talk]]) 17:13, July 29, 2021 (EDT)
|-
|style="background-color:lightblue"|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 &pi; (not starting with the first significant digit) equal to &pi; ?

--[[User:AugustO|AugustO]] ([[User talk:AugustO|talk]]) 17:34, July 29, 2021 (EDT)
|-
|style="background-color:pink"|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]]!--[[User:Aschlafly|Andy Schlafly]] ([[User talk:Aschlafly|talk]]) 22:59, July 29, 2021 (EDT)
|}
Infinity denial indeed! --[[User:AugustO|AugustO]] ([[User talk:AugustO|talk]]) 03:48, August 5, 2021 (EDT)

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

No one '''''knows''''':

*[https://www.youtube.com/watch?v=gg9admPCQhI discovermaths] "Is &pi; normal?"
*[https://mathworld.wolfram.com/Pi.html 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 &pi; 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)

== 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_&pi;_does_not_contain_&pi;]]
*"''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]]! --[[User:AugustO|AugustO]] ([[User talk: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.--[[User:Aschlafly|Andy Schlafly]] ([[User talk:Aschlafly|talk]]) 11:01, August 6, 2021 (EDT)