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Talk:Conservapedia insights (view source)
Revision as of 07:55, July 27, 2021
, 07:55, July 27, 2021→Proof that π does not contain π:: addressing Andy's concerns
--[[User:AugustO|AugustO]] ([[User talk:AugustO|talk]]) 13:18, July 25, 2021 (EDT)
--[[User:AugustO|AugustO]] ([[User talk:AugustO|talk]]) 13:18, July 25, 2021 (EDT)
:Point (3) does not follow from points (1) and (2). A more fundamental flaw is that the above argument incorrectly denies the possibility of an infinite series embedded within an infinite series, which [[pi]] likely contains.--[[User:Aschlafly|Andy Schlafly]] ([[User talk:Aschlafly|talk]]) 15:21, July 26, 2021 (EDT)
:Point (3) does not follow from points (1) and (2). A more fundamental flaw is that the above argument incorrectly denies the possibility of an infinite series embedded within an infinite series, which [[pi]] likely contains.--[[User:Aschlafly|Andy Schlafly]] ([[User talk:Aschlafly|talk]]) 15:21, July 26, 2021 (EDT)
::Thanks for your input! That's how a proof should work: a sequence of steps, each one acceptable for the reader. So, I'll try to do it better:
#Do we agree that π can be represented 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>...... (as any other real number)? I mean, that's the reason that people learn the first hundreds or even thousands of digits of π, and that super-computer race to find ever more digits of the expansion - it is meaningful to say that x<sub>1,000</sub>=9 and x<sub>1,000,000</sub>=1.
#Can we agree that "finding π in &pi" means that there is an index <i>n</i> such that π starts all over again at that place?
If we can agree on the above, then the following should be straight forward:
#If π 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 <k>, i.e.,. <i>k ∈ <b>N</b><sub>0</sub></i>. Andy, you seemed to have no qualms with this step.
#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> = 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 < 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 < n</i> we have: x<sub>w n +k</sub> = x<sub>k</sub>. That makes π periodic, the length of the period is <i>n</i>.
#Therefore, π 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 π is irrational.
#We have a classical contradiction: As our result is wrong, our assumption that π is contained in π has to be wrong.
I hope this proof is acceptable, if necessary, additional steps can be entered. --[[User:AugustO|AugustO]] ([[User talk:AugustO|talk]]) 03:55, July 27, 2021 (EDT)