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→Pi contains pi: no, π does not contain π
--[[User:AugustO|AugustO]] ([[User talk:AugustO|talk]]) 17:41, July 30, 2021 (EDT)
--[[User:AugustO|AugustO]] ([[User talk:AugustO|talk]]) 17:41, July 30, 2021 (EDT)
:Your superb examples demonstrate that if true for "n" significant digits, then one can be found later having "n+1" significant digits. This reinforces the proof by induction. I don't see any support for your assertion "that is impossible." I also don't see any connection to your hypothesis about whether an infinite string of 1s is a "natural number," which is simply a matter of definition.--[[User:Aschlafly|Andy Schlafly]] ([[User talk:Aschlafly|talk]]) 21:30, July 30, 2021 (EDT)
:Your superb examples demonstrate that if true for "n" significant digits, then one can be found later having "n+1" significant digits. This reinforces the proof by induction. I don't see any support for your assertion "that is impossible." I also don't see any connection to your hypothesis about whether an infinite string of 1s is a "natural number," which is simply a matter of definition.--[[User:Aschlafly|Andy Schlafly]] ([[User talk:Aschlafly|talk]]) 21:30, July 30, 2021 (EDT)
::*''Your superb examples demonstrate that if true for "n" significant digits, then one can be found later having "n+1" significant digits.'' Thanks! But a few examples make no proof.
::*''This reinforces the proof by induction.'' No. It does not. As no one has proofed that π is a [[normal number]], your "proof by induction" is on shaky grounds.
::*''I don't see any support for your assertion "that is impossible."'' I put it at [[Talk:Conservapedia insights#Proof that π does not contain π:]] - it is somewhat of an overkill:
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#π 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 π in π" means that there is an index <i>n</i> such that π starts all over again at x<sub>n</sub>
#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 <i>k</i>, i.e.,. <i>k ∈ <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 < 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.
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::*''I also don't see any connection to your hypothesis about whether an infinite string of 1s is a "natural number," which is simply a matter of definition.'' In general, infinity is a difficult concept. But the simplest form of infinity is being "countably infinite". You should make yourself familiar with this concept. Then you would recognize that there is a difference between the statements:
::::#''a sequence contains a countably infinite number of finite sub-sequences with a certain property'' and
::::#''a sequence contains a countably infinite sub-sequence with a certain property''
:::The first statement does not imply the second one. --[[User:AugustO|AugustO]] ([[User talk:AugustO|talk]]) 17:44, July 31, 2021 (EDT)