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Talk:Conservapedia insights (view source)
Revision as of 17:18, July 25, 2021
, 17:18, July 25, 2021→Proof that π does not contain π:
At least both insights are equally valid. --[[User:AugustO|AugustO]] ([[User talk:AugustO|talk]]) 13:01, July 25, 2021 (EDT)
At least both insights are equally valid. --[[User:AugustO|AugustO]] ([[User talk:AugustO|talk]]) 13:01, July 25, 2021 (EDT)
===Proof that π does not contain π:===
1) Let π be represented by the sequence of digits x<sub>0</sub>.x<sub>1</sub>=1, x<sub>2</sub>x<sub>3</sub>x<sub>4</sub>x<sub>5</sub>......
Therefore x<sub>0</sub> = 3, x<sub>1</sub>, x<sub>2</sub>=4, x<sub>1000</sub>=9, and so on.
2) Assume that π contains π: Then there is an index <i>n</i> such that x<sub>0</sub> = x<sub>n</sub> = 3,
x<sub>1</sub> = x<sub>n+1</sub> = 1, x<sub>1000</sub> = x<sub>1000+n</sub> = 9, or in general x<sub>k</sub> = x<sub>n + k </sub> for all natural numbers k.
3) But then, π's representation is periodic: x<sub>0</sub> = x<sub>n</sub> = x<sub>2n</sub> = x<sub>3n</sub> = x<sub>w n</sub> for all natural numbers <i>w</i>.
4) Every number with a periodic representation is a rational number. In this case, we could present π 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)
5) It is well know that π is not a rational number, so we have a contradiction. Ergo, π cannot contain itself.
--[[User:AugustO|AugustO]] ([[User talk:AugustO|talk]]) 13:18, July 25, 2021 (EDT)