Changes
adding two proofs....
I'd like to hear your thoughts. --[[User:AugustO|AugustO]] ([[User talk:AugustO|talk]]) 09:08, 20 November 2016 (EST)
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!AugustO's proof by contradiction that π does not contain π
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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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!Andrew Schlafly's "Proof by Induction" that π contains π
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|style="background-color:pink"|π contains π as one significant digit in a finite representation ("3": 3.14159265'''3'''....)
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|style="background-color:lightblue"|That's your '''base case''': as π does not equal 3, the sentence "π contains π as one significant digit" is wrong on face value. But, I assume you wanted to say something like "π contains a approximation of π of length 1." So, well done, I'll give you that. Now for the two parts of the '''induction step''':
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|style="background-color:pink"|assume π contains π as ''n'' significant digits in a finite representation of π
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|style="background-color:lightblue"|again, your '''induction hypothesis''' is okay: I'd use the phrase "approximation of π of length ''n''", but let's not quibble.
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|style="background-color:pink"|π must also contain π as ''n+1'' significant digits as the number of digits of π is stretched to [[infinity]]
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|style="background-color:lightblue"|'''''What? Why should this be true?''''' Mathematicians have not proofed yet that π is a [[normal number]]! You can easily construct a irrational number which contains an approximation of π of length 10<sup>10,000</sup>, but no approximation of length 10<sup>10,000</sup> + 1 . Perhaps God constructed π this way - just to teach mathematicians a little humility. No one knows yet.
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|style="background-color:lightblue"|But even if your conclusion was true, you would have proofed only that π contains approximations of every imaginable '''''finite''''' length - not of '''''infinite length'''''. That is the same principle which allows you to count to every number, but never reach infinity.
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|style="background-color:pink"|Q.E.D.
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|style="background-color:lightblue"|Yeah, that's funny.
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