JT
Posts:
1,200
Registered:
4/7/12
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Re: Matheology sqrt(-2): WM admits to unlistability of 0/1 sequences
Posted:
Dec 21, 2013 5:31 PM
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Den lördagen den 21:e december 2013 kl. 23:07:22 UTC+1 skrev jonas.t...@gmail.com:
> Den lördagen den 21:e december 2013 kl. 22:45:50 UTC+1 skrev Virgil:
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> > In article <f0e5ddd0-3f73-490d-87bd-d786d931c7ed@googlegroups.com>,
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> > mueckenh@rz.fh-augsburg.de wrote:
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> > > On Saturday, 21 December 2013 21:15:47 UTC+1, Virgil wrote:
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> > > > In article <648b5671-ece9-4998-a412-9766ac3c0c8d@googlegroups.com>,
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> > > > jonas.thornvall@gmail.com wrote:
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> > > > > My very lose understanding of infinity, we can construct an
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> > > > > algorithm that create an unfinite number terms in a list the
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> > > > > simplest would be 1+1+1..., but we can not ever complete the list
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> > > > > and never find the last member.
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> > > > Since infinite lists by definition do not have last members, not
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> > > > being able to find one is a GOOD thing.
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> > > Since infinite sequences by definition have infinitely many terms
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> > > following beyond every term, and every term is belonging to a finite
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> > > initial sequence, not finding nearly all terms is a good thing.
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> > Only in the Wilds of Wolkenmuekenheim.
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> > For any sequence given by formula, one has already found ALL terms,
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> > and THAT is a good thing..
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> > --
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> Without giving a magnitude of ALL, that seem like handwaving to me.
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> Take this computation when will it reach zero.
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> x=1;
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> y=0.9
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> while (x>0) {x=x-y; y=y/10;}
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> According to mainstream math 0.999... is to be equal to 1, although i just proved it never will.
If you claim never you are on the right track, if you claim after infinity where do you place it temporal, because i just can't see that subtraction ever reach one.
So is it possible the answer is never, and could the conclusion be that 0.999... not really equal to 1.
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