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Re: Endorsement of Wolfgang Mueckenheim from a serious mathematician
Posted:
Feb 1, 2013 12:01 AM
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On Thursday, January 31, 2013 8:55:12 AM UTC-8, WM wrote:
> On 31 Jan., 15:27, forbisga...@gmail.com wrote:
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> > The rational 1/3
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> > doesn't have a finite decimal expansion. None the less it is distinguishable
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> > from every rational other than 1/3 at some place in the expansion
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> > and only the infinite expansion can be calculated to be 1/3.
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> If you are so sure about that than you should be able to find out
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> whether 1/3 is among the paths that I have used to construct the
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> complete infinite Binary Tree. Does it in fact differ from the union
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> of all its finite initial segments
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> 0.01
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> 0.0101
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> 0.010101
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> ...
>
Yes it does, under the assumption that the continuation
of all of your initial segments is not the repeating sequence
[01]. If it is then I will give you the repeating sequence
0.[001] and it will not be among your finite initial segments
unless you change your continuation at which point 0.[01] will
not be in your set. The alternative is you allow an infinite
set of continuations for every finite initial segement and we
are back to the point that you have no reason to claim the
continuations countable.
When you multiply 0.[01] by 100 you get 01.[01] I can subtract
0.[01] from that and get 1 then divide both size of the equation
to get x = 1/11 (base 2) = 1/3 (base 10). None of your finite
initial segements will work because you can't eliminate the fractional
part that easily. (0.01 * 100) = 1 not 1.[01]
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