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March 29th, 2014, 10:04 AM   #11
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I already mentioned a book by Rudy Rucker: Infinity And The Mind.
He is a mathematician and writes on his book page 79 that the number K
with decimal expansion 0.99999... is the same as 1, his argument is
10K=9.99999....
- K=0.99999....
------------------------
9K =9
K =1



He continues "maybe this argument is misleading. What if there is some number
call it 1 - 1/ω, that is greater than any finite string of 0.9.....9 of nines, yet
less than 1? If K were equal to 1-1/ω, there is infinitesimal quantity that does not get canceled out,
and the above argument does not work":
10K=10 - 10/ω
- K= 1 - 1/ω
----------------------
9K = 9 - 9/ω
K = 1 - 1/ω

Last edited by TwoTwo; March 29th, 2014 at 10:18 AM.
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March 29th, 2014, 08:54 PM   #12
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You are creating a sequence where


If the sequence approaches a limit, L, we have

So L = 0 or L = 1


So
Thus the sequence is bounded above.


So
Thus the sequence is increasing.

Since the sequence is both bounded above and increasing, it must converge to a limit. And we showed above that the limit must be 1. However, since the sequence is increasing, we can say that there is no N such that

Last edited by v8archie; March 29th, 2014 at 08:59 PM.
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March 30th, 2014, 06:44 AM   #13
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In general, the successive values are approximate, so when the calculator internally obtains, say, 0.999999999896, and cannot display that many digits, it may well display 1 instead. Eventually, even the internally calculated value may become exactly 1.
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