Re: .99999=1 explanation

From: S. Enterprize Company (smart1234_at_aol.com)
Date: 10/28/04 

Date: 28 Oct 2004 15:11:49 GMT

>Luke Valens wrote:
>> Using the same reasoning, we know that .333... + .666...=1, and that
>> .333... + .666...=.999....,
>> so .999... therefore equals 1.
>
>A fine proof woould be:
>.9 = 9*10^{-1}
>.09 = 9*10^{-2}
>..
>..
>.0*{n}*9 = 9*10^{-n}
>
>So, your number is in fine:
>limit when n-> +oo of sum(n=1..n, 9*10^{-n})
>which equals:
>limit when n -> +oo of [9*10^{-1} - 9*10^{-(n+1)}]/[1-10^{-1}] = 1 - 10^{n}
>
>And that limit equals 1!

   Nope, .99999.... is an indeterminate. The 9's repeat forever or infinitely.
This means it never converges in that form. You can use Partial Sums to show a
rounding off effect of a number that just constantly repeats, or you could just
round the number off indicating how accurate it is with respect to the amount
of significant figures used. On my calculator,

2/3 = .6666666666666666667

    Notice the rounding off effect. But, 2/3 =
.666666666666..... in it's most accurate form, but not vise versa. In other
words, you can't cause .666666.... to converge to 2/3 in that form. We can
however say,

.66666666666667 = 2/3

 so that we can use it in math. .

>
>This is very famous, because in fact, if you want to represent every real
>number with a unique 'digital' representation, you must avoid every 'end'
>of a decimalpart like 0.xxxxxx99999999999 as it shall be writen 1 instead.
>
>Romain

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