Re: .99999....=1
- From: Proginoskes <CCHeckman@xxxxxxxxx>
- Date: Sat, 27 Oct 2007 06:58:17 -0000
On Oct 26, 3:34 pm, lwal...@xxxxxxxxx wrote:
On Oct 24, 1:54 pm, "tman_gl" <u38544@uwe> wrote:
I heard that it is mathmatically posible to prove that .9999.....=1
i find it completely irational but im interested at the same time
any answers would be much apreciated
And here we go again with one of the most popular topics on
this newsgroup, namely the number 0.999....
In the standard, Archimedean complete ordered field R, it is
inevitable that 0.999... = 1.
In a nonstandard set of reals, the numbers may be distinct.
They _may_, but it turns out they're not. See http://www.math.wisc.edu/~keisler/foundations.pdf
, pp. 119 (which are numbered as 109; in particular, Proposition 9.11,
which deals with geometric series in particular). (I asked about an
online source for an introduction to the hyperreals, and this file was
recommended. It is a good introduction.)
We
may have that 0.999... = 1 minus an infinitesimal. In the Cantor
threads, various so-called "cranks" have come up with different
infinitesimals with names such as "iota" and "Lil'Un."
It is also possible to use hyperreals with a Transfer Principle. In
the hyperreals every sequence of reals corresponds to a
hyperreal, so why not consider the sequence:
{0.9, 0.99, 0.999, 0.9999, ...}
If we let N be the hyperreal {1, 2, 3, 4, ...}, then we conclude:
0.999... = 1 - 10^N.
I think you mean 1 - (0.1)^N here. Since (0.1)^N is an infinitesimal,
and due to the Transfer Principle, this equation implies that
0.999... = 1.
--- Christopher Heckman
.
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