In sci.logic, Chan-Ho Suh <email@example.com> wrote on Tue, 19 Jul 2005 15:31:44 -0700 <firstname.lastname@example.org>: > In article <ateDe.167016$_o.102331@attbi_s71>, Stephen Montgomery-Smith > <email@example.com> wrote: > >> I'm in the process of writing an article about >> objections to the notion that 0.999... is equal to unity, >> which I plan to contribute to the Wikipedia. I would be interested in >> having some intelligent feedback. Here's the article so far. >> >> > > This post has lotsa potential to go wrong....we'll see.
This thread has already gone very wrong :-) -- and has been rehashed to some extent in other threads, but here's the gist of the problem.
Briefly put, .999... = 1.000... because there are too many problems otherwise, although it's not entirely inconsistent to hypothesize that .999... != 1.000..., but one then has to work with expressions such as the following.
If we still assume that R is a field (and that .999... and 1.000... are members thereof), then we get this strange little number d = 1.000... - 0.999... . (If we extend the concept of field we still have some odd problems with 'd'.)
Briefly put, if d is this difference, then 10d = 10.000... - 9.999... by multiplying by 10 -- but if we subtract 9 - 9 = 0 from the right side we get the rather interesting equality
There are three ways to attempt a path around this.
 d = 0. This gives us standard mathematics.
 Define arithmetic differently for infinitesimals, so that 10d = d/10 = d. I'll admit at this point I've no idea what the details look like, since d - d could be just about anything. One might end up with something akin to little-o notation, but I for one can't be sure.
 Assume 1 - 10d = 0.999...9990. This quickly gets ridiculous, especially if one computes things like (1-d)^2 = 0.999...99800...001 or (1-d)^2 - (1-d)^3 = 0.999...99800...001 - 0.999...99700...00299...999 = ??!?
I can't call it a contradiction as such but it's certainly an abuse of the standard use of the ellipsis.
Some will complain that the above isn't rigorous enough; that is a fair complaint and one can then walk into the realms of calculus (specifically, limits), and go down a path trod long before by Cauchy et al.
Others point out this sequence:
x = .999 ... 10x = 9.999...
10x - x = 9x = 9.000...
therefore x = 1.
The main problem with this "proof" is that 10x-x can be either 9.000... or 8.999..., leaving one a bit lost again with expressions such as 9 - 9d, if x = 1 - d.
And then there are infinite expressions such as d + d + d + ..., which might be the limit of such things as lim(n->+oo) sum(i = 1,n) (i/n)^3 (1/n). (This is equal to 1/4 in standard math, of course.)
Or it might be something completely different. At this point I for one would just want to stick to the standard path, 0.999... = 1.000... .
-- #191, firstname.lastname@example.org It's still legal to go .sigless.