I was thinking the same thing, Kabbes, except I was working in base 3 (0.1+0.1+0.1=1.0) instead of base 12. Kinda points out the fact that the “difference” between .9rep and 1 is simply how it’s represented on paper (or on a Web site). You just can’t precisely represent certain fractions in base 10. Can’t represent other fractions in base 3. 1/2=0.111rep, for example. Add that result to itself, get 0.222rep, which is, indeed, 1.
Alright, I’ll give you that Dave. But goddammit - you didn’t spell it out and some people really need it spelled out 
pan
Spell which out, kabbes? You’re gonna have to spell out what you meant by ‘it’.
[cartoon voice]Oh, pronoun trouble![/cartoon voice]
Yes, you can precisely represent 1/3 in base ten. The decimal representation 0.333… specifies that every decimal position is a three. What is imprecise about that?
I think they don’t really mean “precise”; they mean “without an ellipsis”.
You could go about “proving” this another way, although I am not sure how.
Assume 0.9999… < 1
implies 0.9999… + e = 1 for some e > 0 (a)
(b) 1 - .9999… = e
9/10 + 9/100 + 9/1000 + … + e = 1
9/10 + 9/100 + 9/1000 + … + e = 10/10
9/100 + 9/1000 + … + e = 1/10
.0999…+ e = .10000
.999…+ 10e = 1
10e = 1 - .9999… (from (b))
10e = e which contradicts (a)
Ok, I know this is really just the same proof offered up differently. But it highlights the infinity part, because without it, the bold section would not be correct. (I realize it is a bit sloppy).
DrMatrix wrote:
Um, what ultrafilter said. Sorry about my own lack of precision.