From: utzoo!decvax!harpo!npoiv!eisx!whuxlb!ech
Newsgroups: net.math
Title: Re: .999... = 1 - (nf)
Article-I.D.: whuxlb.967
Posted: Sat Jan 29 10:44:33 1983
Received: Wed Feb  2 02:22:31 1983

#R:ihlpb:-27100:whuxlb:7200014:000:1034
whuxlb!ech    Jan 26 17:19:00 1983

There is a subtle distinction here that I haven't communicated, between
	.999... carried to n places and
	.999... carried to an infinite number of places.
The latter is the LIMIT of the sequence
	.9, .99, .999,...
and is equal to 1, but does not appear anywhere IN the sequence.
Similarly, we can define
	.1, .01, .001,...
which converges to 0 but never reaches it, only the limit being 0.

Now, I will not dispute that all those numbers in the second sequence are
numbers; however, the notion of putting the 1 after INFINITELY many 0's
is simply nonsense: the limit of the second sequence is .000... carried
to an infinite number of places, which is quite clearly 0.

Another, quite simple proof, that .999... is 1 is from the schoolboy method
for resolving repeated fractions into rational number (p/q) form:
	  x = 0.999...
	10x = 9.999...
subracting,
	 9x = 9.0
	  x = 1
The above proof relies only on the idea that .999... must be equal to itself,
which had better be self-evident even in the weird world of real numbers.

=Ned=