[net.math] 1 != .999...

bryan (01/25/83)

I was trying to say that it is possible to add something to
.999... and come up with the number 1. No matter how far you
carry it out, it will not equal 1; since it is always possible to
get closer to something with out reaching it (1/2 1/4 1/8 1/16 etc).

If you take the infinitely smallest portion of 1 and add it to the
infinitely largest portion of 1 you will reach 1. Since .999... is
the infinitely largest portion of 1 and .000...1 is the infinitely 
smallest portion of 1 , when added together you get 1.

P.S. .000...1 may be wrong ;it shoud be 1/100...
     So the equation now reads .999... + 1/100... = 1.0
     

dfz (01/27/83)

Recently there appeared here a refutation of the idea that .999... = 1.
The argument suggested that 1 = 0.999... + 0.000...1; my answer to the
author was that the figure 0.000...1 was in itself a contradiction, since
the dots between the last zero and the one represent an infinite number
of digits, yet the figure implies that it ends in a one followed by no
digits.
There has now appeared another argument in a similar vein, claiming that
one is the sum of the "infinitely largest part of one" and the
"infinitely smallest part of one".  This has merely transformed the
argument into a question of whether the "infinitely smallest part of one"
is zero or not (I will contend that it is).
For example, the infinitely smallest number "terminating" in the digit
one might be expressed as

	1 / ( 10 ^ infinity )

since there are an infinity number of zeros preceding the one.
The number expressed above is exactly zero.

mcewan (01/28/83)

#R:ihlpb:-27000:uiucdcs:28200008:000:312
uiucdcs!mcewan    Jan 28 12:41:00 1983

The problem is that people who think that .000...1 is a non-zero number
don't really believe in infinity! I.e., they are really thinking there
is a "largest" integer (which people call "infinity"). It may be so far
out that they can't count that high, but if you keep counting you're
eventually going to hit it.