[net.math] = .9999 . . . ?

becker (01/25/83)

#N:uiucdcs:28200002:000:748
uiucdcs!becker    Jan 24 23:41:00 1983

my room-mate and i got in an argument over this:

 i said 
          1 = .999999 . . . .
 
and gave a couple of proofs 
    1:    1/9 = .1111 . . . .
          9 * (1/9) = 9 * (.1111 . . . .)
          9/9 = .9999 . . . . 
          1 = .9999 . . . .     qed
 
    2:    10 * (.9999 . . . .) = 9.9999 . . . .
          10 * (.9999 . . . .) - 1 * (.9999 . . . .) 
                  = (9.9999 . . . .) - (.9999 . . . .) = 9
       so 9 * (.9999 . . . .) = 9
          (.9999 . . . .) = 1    qed
 
these may be no news to you.  however, edward did not believe them.  anyone
out there have a good convincing proof he might believe?  i wouldn't want to
say that money was bet on this question, but . . .
 
                thanks,
                 craig

wildbill (01/26/83)

How about:

.999... = 9e-1 + 9e-2 + ..., an infinite geometric series.
The formula for the sum of an infinite geometric series is s = a / (1 - r),
where a is the first term, r is the ratio between any two adjacent
terms, and abs(r) < 1. Applied to this case, with s = 9e-1 and r = 1e-1:

s = 0.9 / (1 - 0.1) = 0.9 / 0.9 = 1.

Now, if he wants you to prove s = a / (1 - r), use the same trick as in
your second proof on the series a + ar + ar^2 + ar^3 + ...
(multiply by r and show that most of the terms match up). If he doesn't
buy that, you will have to resort to the definition of a limit, which,
after all, is what this is. Just convince him that given any number
which he wishes to name, you can find a number N such that all numbers
of the form .99999...9 with n or more digits differ from 1 by less than
the number he gave you, then point to the definition.

If he refuses to believe this, he is either hopeless or is not talking
about real numbers, whose formal definition is in terms of equivalence
classes of sequences of rational numbers. I will be happy to supply more
details to interested parties. To find out which he is, ask him
whether Achilles can catch the tortoise.

					Bill Laubenheimer
					ucbvax!ernie:wildbill
					ucbernie.wildbill@berkeley

mcewan (01/27/83)

#R:uiucdcs:28200002:uiucdcs:28200005:000:99
uiucdcs!mcewan    Jan 27 15:33:00 1983

How about:

	1 - .999... = .000...(infinite number of 0's)...1
		    = 1/(10^infinity)
		    = 0
?