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

FtG (01/27/83)

People are missing the whole idea in "proving" that 1=.999... .
We are dealing with infinite sums here, which have very peculiar
properties.
Example consider S = 1/2 - 1/2 + 1/2 -1/2 .... .
Using the arguments given by some people, I can prove that
S is both 1/2 and -1/2! The "true" answer (as any calculus text will tell
you) is that S = 1/4! The proof is too subtle and complex to give over the
net.  Clearly the manipulation of infinite sums is VERY DIFFERENT from
finite ones. Infinite series cannot be capriciously added, multiplied, etc.

Getting back to what is .999..., the first mistake everyone is making is:

1/9 = .111.... ! This is not true, strictly speaking.
The infite series .1 + .01 + .001 + ... CONVERGES to 1/9. It is not
really 1/9.
Hence 9 times a number that CONVERGES to 1/9 is a number that CONVERGES
1. The mistake people make is in subconciously thinking that the
series and its limit are the same for all practical purposes and
therefore equivalent. REMEMBER: one is a series and the other is a number.
Don't mix apples and oranges.
				Enuf said-
				FtG (rochester)

laurir (01/29/83)

	Example consider S = 1/2 - 1/2 + 1/2 -1/2 .... .
	Using the arguments given by some people, I can prove that
	S is both 1/2 and -1/2! The "true" answer (as any calculus text will
	tell you) is that S = 1/4!

The "true" answer (as any calculus text will tell you) is that the
sequence S does not converge, and so is not "equal" to any particular
number.

	Getting back to what is .999..., the first mistake everyone is
	making is: 1/9 = .111.... ! This is not true, strictly speaking.
	The infite series .1 + .01 + .001 + ... CONVERGES to 1/9.
	It is not really 1/9.  Hence 9 times a number that CONVERGES
	to 1/9 is a number that CONVERGES 1.

This is just a matter of agreeing on notation.  The majority of those
who are so interested in mathematical trivia that they didn't "n"
this article seem to agree that ".999..." is just shorthand for

	 inf
	sigma  (9*10^-i)
	 i=1

which is the limit of the sequence, which *is* a number, equal to 1.
Hence, 1/9 = .111... and 1 = .999...

  -- Andrew Klossner (decvax!tektronix!tekmdp!laurir)

dap1 (01/30/83)

#R:rocheste:-52200:ihlpb:6200009:  0:778
ihlpb!dap1    Jan 29 16:42:00 1983

Whoooaaa!

I hate to argue, but your S series most definitely does NOT converge to
1/4!  You are thinking of Abel summability which is not the classical
definition of convergence.  The classical definition requires you to
supply me a number N such that after N terms your series does not stray
from 1/4 by more than an arbitrary epsilon.  For epsilon < 1/4 you
obviously cannot do this.  Your series is Abel summable to 1/4.  This
result is gotten by representing your sum as a power series, summing the
power series to obtain a formula and then using that formula outside the
radius of convergence.  This has it's uses, but doesn't represent
classical convergence.  For more info, see "Methods of Real Analysis"
by Richard R. Goldberg, page 271.
					Darrell Plank
					BTL-IH

ken (02/02/83)

It was asserted that the following series converged to 1/4.  Hogwash!

	S = 1/2 - 1/2 + 1/2 - 1/2 + ...

It is a divergent series.  Even though it is an alternating series,
the terms all have the same magnitude.  Group adjacent terms pairwise
and they cancel out.  It is like a square wave which DOES NOT CONVERGE!