Re: Proof 0.999... is not equal to one.

On May 31, 2:16 am, chaja...@xxxxxxxx wrote:
I have written a proof that 0.999... cannot be equal to one in the
system of real numbers.

While at the end of it all you may not fully agree with my proof, much
I as have never seen a proof asserting they were equal that I was able
to consider valid, I'm sure you will agree that the
ideas I present are not a simply rehashing of basic objections of
others before me.

It is available in several formats:http://www17.brinkster.com/chajadan/Math/Proofs/Proof1.dochttp://www17.brinkster.com/chajadan/Math/Proofs/Proof1.odthttp://www17.brinkster.com/chajadan/Math/Proofs/Proof1.txt

--Charles J. Daniels
chaja...@xxxxxxxx



You spend a lot of time on what you call the "classic proof".
However, what this is, is not a proof but a plausability
argument.

The stadard approach start out


1. define the real numbers (e.g. Dedekind cuts)

2. let a = 0.999... be a real number. We do not need
to give a full definition at this point
a<=1
and
a>(1-(1/10^n) for any natural number n
is sufficient.

3. Note that since the real numbers form a field, 1-a
must be a real number, call it d. Clearly d>=0,
and d<(1/10^n) for any natural number n.
The only real number with these properties is 0.


We conclude if 0.999... is a real number, then
0.999... = 1.

Note: under the standard "limit" definition of
0.999... = 1

You wish to call d a "dubious" number. However,
if you use the usual definition of a real number
then a dubious number is not a real number, nor is
0.999... = 1-d
a real number. To make things work
you have to abandon the usual real numbers
and work with a system of "numbers" that is
not a field.

- William Hughes



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