
Re: Proof 0.999... is not equal to one.
Posted:
May 31, 2007 4:02 AM


On May 31, 12:33 am, William Hughes <wpihug...@hotmail.com> wrote: > On May 31, 2:16 am, chaja...@mail.com 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://www1... > > > Charles J. Daniels > > chaja...@mail.com > > 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, 1a > 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... = 1d > 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
I disagree with the definition given:
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
You have defined 0.999... to be a real number without jusitification. I can make no such assumption. Each position within 0.999... can be expressed as a real number, but the totality, the very infinite nature of it, seems to render it a never ending relation more than a specific explicit location on the real number line.
I do agree with your statement about real numbers being a closed field under subtraction. Real numbers are a form of increment or metric from 0 is a certain sense. I do not believe the distance between 0.999... and 1 to be such a real number increment  it would take an infinity of them to equal one. No other real number can keep from exceeding one a finite number of its increment, let alone an infinity of them.
You have said a > (1(1/10^n)) for any natural number n and this is indeed the case. I agree that 1(1/10^n) is a real number for any real number n. Therefore 1  a will be a real number. 1  a will also always be less than 0.999... for any paricular real number n. Therefore, the needed subtraction to obtain 0.999... from one, should a relation exist, may not itself be a real number.
The fact that 0.999... is or is not a real number is of little concern to my proof  it makes no assumption and has no need one way or the other. What it demonstrates is that if x = 0.999... then 10x  x must be less than 9 within the confines of multiplication upon real numbers. What the proof demonstrates is that subtraction is required and omitted. It would not be surprising that the needed subtraction is not real. Real numbers simply will not allow 10x  x to be 9, they need it to be somehow less, whether the lessness is describable in real numbers or not.
I know this may sound like pure assertion on my part, as I am relying on my proof itself to make this needed subtraction clear, rather than rerelating my proof in form. But I would love and be interested in discussing ideas from the proof that are of "dubious" nature =)
charlie

