Virgil
Posts:
7,760
Registered:
5/14/06


Re: Proof 0.999... is not equal to one.
Posted:
May 31, 2007 4:29 AM


In article <1180598532.293544.34870@q19g2000prn.googlegroups.com>, chajadan@mail.com wrote:
> 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.
You seem to think that the real numbers derive their justification from geometry ( the real number line?). There are at least two totally nongeometric justifications, via Cauchy sequences of rationals or via Dedekind cuts of the rationals, neither of which depends on geometry, and in both of which the proof that 0.999... and 1.000... must be interpreted as the same real number. > > 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.
Then you are not talking about the real real numbers, which your interpretation of 1  0.999... is prohibited by the Archimedean property of the reals.
> 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.
So suddenly, the reals are not closed under subtraction? That is another property impossible in the standard real number system. > > 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. But it is trivially imperatively that 0.999... is real in the Cauchy sequence model, and while somewhat less trivial, it is also imperative in the Dedekind model.
> 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.
The difference, whatever "size" it may be, must be a real number, as both terms of the difference are, and according to you, must simultaneously not be a real numbers.
> What the proof demonstrates is that subtraction is required > and omitted.
it is clear by now that your proof does not hold for any standard model of real numbers, so you must be talking about some private number system of your own that no one else uses.
> It would not be surprising that the needed subtraction is > not real.
Not to me.
> Real numbers simply will not allow 10x  x to be 9,
If x = 0.999... = Sum(9/10^n: n in N), the real numbers I use will not allow it to be anything else.
they > need it to be somehow less, whether the lessness is describable in > real numbers or not.
The difference between any two reals must be a real, at least if one is to have the standard field of reals. > > 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 =)
Well, you have certainly presented us with one that is quite dubious enough to go on with.

