0.9999… is not a number in the same way that infinity is not a number.
For any physical manifestation of it, 0.99999… will behave exactly the same as 1.0, yet, their mathematics can be different.
Frantic’s Law of infinitesimals:
There is no difference between 0.9999… and 1.0, but they are not the same.
(pun intended)
FranticMad, could you give us an example of how the mathematics of .9999… and 1 are different?
Ooh boy, I can’t wait to hear this one. Didja know that there are no infinitesimals in the real number system?
No.
Except that 1 is a real number, no question about it.
But, .9999… is, um, not quite the same kind of number. Since there are no infinitesimals in the real number system, then 0.999… is not a real number since it must include the idea of an infinitesimal. Hence, they are not the same.
I have a feeling that I’ve not answered your question, though I don’t know why. Could you be more specific?
.9… is a real number, in exactly the same way that 1.0… is. They’re both the limits of sequences of real numbers, and by the completeness of the real number system, they are real numbers.
So far, every person who understands math has said flatly, without equivocation, that 1 and .999999… are just two different representations of exactly the same number, just as 2/2 and 1/3 + 1/3+ 1/3 are.
This happens to be completely correct.
“But, .9999… is, um, not quite the same kind of number.” is not exactly a mathematical expression. Nor is it in any way correct. What it is about the concept of infinitely repeating numbers that you don’t understand?
Ah ha!! (says FranticMad, knowing that the ultrashark smells blood and is swimming fast to make the kill), but does the completeness of the real number system also specify that real numbers A, and B can be subtracted such that
A - B = C
where C is not a real number (i.e. an infinitesimal)??
(fish darts away from the shark, looking in vain for the rest of his school of fish which have disappeared for some reason)
I don’t follow you here. Would you also claim, for example, that pi is not a real number? After all, it also has a never ending decimal expansion.
Or perhaps you are claiming that the infinitesimal is introduced by the idea that, if .999… is a real number, we can subtract 1 - .999…, and in that way produce a real infinitesimal. Yes, that would be a problem if we could get a real infinitesimal that way. However, only you (and perhaps a few others) are claiming such an infinitesimal; the rest of us claim that 1 - .999… is exactly zero, not an infinitesimal.
Also:
Ahem. You said:
I was merely asking for an(y) example of how their mathematics can be different. Apparently it’s you who need to be more specific.
Argumentum ad verecundium. A logical fallacy, which negates the validity of your argument.
Implied argumentum ad hominem. Invalid.
I made no claims about pi. It was ultrafilter who said infinitesimals (not pi, BTW) are not part of the real number system.
No, in fact you support my point. My question was whether such a subtraction can create a real number. If not, then how can 1.0 - 0/999 be a real number, i.e. 1? You are shadow boxing with your own shadow, not with me.
I posed a specific example and question to ultrafilter, who is more capable than you of arguing me into oblivion. Await his answer and learn how one truly destroys an imbecile, oh student.
And that would be Argumentum ad logicam.
Bippy the Beardless
If you have a set in which all of the elements are equal, how can there be an infinite number of elements? Isn’t there only one element? Aren’t you just creating an infinite number of names for the same thing?
ultrafiller:
I’m a nitpickaholic. You can’t blame me; it’s a disease.
Derleth:
This proof is technically incomplete, as it relies on aspects of the real number ordering system which are implied but not stated. There are number systems in which there is no number between X and Y, but X !=Y.
                                                                   _
PS re putting line over numbers: how about this? .9
No. The real numbers are closed under subtraction. If A and B are real, then A-B is real also.
For an explanation of Completeness see Real Number Wikipedia (From That page under the topic Construction from the Rationals): “Notice that the sequence (0,0.9,0.99,0.999,0.9999,…) is equivalent to the sequence (1,1.0,1.00,1.000,1.0000,…); this shows that 0.9999… = 1.”
Oh, and FranticMad, Cabbage can hold his own in a math discussion. You could possibly learn from him.
This isn’t a debate. It’s about math facts. 0.999… is exactly equal to 1.000.
No. That’s closure, which is something different. The real numbers are closed under addition, subtraction, multiplication, and division. Completeness means that every convergent sequence of real numbers converges to a real number. I trust it’s obvious that .9, .99, .999, … converges to .9…
And because I know you’re going to ask, division by zero is not a counterexample. In order that the real numbers not be closed under division, you would have to find a pair r, s such that r and s are real numbers, r/s is defined, and r/s is not a real number.
The Ryan: 
ultrafilter
I disagree with your definition of closure. The reals are not closed under division. S is closed under an operation we’ll call @, if for every r, s in S, r@s is in S.
If we haven’t yet defined negative integers, would you say the non-negative integers are closed under subtraction?
FranticMad:
Of course not, I never claimed you did. I brought up pi in the hopes that you might use that example to clarify your own point. I was wrong.
Perhaps you have not read my posts closely if you believe I support your point. To answer your question, yes, the subtraction of two real numbers is always a real number.
I’m claiming that .999… is a real number (in fact, it is equal to 1). You, as far as I can tell, are claiming that .999… is not a real number. Furthermore, you seem to claim that since .999… is not a real number, 1 - .999… is also not a real number (and possibly an infinitesimal). Note, however, that this is all based on your original unfounded claim that .999… is not a real number. Nothing you have said supports this claim; all you have presented is a string of reasoning which follows from your claim.
Would you care to begin supporting your claim now?
ultrafilter:
I believe you actually meant to say, “Completeness means that every Cauchy sequence of real numbers converges to a real number.”
Oh, and thanks, DrMatrix! 
I see my nitpicks have been preempted. “Every convergent sequence converges” certainly is true, but it’s rather trivial.
And one can simply define alpha to be 1/0 (and vice versa) and you have a number r/s which is defined but not part of R. No less valid than defining alpha to be a root of an irreducible polynomial.
For the record, argument from authority is a fallacy only when the authority is illegitimate.
Just for kicks, following up about repeating units in different base systems, you have in binary:
0.111111… = 1
I always knew that there was some sort of ulterior motive! 
Wow, those old threads get all balled up don’t they?