We convince children that 1+1 =2 without delving into the Peano axioms. It's ok to not delve too deeply into the axiomatic structure of the reals.
There is nothing to do with completeness here; 0.999… = 1 is a statement about a series of rational numbers summing to a rational number, and the convergence of the series is established by the fact that it sums to the right-hand side, not by an abstract appeal to completeness.
0.99999999...1
This denotes the abstract idea that we take 0.999... (infinite number of 9's) and add another digit.
This is no more or less abstract than 0.999... to begin with.
0.999... is a unicorn, and 0.999...9 is a unicorn with a pink ribbon on the tip of its horn.
But you can't. All of the places where you might want to add a digit are already occupied by 9's.
You must then disbelieve concepts such as that the even integers can be put in 1:1 correspondence with all integers; i.e. that there are exactly as many even integers as there are integers.
No. I'm saying that a countable infinity doesn't have room for one more at the end. There is no end.
You can obviously stick a 1 in the middle, but then you have a number strictly less than 0.999...
(This is of course flawed, but I think it illustrates that the question isn't completely trivial. It requires us to carefully distinguish between the notion of an ordered set and a sequence and even then we'll have to deal with the fact that the rationals can be made into a sequence, but not with the same ordering.)
Eventually, yes, you're going to be able to poke a hole in my argument. It is definitely flawed. I don't know how long it'd take us to get there, but it doesn't really matter. But we're already at the point where this cannot be considered "basic", and that is my true point here.
Attempting to demonstrate that 0.999... = 1 while meticulously avoiding any rigorous definition of what 0.999... means is not very easy and will require you to fend off all sorts of potential jabs from various directions. It's much easier to just talk about infinite sums and be done with it.
Indeed. But it's a lot less fun :-)
You're spot-on about needing to understand there's a distinction between a number and its decimal representation, though.
It meant that I understood the topological idea of limits before I had to do proofs using just the epsilon (for sequence) or epsilon-delta (for functions) definition, and so could translate the logic of showing things about neighborhoods in to the terminology of (real analysis) limits.
Limits, in the abstract, are a fairly simple concept: in the case of sequences, for any neighborhood of the limit, the entire tail of the sequence (past some point) is contained in the neighborhood; in the case of functions, for any neighborhood of the limit at f(x), there's a neighborhood around x, such that every point in that neighborhood maps to the neighborhood around the limit.
Up to that point, I had only a fuzzy notion in my mind of what 0.33333... even was or how it was defined. But the professor helped clear this up for me: that "infinitely-repeating decimals" was actually just shorthand for a limit definition, i.e. "0.333..." is defined to be the limit as N approaches infinity of (3/10^1)+(3/10^2)+...+(3/10^N).
A "proof" that just plays a trick on people's logically inconsistent assumptions to derive a result isn't very satisfying. You're not really uncovering anything fundamental through that proof, just playing games.
Edit: what I mean to say is there are reasonable things "..." could mean such that .99... and .33... are both not equal to 1 and 1/3, respectively. But there are none for which one pair is equal and the other not, as you rightly point out. So all your proof does is show that the other person's viewpoint is inconsistent, but it doesn't give any evidence for one of the two consistent viewpoints over the other.
What slips people up is that ignoring everything but the total pie consumed (taking the limit) is embedded in the definition of real numbers.
There's an analogous story with rationals: Suppose x1 = 1, y1 = 3, x2 = 2, and y2 = 6. If we plot them, (x1, y1) and (x2, y2) are clearly different points, but x1/y1 "equals" x2/y2 because they lie on the same line through the origin. We decide that we don't need to know about those individual points.
Let 10X = 9.9. Then 10X - X = 9.9 - 0.9 = 9 = 9X. Hence X = 1, but X is actually 0.99 in this case (not 0.9). You need 0.99 = 0.9 for this to work with the exact same structure as your version.
Your proof only works because appending a 9 to an infinite expansion of 9s does not actually add a 9. But at this point you're forced to establish meaning for an infinite expansion of 9s, at which point this is really not just algebra anymore.
This is always wrong except in the case of infinitely many repeated digits, and the proof does not explain this.
More rigorously, let 9.999{n} denote an expansion with n 9s after the decimal point, where n can also be infinity. The subtlety with the argument is that it needs X to be the same as everything after the decimal point (so that the result of the subtraction is just 9). This is never true for finite values of n, and the proof does not establish that it's true for an infinite value of n -- indeed, it can't do so without supplying a meaning in the first place.
Another way of phrasing it is that it assumes that if X = 0.999..., then 10X = 9.999..., where there are the "same number" of 9s after the decimal point in 10X as there are in X. This seems intuitive for an infinite repeating sequence of 9s, because "one less than infinity" is still infinity, but it's not very rigorous, and the argument as written certainly doesn't explain this.
[And hence 10X - X = 9.9 - 0.99 = 8.91 = 9X implying that X = 0.99]
Following, as delineated above, from there, you'll see there's no contradiction. It's not so easy to break arithmetic that easily without dividing by zero :)
Sorry for dragging on this meaningless thread.
On the other hand, for any finite value of N, the sum [1] is equal to 0.9 * (1-(0.1)^N) / (1-0.1) = 1 - (0.1)^N. This value goes to 1 as N goes to ∞.
More rigorously, for any positive value, ε, there is a value N, such that the value of the finite sum is within ε of 1.
For one thing, 1 - 0.999... = 0.000... because you never get to have any remainder since 0.999... is infinite.
Or here's another proof:
x = 0.999...
10x = 9.999...
10x - x = 9.999... - 0.999...
9x = 9.000... = 9
9x = 9
x = 1
9.999.. - 0.999 is 9 no matter how we define .999... just as long as two or more occurrences of the 0.999... notation all denote the same entity, and we understand that the syntax 9.999... is 9 + 0.999...
For example, if we define 0.999... as "rubber duck" then 9.999... stands for 9 + "rubber duck", and 9.999... - 0.999... stands for 9 + "rubber duck" - "rubber duck" = 9.
0.999... * 10 = 9.999...
Because "one less 9 than infinity is still infinity" it's what really closes the loop on the proof.
You can't represent 0.2 exactly using IEEE floats, either, but that doesn't mean the representation 0.2 is not exactly equal to 1/5th.