
Re: Proof 0.999... is not equal to one.
Posted:
May 31, 2007 11:34 PM


> Near the bottom of page 1 you write: > Let S be the set of all real numbers in the interval [0,1] > Let T be the set of all real numbers in the interval (1,0] > > If you applied an operator '+' to these two sets that sums the > elements of both sets, > you would get: S + T = 1 > > You have never defined this operator '+' for intervals. > The standard use (derived from the '+' for numbers) is to take the set > of all possible sums x+y where x is in S and y is in T. > However, this would result in S + T = (1,1].
The operator is not applied to the intervals. It is applied to two sets whose elements are all real numbers contained within these intervals.
> To save my own sanity, I scrolled ahead until I found this example > of downright nonsense: > > However, defining it as a limit is all but directly saying that it > may > not be exactly equal to one.
I'm surprised you see this as nonsense. A limit is a bound, that which can be infinitely approached. A limit does not indicate that this limit can be reached, only that any real number between this limit and our entity will be exceeded.
It's the classic story of crossing half a distance, then half the remaining distance, on and on. The limit is 1. But you can never reach the other side, even should you attempt to do so infinitely. I'm always boggled why people say that a journey taken in ninetenths the remaining distance would in fact reach 1...
> [0, 1], [0.9, 1], [0.99, 1], ..., [0.999..., 1] > > But what about the closed set [1]? How could we have felt sure > 0.999... > would be in this set as well? The nested set theorem only guaranteed > you'd get to a single value. > > Indeed, you'd get only a single value AND you'd get 0.999... AND you'd > get 1. This implies 0.999...=1.
You can only be sure that 0.999... will be in the final nested set if you define 0.999... as a limit, which again, I do not do. Recall, that 0.999... is strictly equal to 1 is taught to students everyday outside of a concept of limits. This brings up a natural reservation that they are being taught a untruth. Within the definition of limit, I do not argue it yields 1.
> Thus one can hope to define sum(S) and in fact whatever you claim > 0.999... to be is what you define as sum in this case. > > If r is a positive rational number, one can define the set T > containing > all elements of S multiplied by r, i.e. the numbers r*9/10^n. > > It should definitely be true that, whenever either sum(S) or sum(T) > is defined so is the other and the equality > sum(T) = r * sum(S) > holds, shouldn't it?
Agreed.
> > If s is a rational number not contained in S, we can define the set > U = S u {s} containing all elements of S and also s. > It should definitely be true that, whenever either sum(S) or sum(U) > is defined so is the other and the equality > sum(U) = s + sum(S) > holds, shouldn't it?
Agreed.
> With your set S defined as above, I choose r=10 to form T and I choose > s=9 > to form U. > If you look at it closely, you will note that T=U holds. > It should definitely be true that the equality > sum(U) = sum(T) > holds in this case, shouldn't it?
I disagree. Your claim is that 0.999... * 10 = 9.999...
S is an infinite set. Should you lay two copies of S upon itself they are onetoone mappable  you cannot point to an element in one copy that does not directly correspond to an element in the other copy. Should you point to all elements within one copy, no element in the other does not correspond in value to something you have pointed out in the first. They are infinitely pairable.
T is also onetoone mappable with S. If you lay S upon T, there is no element that in one set does not directly correspond to an element of the other. Should you point to all elements at once within one, there is no element within the other that does not correspond to all that you pointed out in the other. Indeed the very elements of T are generated one per each element in S. Should you point out all of S, no T exists that is not ten times something you have pointed out.
U is not infinitely mappable in a onetoone corresponance with either S or T. Should you remove s from U, it becomes S and is onetoone mappable with either S or T. In this respect, U has more in it than S or T. We could remove elements in pairs, one from U and one from S or T, and have one left over, not having omitted any other element from S or T that we attempted to map with U.
Note that both intervals [0, 1) and [0, 1) are infinite, and mappable in onetoone correspondance of real number. [0, 1) and [0, 1] are not mappable in onetoone correspondance in infinite completion. If you attempted to map all values from one onto values in the other of equal value, the 1 would be left unpaired.
For Sum(U) to equal Sum(T), we would need to be able to remove equal elements in a onetoone mapping, thus yielding 0. Should we remove s, or 9, from U, the result (or S) is directly mappable in a one to one correspondace with T as expressed earlier. The correspondance is not in value. It is of an element of S with ten times this element in T. But one exists for each.
If we remove 9 from both U and T, the resulting sets are still not one toone mappable. There will always remain an unpairable element in U with the rest directly mappable. Should we remove a count of elements approaching infinite from both sets in pairs of equal value, still U lest these elements will not be directly mappable to T lest these elements.
To be able to assert that U is equal to T, we would need to be able to map all values in onetoone correspondance. Traversing their elements from left to right, this appears possible, but we know that U will always have some aspect that is left unpairable. This makes U larger, for all else lest some nature can be removed in pairs.
This concept is difficult to relate as our ideas of infinity have not been honed and trained to the extent that our arithmetic has been.
9 + 0.999... = 9.999... 9.999... > 10*0.999...
charlie
> > But then > r * sum(S) = sum(T) = sum(U) = s + sum(S) > i.e. > 10 * sum(S) = 9 + sum(S). > > In my math, this can be transformed to > sum(S) = 1. > Your mileage my vary, but then you have not given > adequate definitions of '+', '*', '='. > > hagman

