
Re: Proof 0.999... is not equal to one.
Posted:
Jun 1, 2007 2:10 PM


On Jun 1, 1:24 am, hagman <goo...@voneitzen.de> wrote: > On 1 Jun., 03:28, chaja...@mail.com wrote: > > > > > > > > No, by "are they real" I mean much more. As well as an order you also > > > need > > > the field operations (addition, subtraction, multiplcation and > > > division) > > > and you need them to preserve order. > > > Makes sense to me and I agree. > > > > No. If we use the standard definition of real numbers and > > > the standard way of talking about decimals (infinite decimals > > > are defined in terms of limits), then > > > decimals can express any real number (including > > > sqrt(2) and pi). > > > I find this quite simply impossible to accept. For any decimal that > > tries to express a number beyond zero, a whole infinity of real > > numbers lie in between that and zero. Decimal can never point out a > > value beyond zero that has no intervening infinity. To me the > > consequence is that is cannot point out them all, for to point one out > > only shows there are others. > > > > It was to avoid nonsense like > > > "would be able to comply in infinite completion" > > > that the concept of limits was developed. > > > >  William Hughes > > > You reject the idea of infinite completion seemingly because it is > > counterintuitive to you, and indeed to many people, as infinity is > > unbounded it our most general concepts. What about this: if you take > > the sets of all real numbers in the intervals [0, 1] and [1, 0] they > > can be paired in cancelling values in infinite completion. The sets of > > all real number in the intervals of [0, 1) and [1, 0] could not be > > paired in cancelling values in infinite completion, we could only > > traverse infinity and yield pairs, refusing to complete the operation > > and acknowledge that one value will not pair. > > Take any element x of [0,1). > If x is the reciprocal of a natural number, map x to x/(1x). > Otherwise, map x to x. > > Which value will not pair? > > hagman Hide quoted text  > >  Show quoted text 
Let A be the set of all numbers of the form 1/n for every n in the set of natural numbers >= 2 contained in the interval [0,1) in descending order of value, ie {1/2, 1/3, ... 1/n} Let B be the set of all numbers of the form 1/n for every n in the set of natural numbers >= 2 contained in the interval (1, 0] in ascending order of value, ie {1/2, 1/3, ... 1/n} Let a(n) be the nth element of A for every n in the set of natural numbers >= 1 Let b(n) be the nth element of B for every n in the set of natural numbers >= 1 a(n) + b(n) = 0
Sets A and B have a onetoone mapping and are infinitely pairable in completion.
Let C = B + {1}
Sets A and C do not have a onetoone mapping and are not infinitely pairable in completion. If A and B can be infinitely paired in completion, adding or removing any number of elements from one set without the same count added or removed from the other will leave one unable to pair in infinite completion between these sets.
Through my use of the term infinitely complete, I would say that the pairing you offered of how to map x values in not infinitely completable and here's why:
for every x of the form 1/n for each n in the set of natural numbers >= 2 in the interval [0, 1), x maps to 1/(x1) in the interval [1, 0], that is, 1/2 maps to 1, 1/3 maps to 1/2, etc.
Each x introduces the necessity that x exists. To this respect, we map 1/2 to 1, thus mentioning the 1/2 value, so we must create another pair, namely 1/3 to 1/2. But now we have mentioned 1/3, so we must create a pair to account for it's negative counterpart in the other interval, namely 1/4 to 1/3. Each time we make such a pair, we only point out that there is yet one more pair we must make. At no point can we stop making pairs and be done  this pairing operation is not infinitely completable. No matter how many pairs we have assigned, approaching infinity, any set of such pairs paired though the method you presented would be incomplete, in that a value has been left behind in the set [1, 0] that would need to be paired. Through your method, you could pair values infinitely, but no set of such pairs could every reach completion.
If we were to pair elements of sets A and B through an operation that maps a(n) to b(n) for every natural number n >= 1, this operation is infinitely completable. You could fully yield an infinity of pairs that map every value to something else.
The pairing operation you described is traversive (noncompletable). It would not allow an infinite mind to map all values in a single whole completion of pairs that is complete, since any such set of pairs must leave out a value that would need pairing.
If we applied your pairing operation to the intervals (1, 0] and (1, 0], still it would not be infinitely completable in and of itself, even though the intervals have a onetoone correspondance. We would have to avoid 1/2 at first as it would map to 1 which is no longer present, but! we'd at least have a value set aside  we'll need it later. Should we apply your pairing algorith we begin to get pairs (1/3, 1/2), (1/4, 1/3), (1/n, 1/(n1)). At no point can these pairs be complete, but for any n at which we stop applying your algorithm, we can make a pair (1/2, 1/(n2)), and the rest can be paired in infinite completion. So a traversive (noncompletable) pairing operation can be applied to a onetoone mapping for any length of terms, but then must at some point yield. Should we be able to infinitely complete your algorithm to the end of infinity, we would be left with a single value that would pair with 1/2 which we left behind. This is necessary if the sets have a onetoone correspondence.
Do you understand what I mean, even if you disagree?
charlie

