An uncountable countable set
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From: Virgil on 14 Sep 2006 14:59 In article <45097a95(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Dik T. Winter wrote: > > But to get more in your field. Consider Newton-Raphson for the calculation > > of the square root of a floating point number in the range [1/4, 1]. > > Also consider that the arithmetic is truncating (in the direction of 0) to > > the precision used. Now consider the stopping criterium to be that two > > successive iterands are equal. Does that algorithm stop? The answer is: > > in general it will stop. There are precisely two input numbers in the > > complete range for which it will not stop. > > Those being 1 and oo? I am not intimately familiar with the > Newton-Raphson solution, but a cursory examination of the method > indicates it will not stop, in terms of actually identifying sqrt(2) in > its entirety. It will produce the same value repeatedly if if one's original guess is correct. As that has nothing to do with the problem actually posed, it seems as though TO is incompetent even in the area of his supposed speciality, computing. In case TO is incapable of Googling "Newton-Raphson", the method is an iterative aproximation to a value, x_r, for which a differentiable function f(x) satisfies f(x_r) = 0. Under suitable constraints on the function and the initial value x_0, the iterative procedure x_{n+1} = x_n - f(x_n)/f'(x_n) will produce a sequence converging to the desired root, x_r. And under nice enough conditions, iteration by a computer program with limited precision will eventually produce a constant sequence. However, if one is forced to do the iteration in a computer system of limited precision with automatic truncation of values to the nearest value of allowable precision, Dik states that there are two values, y between 1/4 and 1, inclusive, such that for f(x) = sqrt(x) - y, the iteration will eventually oscillate between two values instead of converging to a single value. > But, it will stop once it gets within any finite distance > of sqrt(2). This bijective function between iterations and precision > holds in general, given certain constraints. > > Thanks for your input. > > :) Tony
From: Virgil on 14 Sep 2006 15:14 In article <45097df8(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Dik T. Winter wrote: > > In article <4506dbd2(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> > > writes: > > ... > > > Yes, it should, and Wolfgang's and my position is that N is unbounded > > > but finite, given that it only contains finite values, and only one per > > > unit of value range. > > > > By the definition of "finite", "unbounded" and "finite" are in > > contradiction > > with each other. So a number can not be both. > > Given the standard definition of an infinite set, and its "equivalence" > with finite quantities, that's true. No one except TO claims any infinite set is equivalent to any finite set. > > Not the Dedekind definition, but the standard definition. There are > > models with sets that are infinite, but Dedekind finite. (But in that > > case you do not have AC, because AC implies the two kinds are the same. > > Do you mean finite but Dedekind infinite? That's what I'm proposing. In order to have any certain difference between ordinary infinite and Dedekind infinite, one must claim the axiom of choice false, because if it is true, there cannot be any difference, and if its truth is unknown, one can't tell whether there is any difference. > > > At which bit position can the string achieve an infinite value? None > > > that exists in that string. > > > > Indeed. And that is why you need to do something to give actual meaning > > to such strings. In the 2-adics such is done, and in that case, > > ...111 is sum{n = 0 -> oo} 2^n = lim{n -> oo} (1 - 2^n)/(1 - 2) = -1 > > because in 2-adic metric lim{n -> oo} 2^n = 0. (The metric is defined > > as d(a, b) = 1/2^n if the n-th digits of a and b are the highest order > > digits where they differ.) > > That would appear to support my position that the number line CAN be > viewed as a circle, that there is a generalization from 2's complement > to n-base systems where numbers form a circle. Having looked up p-adic > arithmetic, I gleaned that using prime numbers solves some problems. > But, the T-riffics, solve some additional problems. :) But if the number line is a circle, then it is not an ordered set, and all TO's imaginings that require it to be, like that inverse function garbage, are down the tubes, > That depends on the acceptance of the limit ordinals as a model for the > naturals, really, when it comes right down to it. To say the least > ordinal for all finites is infinite is a mistake, in my opinion. And just how much is TO's opinion worth on matters mathematical? Less than a tinker's dam.
From: Virgil on 14 Sep 2006 15:17 In article <45098084(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Dik T. Winter wrote: > > In article <4506d1ae(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> > > writes: > > > David R Tribble wrote: > > ... > > > > Start with a simple proof: Take the set of naturals, N = {0,1,2,3,...} > > > > and remove one element to get set S = {1,2,3,...}. Now show that > > > > the "T-size" of N is exactly one less than the T-size of S. In other > > > > words, find a way to show that every counting of S versus every > > > > counting of N always leaves one element of N (0) left over. > > > > > > Use IFR. N maps to S using f(n)=n+1. The inverse of that function is > > > g(x)=x-1. So, over the range of 0 to N, |S|=|N|-1. Map N to the Evens E > > > using f(n)=2n. The inverse function is g(x)=x/2, so over the range of N, > > > the evens have |N|/2 elements. Isn't that intuitively satisfying? And > > > gee, it works for finite sets accurately too!! > > > > How many elements has the set of primes? > > There is no well-known function that maps n to the nth prime, so IFR > doesn't apply. Do you have an inverse function that specifies the nth > prime for all neN? Didn't think so. Then there are sets whose size TO cannot measure, which makes his "measure" less useful that cardinality, at least cardinality in ZFC and NBG.
From: Virgil on 14 Sep 2006 15:22 In article <450984a0$1(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Dik T. Winter wrote: > > In article <45070bcb(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> > > writes: > > ... > > > What about an unboundedly large but finite number? > > > > Let me ask you to, before I can answer such a question. What is your > > definition of "number"? (I asked the same from Wolfgang Mueckenheim, > > but his answer was not satisfactory, also not to himself, I think, > > because he never answered to questions about it.) > > Let me start with a geometrical expression of quantity: distance. TO conflates counting with measuring. They are inherently different processes. Counting is, at least in principle, exact. Measurement is, even in principle, inexact. So TO is trying to make an exact process inexact by embedding counting in measuring. > > As far as I understand, for any kind of numbers, they are fixed, so > > by definition not unbounded. > > Even if they have infinitely-indexed bit positions? There are no such "numbers" in standard mathematics, and as TO has no system in which they can exist, he does not have them either.
From: Virgil on 14 Sep 2006 15:25
In article <45098849(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > > I have renewed sense of faith in what you are trying to accomplish, > given your faith to Cantor's original vision. I do think he grappled, in > the face of fierce opposition, with fundamental concepts of infinity. I > think he suffered for even attempting to answer such questions. Now that > he has laid a groundwork, it's not unreasonable that it be improved. In order to improve on any of Cantor's theories, one would have to be at least as competent mathematically as Cantor himself, which neither "Mueckenh" nor TO are even close to being. |