Give me ONE digit position of any real that isn't computable up to that digit position!
in [Math]
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From: Ostap Bender on 4 Jun 2010 23:23 On Jun 4, 8:13 pm, Charlie-Boo <shymath...(a)gmail.com> wrote: > On Jun 4, 8:22 pm, "dannas" <inva...(a)invalid.com> wrote: > > > "|-|ercules" <radgray...(a)yahoo.com> wrote in message > > > A computer can calculate ANY digit sequence up to INFINITE length. > > > not infinite, as that would take infinite power, and the power company would > > send you an infinite bill. > > > that's where the dark numbers lie! > > > how did they get dark? > > They didn't pay their infinite power bill. They don't have to pay now. They can put these infinite charges on their BoA credit card and pay them at an infinite time from today, with a continuously compounded 20% annual interest of course.
From: |-|ercules on 5 Jun 2010 05:54 "Sylvia Else" <sylvia(a)not.here.invalid> wrote > On 5/06/2010 7:55 AM, |-|ercules wrote: >> According to sci.math, the number line consists mostly of "dark numbers". >> >> You see, 99.999999.... nearly % of real numbers are not computable, >> there is >> no program that represents them. >> >> A computer can calculate ANY digit sequence up to INFINITE length. >> >> Sci.math will make a minor correction there, a computer can only calculate >> all digit sequences up to ALL finite lengths. >> This is like saying a computer can only calculate all digit sequences up >> to INFINITE finite lengths. > > A computer is a finite state machine, so although it may continue to > compute forever, its outputs must start to repeat. Any number whose > sequence of digits does not appear in that recurring output will never > be output by that computer no matter how long you wait. > > If you don't constrain the computer's storage space, letting it be > "inifinite" you still have the problem that the set of storage locations > will be countably infinite, and thus the set of all of its states will > also be countably infinite, as will the set of all of its possible outputs. > > Since the reals are not countable, the computer will still omit most of > them from its output. That is, it will not be able to compute them. > > Of course, you can hypothesise a computer that is not subject to those > constraints, but that's really just assuming the answer. > > Sylvia. Your argument for uncountable infinity is based on the fact that reals are not countable? Perhaps you should freshen up with a nice poll. >> The proof of higher infinities than 1,2,3...oo infinity relies on the fact that there >> is no box that contains all and only all the label numbers of the boxes that >> don't contain their own label number. True or False? Herc
From: William Hughes on 5 Jun 2010 07:34 On Jun 4, 6:55 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: where the dark numbers lie! > > The missing numbers are different from computable numbers at the ... err..... Xth digit! > Of course not. There is no such digit. Simple proof. Let y be a non-computable number Let y_k have a decimal expansion that agrees with y to the k th decimal place and is 0 for all other places. Then, for every k, y_k is computable, that is there is a finite program f_k which produces y_k. So there is no digit position for which y differs from all computable numbers. However, y is not computable. There is no way to combine all the finite f_k into a finite f which produces y. - William Hughes
From: |-|ercules on 5 Jun 2010 08:41 "Sylvia Else" <sylvia(a)not.here.invalid> wrote > On 5/06/2010 7:54 PM, |-|ercules wrote: >> "Sylvia Else" <sylvia(a)not.here.invalid> wrote >>> On 5/06/2010 7:55 AM, |-|ercules wrote: >>>> According to sci.math, the number line consists mostly of "dark >>>> numbers". >>>> >>>> You see, 99.999999.... nearly % of real numbers are not computable, >>>> there is >>>> no program that represents them. >>>> >>>> A computer can calculate ANY digit sequence up to INFINITE length. >>>> >>>> Sci.math will make a minor correction there, a computer can only >>>> calculate >>>> all digit sequences up to ALL finite lengths. >>>> This is like saying a computer can only calculate all digit sequences up >>>> to INFINITE finite lengths. >>> >>> A computer is a finite state machine, so although it may continue to >>> compute forever, its outputs must start to repeat. Any number whose >>> sequence of digits does not appear in that recurring output will never >>> be output by that computer no matter how long you wait. >>> >>> If you don't constrain the computer's storage space, letting it be >>> "inifinite" you still have the problem that the set of storage >>> locations will be countably infinite, and thus the set of all of its >>> states will also be countably infinite, as will the set of all of its >>> possible outputs. >>> >>> Since the reals are not countable, the computer will still omit most >>> of them from its output. That is, it will not be able to compute them. >>> >>> Of course, you can hypothesise a computer that is not subject to those >>> constraints, but that's really just assuming the answer. >>> >>> Sylvia. >> >> Your argument for uncountable infinity is based on the fact that reals are >> not countable? >> >> Perhaps you should freshen up with a nice poll. >>>> The proof of higher infinities than 1,2,3...oo infinity relies on the >>>> fact that there >>>> is no box that contains all and only all the label numbers of the >>>> boxes that >>>> don't contain their own label number. >> >> True or False? > > False. Does the proof of higher infinities than 1,2,3...oo infinity rely on this verbage being true? http://en.wikipedia.org/wiki/Cantor's_theorem Suppose that N is bijective with its power set P(N). Let us see a sample of what P(N) looks like: P(N) = { {}, {1,2}, {1,2,3}, {4}...} P(N) contains infinite subsets of N, e.g. the set of all even numbers {2, 4, 6,...}, as well as the empty set. Now that we have a handle on what the elements of P(N) look like, let us attempt to pair off each element of N with each element of P(N) to show that these infinite sets are bijective. In other words, we will attempt to pair off each element of N with an element from the infinite set P(N), so that no element from either infinite set remains unpaired. Such an attempt to pair elements would look like this: 1 <-> {} 2 <-> {1,2} 3 <-> {1,2,3} (different example than text refers to, see wiki) Given such a pairing, some natural numbers are paired with subsets that contain the very same number. For instance, in our example the number 2 is paired with the subset {1, 2, 3}, which contains 2 as a member. Let us call such numbers selfish. Other natural numbers are paired with subsets that do not contain them. For instance, in our example the number 1 is paired with the subset {4, 5}, which does not contain the number 1. Call these numbers non-selfish. Likewise, 3 and 4 are non-selfish. Using this idea, let us build a special set of natural numbers. This set will provide the contradiction we seek. Let D be the set of all non-selfish natural numbers. By definition, the power set P(N) contains all sets of natural numbers, and so it contains this set D as an element. Therefore, D must be paired off with some natural number, say d. However, this causes a problem. If d is in D, then d is selfish because it is in the corresponding set. If d is selfish, then d cannot be a member of D, since D was defined to contain only non-selfish numbers. But if d is not a member of D, then d is non-selfish and must be contained in D, again by the definition of D. This is a contradiction because the natural number d must be either in D or not in D, but neither of the two cases is possible. Therefore, there is no natural number which can be paired with D, and we have contradicted our original supposition, that there is a bijection between N and P(N). Herc
From: William Hughes on 5 Jun 2010 08:58
On Jun 5, 9:41 am, "|-|ercules" <radgray...(a)yahoo.com> wrote: > > Does the proof of higher infinities than 1,2,3...oo infinity rely on this verbage being true? > > http://en.wikipedia.org/wiki/Cantor's_theorem Nope. There are other proofs. - William Hughes |