Prev: Posting Msgs with Math Symbols?
Next: A semigroup is not isomorphic to the semigroup of its machine
From: |-|ercules on 12 Jun 2010 06:29 Every possible digit sequence is computable to ALL (an INFINITE AMOUNT of) finite initial substrings. ___________________________________________________________________________________________________ 2 or 3 posters on sci.math have agreed with that statement. BTW: ALL (an INFINITE AMOUNT of) natural numbers are in "all natural numbers" incase any of you want to feign comprehension disability. Is it too much for you to fathom that I disagree whether no box containing the box numbers that don't contain their own box numbers is self referential on some level? MAYBE, just MAYBE there is another explanation than sets larger than infinity? Do you believe there are numerous different digits (at finite positions) all along the expansion of some reals that are not on the computable reals list? What does that mean? The only interpretation is "There is a finite substring between 2 digits (inclusive) that is not computable" which is a clear contradiction. Herc -- the nonexistence of a box that contains the numbers of all the boxes that don't contain their own box number implies higher infinities. - Cantor's Proof (the holy grail of paradise in mathematics)
From: hagman on 12 Jun 2010 07:51 On 12 Jun., 12:29, "|-|ercules" <radgray...(a)yahoo.com> wrote: > Every possible digit sequence is computable to ALL (an INFINITE AMOUNT of) finite initial substrings. > ___________________________________________________________________________________________________ > > 2 or 3 posters on sci.math have agreed with that statement. > > BTW: ALL (an INFINITE AMOUNT of) natural numbers are in "all natural numbers" > incase any of you want to feign comprehension disability. > > Is it too much for you to fathom that I disagree whether no box containing the box numbers > that don't contain their own box numbers is self referential on some level? MAYBE, just MAYBE > there is another explanation than sets larger than infinity? > > Do you believe there are numerous different digits (at finite positions) all along the expansion > of some reals that are not on the computable reals list? > > What does that mean? > > The only interpretation is "There is a finite substring between 2 digits (inclusive) that is not computable" > which is a clear contradiction. > > Herc > -- > the nonexistence of a box that contains the numbers of all the boxes > that don't contain their own box number implies higher infinities. > - Cantor's Proof (the holy grail of paradise in mathematics) What's wrong with self-reference in a statement? The set of all sets not containing themselves would be a self- referential (aka. impredicative) definition and hence leads to a contradiction (i.e. there cannot be an object satisfying the definition). OTOH, given a collection F of boxes, where each box has a number and contains some numbers, whether or not one of these boxes contains its own number is a precisely defined predicate. For a single box this can be checked because the number n of the box can be determined by assumption and for each number, including that n, it can be determined by assumption, wehteer or not n is in the box. Finally, it is by means of quantification also possible to talk about whether or not the collection of F has the property that any or none of its boxes contains its own number.
From: |-|ercules on 12 Jun 2010 08:21 "hagman" <google(a)von-eitzen.de> wrote... > On 12 Jun., 12:29, "|-|ercules" <radgray...(a)yahoo.com> wrote: >> Every possible digit sequence is computable to ALL (an INFINITE AMOUNT of) finite initial substrings. >> ___________________________________________________________________________________________________ >> >> 2 or 3 posters on sci.math have agreed with that statement. >> >> BTW: ALL (an INFINITE AMOUNT of) natural numbers are in "all natural numbers" >> incase any of you want to feign comprehension disability. >> >> Is it too much for you to fathom that I disagree whether no box containing the box numbers >> that don't contain their own box numbers is self referential on some level? MAYBE, just MAYBE >> there is another explanation than sets larger than infinity? >> >> Do you believe there are numerous different digits (at finite positions) all along the expansion >> of some reals that are not on the computable reals list? >> >> What does that mean? >> >> The only interpretation is "There is a finite substring between 2 digits (inclusive) that is not computable" >> which is a clear contradiction. >> >> Herc >> -- >> the nonexistence of a box that contains the numbers of all the boxes >> that don't contain their own box number implies higher infinities. >> - Cantor's Proof (the holy grail of paradise in mathematics) > > What's wrong with self-reference in a statement? > The set of all sets not containing themselves would be a self- > referential (aka. impredicative) definition and hence leads to a > contradiction (i.e. there cannot be an object satisfying the > definition). > OTOH, given a collection F of boxes, where each box has a number and > contains some numbers, > whether or not one of these boxes contains its own number is a > precisely defined predicate. > For a single box this can be checked because the number n of the box > can be determined by > assumption and for each number, including that n, it can be determined > by assumption, wehteer or not n is in the box. > Finally, it is by means of quantification also possible to talk about > whether or not the collection of F has the property that any or none > of its boxes contains its own number. What's wrong is the wacky conclusions mathematicians get every time they self reference a collectic set and negate itself. You only make a point with Russell's set by putting it in perspective. i.e. does it mean that there is no set of all sets? Why accept such a naively defined set at all? The smarter definition would be "the set of all sets, that don't contain themselves, barring the set of all sets that don't contain themselves". Obviously it can't contain itself and be viable, so DECLARE THAT FACT. Collective sets come with an E&OE! Herc
From: Colin on 12 Jun 2010 12:15 On Jun 12, 7:21 am, "|-|ercules" <radgray...(a)yahoo.com> wrote: > [snip] Why do you insist on flogging a dead horse? Even wikipedia writes "Although the set of real numbers is uncountable, the set of computable numbers is countable and thus almost all real numbers are not computable. The computable numbers can be counted by assigning a Gödel number to each Turing machine definition. This gives a function from the naturals to the computable reals. Although the computable numbers are an ordered field, the set of Gödel numbers corresponding to computable numbers is not itself computably enumerable, because it is not possible to effectively determine which Gödel numbers correspond to Turing machines that produce computable reals. In order to produce a computable real, a Turing machine must compute a total function, but the corresponding decision problem is in Turing degree 0â² â². Thus Cantor's diagonal argument cannot be used to produce uncountably many computable reals; at best, the reals formed from this method will be uncomputable." So, the computable reals are countable, and Cantor's diagonal argument won't show they're uncountable. No one disputes this. Why do you keep insisting that there are people who do dispute it and keep trying to argue with them? You're arguing with people that don't exist.
From: |-|ercules on 12 Jun 2010 12:44 "Colin" <colinpoakes(a)hotmail.com> wrote >. Thus Cantor's diagonal argument cannot be used to produce > uncountably many computable reals; at best, the reals formed from this > method will be uncomputable." OK, but you can make a superset including the computable reals with the godel index. It's not conceptually difficult to question Cantor's argument on a hypothetical list. But there's little point going into detail, perhaps if it was acknowledged modifying the diagonal results in a new digit sequence that is not computable, then we could increase the scope to work out the mechanics. But it wouldn't demonstrate anything at this point, rather the opposite! Herc
|
Next
|
Last
Pages: 1 2 3 Prev: Posting Msgs with Math Symbols? Next: A semigroup is not isomorphic to the semigroup of its machine |