Why can no one in sci.math understand my simple point?
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From: Virgil on 15 Jun 2010 18:35 In article <87ocfchwa7.fsf(a)dialatheia.truth.invalid>, Aatu Koskensilta <aatu.koskensilta(a)uta.fi> wrote: > Virgil <Virgil(a)home.esc> writes: > > > A real is uncomputable if it is know to be one of a set of 2 or more > > known reals, but it cannot be determined which one. > > Not in classical mathematics. Then how does one compute it? According to http://en.wikipedia.org/wiki/Computable_number Such a "number" would have no turing machine if it would require infinitely many steps do determine that one ambiguous digit. So would it be, in that sense uncomputable?
From: Virgil on 15 Jun 2010 18:39 In article <87iq5khv47.fsf(a)dialatheia.truth.invalid>, Aatu Koskensilta <aatu.koskensilta(a)uta.fi> wrote: > Virgil <Virgil(a)home.esc> writes: > > > There are undecidable propositions in mathematics, so if P is one of > > them then "x = 1 if P is true otherwise x = 0" defines an uncomputable > > number. > > Not in classical mathematics. In classical mathematics are there propositions that are undecideable only because they would require infinitely many steps to decide the value of a particular digit? Probably!
From: |-|ercules on 15 Jun 2010 19:16 "Daryl McCullough" <stevendaryl3016(a)yahoo.com> wrote ... > |-|ercules says... >> >>"Daryl McCullough" <stevendaryl3016(a)yahoo.com> wrote... > >>> That's *all* that matters, for Cantor's theorem. The claim >>> is that for every list of reals, there is another real >>> that does not appear on the list. >> >> >>Yes but HOW does Cantor show that? > > You've been told many times. He shows that for every > list L of reals, there is another real antidiag(L) that > is defined in such a way that > > forall n, antidiag(L) differs from the nth real in L at > the nth decimal place. > > From this, it follows: > > forall n, antidiag(L) is not equal to the nth real. > > From this, it follows: > > antidiag(L) is not on the list L. > Your argument is like this. ----------------------------------------- <Daryl> Consider the infinite series 1/2 + 1/4 + 1/8 + ... At every element in the series, the sum =/= 1. Therefore the sum of the series =/= 1. </Daryl> --------------------------------------------- But you should consider the entire infinite series, as you should consider the entire infinite list of computable reals, which has this property. >the list of computable reals contain every digit of ALL possible infinite >sequences (3) Herc
From: Marshall on 15 Jun 2010 20:05 On Jun 15, 12:58 pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > > In what form does the barbecue pork bun I'm eating exist? Damn I love those things. Maybe I'll order delivery Chinese tonight. Now I'm hungry. Marshall
From: Peter Webb on 15 Jun 2010 20:39
> Nevertheless your "definition" belongs to a countable set, hence it is > no example to save Cantors "proof". > > Either all entries of the lines of the list are defined and the > diagonal is defined (in the same language) too. Yes. If you provide a list of Reals, then the diagonal is computable and does not appear on the list. > Then the proof shows > that the countable set of defined reals is uncountable. No, it shows that all "definable" (computable) Reals cannot be explicitly listed. This is *not* the same as being uncountable. > Or it does not > show anything at all. > It shows that all "definable" (computable) Reals cannot be explicitly listed. This is a well known proof in set theory. This is *not* the same as being uncountable. > To switch "languages" is the most lame argument one could think of. > The diagonal argument does not switch languages. And it cannot be > applied at all because the list of all finite defiitions does not > contain infinite entries. Those however are required for the diagonal > argument. > No, that paragraph above is close to gibberish. Cantor said and proved that any purported list of all Reals cannot contain all Reals. His proof is simple and clear, provides an explicit construction of at least one missing Real, and does contain or require any concepts of uncomputable numbers, or use of the Axiom of Choice. Perhaps if you were to identify the step in Cantor's proof that you consider wrong, then we might gain some idea as to what you are actually objecting to? > Regards, WM |