Why can no one in sci.math understand my simple point?
in [Logic]
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From: George Greene on 15 Jun 2010 18:29 On Jun 15, 5:02 pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > So WM calls me a crank. Big deal. Indeed. My medal is a brighter color than yours because WM NEVER CALLS ME ANYthing. Seriously, I rebutted his bullshit so well the first time that he was forced to simply stop replying to me ALTOGETHER as long as other people were in the room. This is why you will occasionally see me bullying not only the morons on the other side but even the less competent ON OUR side, simply because WM is smart enough to make hay out of objections that are not actually on point.
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 |