CANTOR DISPROOF <<<<<<<<<<<<<<<<
in [Logic]
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From: George Greene on 27 Jun 2010 00:32 On Jun 26, 6:05 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: > Your 'IS THEN ALSO computable' is as far as we can comment. It's a hypothetical list > so the computability of the diagonal is irrelevant. Oh, believe me, it's very relevant. If the anti-diagonal were not computable then there are a lot of people who would say that we didn't need to worry about it, that it didn't exist. After all, it would not be on the list of all computable reals if it were not computable, right? And we are SAYING that it's not on the list, right? So if it's not computable, then THERE IS NO contradiction and YOU WIN! > It is perplexing if outputs of all computer programs are listed, how do you find a program > to compute the diagonal digit at the position that is contradictory? YOU DON'T! That's the whole point! This is the first intelligent question you have asked in a long time! The list of all computable numbers IS ITSELF NOT computable! If it were, we would have precisely the contradiction you just almost stumbled onto! There IS NOT a computer program that lists the outputs of all computer programs! The LIST exists, but howEVER you got it, you DIDN'T get it from a computer. Once you have it, however, you could run any computer program on it you liked, as long as it was looking for a list (of that size and shape) as input. That list still has an anti-diagonal and its anti-diagonal WOULD be computable IF the original list were (BUT IT ISN'T).
From: George Greene on 27 Jun 2010 00:41 On Jun 26, 6:05 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: > Your 'IS THEN ALSO computable' is as far as we can comment. > It's a hypothetical list > so the computability of the diagonal is irrelevant. DAMN, you're stupid. If I said, "do you know how to multiply a number by ten, in decimal?" You COULD say, if you only had a brain, "Oh, that's trivially computable; you just tack a zero on the (right) end." That is TRUE, but BECAUSE YOU ARE STUPID, you would disagree with and say, "It's a hypothetical number, so the computability of 10-times-it is irrelevant". DUMBASS: WHEN YOU ARE DEFINING A FUNCTION, *IT'S*ALWAYS* A HYPOTHETICAL number/list/argument whatEVER! If I tell you that f(x)=x^2, the x IS A HYPOTHETICAL NUMBER, but that does NOT make *f* any LESS well-defined!! f IS A COMPUTABLE FUNCTION! And for (square) lists, the anti-diagonal is a computable FUNCTION. YOU CAN ALWAYS compute the anti-diagonal IF YOU HAVE THE LIST ALREADY. It is IMPORTANT that the list be purely hypothetical! If the list were something specific then you might face the problem of knowing how to compute the anti- diagonal for THAT list, but NOT for other kinds! Here, WE DON'T HAVE that problem! I have EXACTLY ONE personal favorite computer program, EXACTLY ONE TM, that works ON ALL square lists (and halts/rejects if the list is not square).
From: George Greene on 27 Jun 2010 00:43 On Jun 26, 6:05 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: > It is perplexing if outputs of all computer programs are listed, how do you find a program > to compute the diagonal digit at the position that is contradictory? You just write the program that says output(n) = 9 - L(n,n) AND YOU'RE DONE, VOILA, WHOOT, THERE IT IS! It is a VERY SIMPLE program. But there is no "contradictory" position. The number being computed simply IS NOT ON the list. You ONLY get a contradiction IF YOU ASSUMED that ALL the numbers were originally on the list. There is a list of all computable reals but IT is NOT computable.
From: Sylvia Else on 27 Jun 2010 00:49 On 27/06/2010 2:15 PM, |-|ercules wrote: > "Tim Little" <tim(a)little-possums.net> wrote >> On 2010-06-27, Sylvia Else <sylvia(a)not.here.invalid> wrote: >>> I've already shown that your induction doesn't produce a list of all >>> infinite sequences, because 1/9 isn't in the list - it would have >>> infinitely many predecessors. > > huh? this is total garbage, the Complete Permutation Sets are sets of oo > long reals. > I've explained this 5 times already. It is the prefixes of complete > permutations that are 'finite', > not the reals themselves. > > >> >> Yes, and what's worse it doesn't even have a direct predecessor. >> Induction starts with a base case "P(0) is true" and uses implication >> "P(n)->P(n+1)" to prove "for all natural numbers n, P(n) is true". >> There is no sequence length "n" for which an infinite sequence appears >> in length "n+1", so induction doesn't apply. > > > The proof is so simple I am surprised nobody can spot the inductive step. > > The initial step is a trivial case of all finite prefixes of digit width > 1, e.g. > > CPS_1 = { > 0.04748574849.. , > 0.14848498483.. , > 0.24848439494.. , > 0.34505940940.. , > 0.45059594040.. , > 0.54584848483.. , > 0.60606060606.. , > 0.73748747484.. , > 0.83834838484.. , > 0.94834748484.. , > } > > Back to Sylvia, you just made a lucid post that my proof COULD work No, I didn't say that. , and > the attacks > based on ZFC are assumptive. > > This contradicts your other claims that the proof is not valid. > > When you say "YOU HAVEN'T PROVEN THAT" > you should word it "YOU HAVEN'T PROVEN THAT TO ME". How about, "THERE IS NO PERSON OTHER THAN HERC WHO ACCEPTS THE PROOF AS VALID" Sylvia.
From: |-|ercules on 27 Jun 2010 01:09
"Sylvia Else" <sylvia(a)not.here.invalid> wrote ... > On 27/06/2010 2:15 PM, |-|ercules wrote: >> "Tim Little" <tim(a)little-possums.net> wrote >>> On 2010-06-27, Sylvia Else <sylvia(a)not.here.invalid> wrote: >>>> I've already shown that your induction doesn't produce a list of all >>>> infinite sequences, because 1/9 isn't in the list - it would have >>>> infinitely many predecessors. >> >> huh? this is total garbage, the Complete Permutation Sets are sets of oo >> long reals. >> I've explained this 5 times already. It is the prefixes of complete >> permutations that are 'finite', >> not the reals themselves. >> >> >>> >>> Yes, and what's worse it doesn't even have a direct predecessor. >>> Induction starts with a base case "P(0) is true" and uses implication >>> "P(n)->P(n+1)" to prove "for all natural numbers n, P(n) is true". >>> There is no sequence length "n" for which an infinite sequence appears >>> in length "n+1", so induction doesn't apply. >> >> >> The proof is so simple I am surprised nobody can spot the inductive step. >> >> The initial step is a trivial case of all finite prefixes of digit width >> 1, e.g. >> >> CPS_1 = { >> 0.04748574849.. , >> 0.14848498483.. , >> 0.24848439494.. , >> 0.34505940940.. , >> 0.45059594040.. , >> 0.54584848483.. , >> 0.60606060606.. , >> 0.73748747484.. , >> 0.83834838484.. , >> 0.94834748484.. , >> } >> >> Back to Sylvia, you just made a lucid post that my proof COULD work > > No, I didn't say that. > > , and >> the attacks >> based on ZFC are assumptive. >> >> This contradicts your other claims that the proof is not valid. >> >> When you say "YOU HAVEN'T PROVEN THAT" >> you should word it "YOU HAVEN'T PROVEN THAT TO ME". > > How about, "THERE IS NO PERSON OTHER THAN HERC WHO ACCEPTS THE PROOF AS > VALID" > > Sylvia. That's speculation. But you appear to have no idea if it's proof of anything. Herc |