A BLATENT FLAW in Cantor's diag proof
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From: Daryl McCullough on 9 Jun 2010 09:08 |-|ercules says... >It is your aversion to express your fundamentally flawed mathematical >assertions into ENGLISH that causes you to MISS blatant errors in >those assertions. The whole reason for developing mathematics as a language (as opposed to doing everything in English, or Latin, or whatever) is because the language of mathematics is much more precise, and resolves ambiguities in natural language. Before answering a question expressed informally in a natural language, the mathematician must first figure out what the question means, precisely. Casting it in more formal language is one common approach to clarifying a question. For some reason, you always do the opposite. Whenever you see a statement that is clear and unambiguous, you want to paraphrase in a form that is murky, unclear and ambiguous. You seem to like incoherence. That's fine for entertainment purposes, but it is not fine for meaningfully discussing mathematics. Unless you can express what you are talking about clearly, you cannot do mathematics >There is NOTHING wrong with this statement. Nothing except that it is an incoherent mess. -- Daryl McCullough Ithaca, NY
From: Tim Little on 9 Jun 2010 20:21 On 2010-06-09, |-|ercules <radgray123(a)yahoo.com> wrote: > Here's my computable reals list. > > The pos digit of the index real is > UTM(index, pos) mod 10 The problem with your proposal: many Turing machines never halt, so how do you propose to define (let alone compute) the mod-10 value of their nonexistent output? - Tim
From: |-|ercules on 9 Jun 2010 20:57 "Tim Little" <tim(a)little-possums.net> wrote > On 2010-06-09, |-|ercules <radgray123(a)yahoo.com> wrote: >> Here's my computable reals list. >> >> The pos digit of the index real is >> UTM(index, pos) mod 10 > > The problem with your proposal: many Turing machines never halt, so > how do you propose to define (let alone compute) the mod-10 value of > their nonexistent output? If the diag argument doesn't work on a hypothetical list it doesn't work at all. The hypothetical list of all computable reals is a subset of my list. If no new sequence can be found that is not on a subset of my list, then it's not on my full list either, blanks and all! Herc
From: George Greene on 10 Jun 2010 00:27 On Jun 8, 11:43 pm, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > I suggest you revisit Turing's 1936 paper on computable numbers. You may do as you like, but anything you do in public has the property that you must surely have thought it might have some effect on the outside world, an effect somehow causally connected to your enjoyment of having done it. Unfortunately, if you thought that in this case, you were mistaken.
From: George Greene on 10 Jun 2010 00:38
> George Greene <gree...(a)email.unc.edu> writes: > > If you order the list properly then the list will be computable as > > well. On Jun 8, 11:43 pm, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > I suggest you revisit Turing's 1936 paper on computable numbers. One of the reasons why I'm better at this than you are is that I am not foolish enough to make this sort of suggestion. The actual correct reply was "No, it won't." It (still, after 20 years) astonishes me that you would suggest that somebody read a paper, regarding a point that can be made in 3 sentences. |