Why can no one in sci.math understand my simple point?
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From: WM on 16 Jun 2010 11:43 On 16 Jun., 14:05, "J. Clarke" <jclarke.use...(a)cox.net> wrote: > On 6/15/2010 11:59 AM, Aatu Koskensilta wrote: > > > > > > > WM<mueck...(a)rz.fh-augsburg.de> writes: > > >> We should not use oracles in mathematics. > > > Nonsense. Using orcales we can show for example that the P = NP problem > > can't be solved using any technique that relativizes. This is a useful > > result. > > >> A real is computable or not. My list contains all computable numbers: > > >> 0 > >> 1 > >> 00 > >> ... > > > Your list doesn't appear to contain any real at all, just finite binary > > sequences. > > Did someone redefine the real numbers to exclude all numbers that > consist only of the digits 0 and 1? All definable real numbers and all their possible representations are contained in that list. This list is a list of everything. The alphabets and the languages and the dictionaries are given in later lines. It is very probable that every line has many different meaning, but no line has uncounatbly many meanings. Therefore the list contains only countably many finite definitions. And obviously the list cannot be diagonalized - like every meaningful list of meaningfuls words. Regards, WM
From: MoeBlee on 16 Jun 2010 11:46 On Jun 15, 10:37 pm, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > Virgil <Vir...(a)home.esc> writes: > > But until you can determine which of those 10 cases, how can you > > compute the number? > > You can't. The number is computable nonetheless, in the sense that there > exists an effective procedure for churning out its decimal expansion. > > As noted, computability is a purely extensional notion. Recall the > classical recursion theory exercise, which we find, in some form or > other, in pretty much any text on the subject: > > Let f : N --> N be a function such that > > f(x) = 0 if Goldbach's conjecture is true, and 1 otherwise. > > Is f computable? Hey, even I, a pathetic tyro, knows that one (I think)! If GC is true, then f is the constant function whose value is always 0 If GC is false, then f is the constant function whose value is always 1. So, in either case, f is recursive. I'm ready for my PhD! MoeBlee
From: Aatu Koskensilta on 16 Jun 2010 11:55 WM <mueckenh(a)rz.fh-augsburg.de> writes: > It is very probable that every line has many different meaning, but no > line has uncounatbly many meanings. Every line has an indefinite and indeterminate number of possible meanings. It makes no sense to speak of the cardinality of the totality of all possible meanings of a string of symbols. > Therefore the list contains only countably many finite definitions. The list contains just random, meaningless strings. Whenever we instill, with mathematical precision, these strings with some definite meaning, so that they become definitions of reals, we find there are definitions not included in the list. There is an absolute notion of computability in logic. There is no absolute notion of definability. -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: MoeBlee on 16 Jun 2010 11:59 On Jun 16, 12:34 am, Virgil <Vir...(a)home.esc> wrote: > In article <874oh3ir6m....(a)dialatheia.truth.invalid>, > Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > > Let f : N --> N be a function such that > > > f(x) = 0 if Goldbach's conjecture is true, and 1 otherwise. > > > Is f computable? > > Not yet! There exists an algorithm to compute f. We don't know what the algorithm is (we don't know whether it is to set f(x) = 0 for all x, or to set f(x) = 1 for all x), but still the algorithm exists - either the algorithm is to set f(x) = 0 for all x, or the algorithm is to set f(x) = 1 for all x.. In ordinary study of computability, that is what it means to say that a funciton is computable - that there exists an algorithm to compute it. If you wish to include also that the algorithm is KNOWN (known to whom? and known when?) then you need to find some treatment that supports that notion or invent your own. But then would you say that each function becomes computable only historically as certain people, through history, figured out algorithms for various functions? So whether a function is computable is relative to what certain people know at certain times (and even as some people know at a certain time but other people don't)? MoeBlee
From: Aatu Koskensilta on 16 Jun 2010 12:07
Herman Jurjus <hjmotz(a)hetnet.nl> writes: > Also many classical mathematicians appreciate this as an example > showing that the extensional notion 'Turing computable' is a slight > distortion of the intuitive notion 'computable'. Possibly, but I don't think this is quite the right diagnosis. The issue is more subtle. It's a well known phenomenon that many classically minded mathematicians who have had little practice in constructive thinking are unwittingly inconsistent in their reading of intuitionistic quantifiers. It's an equally striking phenomenon that classically minded mathematicians in certain contexts naturally adopt an intuitionistic reading of classical quantifiers. In addition to the example provided by Virgil it's not uncommon for people to mistakenly think, in a classical context, that countable sets come equipped with a designated bijection witnessing their countability. -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus |