Why can no one in sci.math understand my simple point?
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From: stevendaryl3016 on 19 Jun 2010 09:31 Peter Webb says... >Well, there are lots of definitions of a computable Real, I will use >Wikipedia's most intuitive definition: "computable reals, are the real >numbers that can be computed to within any desired precision by a finite, >terminating algorithm." That's too loose. It should be a an algorithm with *one* argument, the degree of precision required. An algorithm with two arguments doesn't count. You can define an algorithm that takes two arguments, a list of computable reals, and a natural n, which returns a computable real that is not on the list. But that real is not necessarily computable by an algorithm that only has one argument. >1. Given a list of computable Reals, we can identify the nth >computable Real on the list by simply counting down to the nth entry. > >2. Given the nth computable Real on the list, we can count off and identify >the nth digit of this Real. > >3. With the nth digit of the Real, we can use Cantor's construction to >identify the nth digit of the anti-diagonal. > >4. As we can specify every digit in the anti-diagonal explicitly and to any >desired degree of accuracy, it is therefore computable. You've described a computable function of *two* arguments: (1) the list, and (2) a natural n. For the antidiagonal to be computable, you must show that there is an algorithm that only takes one argument, a natural number n, and returns the nth decimal place of the antidiagonal. >5. But the anti-diagonal number does not appear on the list. > >6. Therefore, the list could not have included all computable Reals. > >Exactly the same as Cantor's proof that the Reals cannot be listed. The difference is that the antidiagonal of a list of reals returns a real that is not on the list. The antidiagonal of a list of computable reals does *not* necessarily return a computable real. >The only difference is that I have to prove that the anti-diagonal >is also computable which you haven't done. To prove that it is computable, you have to have an algorithm that takes one argument, n, and returns the nth decimal place of the real. You haven't shown that there is such an algorithm. You've shown that there is an algorithm that takes two arguments and produces the nth decimal place. >but because Cantor's proof explicitly constructs the >anti-diagonal it is clearly computable It's a computable function of *two* arguments. That doesn't mean that there is a computable function of *one* argument. -- Daryl McCullough Ithaca, NY
From: WM on 19 Jun 2010 09:54 On 19 Jun., 15:31, stevendaryl3...(a)yahoo.com (Daryl McCullough( wrote: > Peter Webb says... > > >but because Cantor's proof explicitly constructs the > >anti-diagonal it is clearly computable > > It's a computable function of *two* arguments. That doesn't > mean that there is a computable function of *one* argument. What a ridiculous discussion! Whatever Cantor's argument produces, it is something more or less definite. But everything that is definite enough to be distinguished from every other real number belongs to a countabe set. And everything less definite cannot be distinguished from other less definite objects and is certainly not a number. Cantor's "proof" shows either that a countabke set is uncountable or that a number that can be distinguished from every other number cannot be distinguished from every other number. It is a pity that you were told that many years ago and have not yet understood it! Regards, WM
From: WM on 19 Jun 2010 09:59 On 19 Jun., 15:06, stevendaryl3...(a)yahoo.com (Daryl McCullough( wrote: > Peter Webb says... > > >I agree that computable reals are countable. But I do not agree this means > >they can be listed. In fact, I can easily prove they are not. If you give me > >a purported list of all computable Reals I can use a diagonal argument to > >form a computable Real not on the list. > > You can use a diagonal argument to form a *real* that is not on the list. > For that real to be *computable*, you need to show that you can compute > that real *without* using the list. > > If the list were a computable list, then you could reconstruct it yourself, > so the antidiagonal would be computable. If the list is not computable, > then neither is the antidiagonal. All finite definitions of numbers map on infinite sequences of digits. But infinite sequences of digits do not map on finite definitions. It is not possible to define a number by an infinite sequence of digits without having the finite definition. Cantor's list consists of infinite sequences of digits. They are not defined and they do not define anything. Only finite definitions define numbers. But an infinite list of finite definitions cannot be diagonalized. Reagrsd, WM
From: WM on 19 Jun 2010 10:09 On 19 Jun., 05:10, Virgil <Vir...(a)home.esc> wrote: > In article > > It is. Every real number that is defined is defined by a finite word > > (definition or formula). It is impossible to define a number by an > > infinite sequence, because the sequence never ends and the definition > > is never known. > > F:N -> R: n -> 3/10^n is an infinite sequence defining a real number. No, "F:N -> R: n -> 3/10^n" is a finite rule *producing* an infinite sequence! > > So as one can easily see it is possible to have an infinite sequence > defining a real number. > > In real mathematics, an infinite sequence is merely a function whose > domain is the set of natural numbers, and there are lots of them which > define real numbers. There is none. But you are not able to grasp the difference between an infinite sequence and the rule producing that sequence. Sorry, but if you cannot think so far, then further discussion is meaningless. Regards, WM
From: Sylvia Else on 19 Jun 2010 10:13
On 19/06/2010 7:07 PM, |-|ercules wrote: > "Sylvia Else" <sylvia(a)not.here.invalid> wrote ... >> On 19/06/2010 4:11 PM, |-|ercules wrote: >> >>> To support your argument you should at least show that you've formed a >>> new sequence of digits. >> >> I'll explain it simply then. The first digit of the created number >> differs from the first digit of the first number in the list. The >> second digit differs from the second digit of the second number in the >> list. >> >> In general, digit n differs from digit n of the nth number in the list. >> >> So for all n, the created number differs from number n. Therefore the >> created number is not in the list - it is a new sequence of digits. > > No I've told you all 20 times that does not create any new sequence at all. > > All you've done is > CONSTRUCT a digit sequence like so > An AD(n) =/= L(n,n) > > And then you say, it's different to each number like so > > PROOF > An AD(n) =/= L(n,n) > > But you have not demonstrated a NEW SEQUENCE OF DIGITS. How can it not be a new sequence of digits if it's not in the list? > > All you've done is this > > [ An AD(n) =/= L(n,n) -> An AD(n) =/= L(n,n) ] -> Superinfinity > > Your actual 'proof' is a specific example of the above 'proof'! > > [ An AD(n) = (L(n,n) + 1) mod 9 -> An AD(n) =/= L(n,n) ] -> Superinfinity > > Do you agree with the above version of Cantor's proof? That is not a statement of Cantor's proof. For a start, it leaves out the assumption that the list of numbers is a list of all the reals. >>> >>> If you actually read my derivation of herc_cant_3 instead of blindly >>> dismissing it, >>> you'll see it holds, just like all digits of PI appear in order below >>> this line, if interpreted >>> correctly. >>> >>> Herc >>> >>> ___________________ >>> >>> 3 >>> 31 >>> 314 >>> 3141 >>> ... >>> >>> >> >> herc-cant-3 is not a derivation. It's a wild leap of faith. Nothing is >> proved therein. >> >> Sylvia. > > > Then which step do you disagree with? > > > defn(herc_cant_3) > The list of computable reals contains every digit (in order) of all > possible infinite sequences. > > Derivation > > Given the increasing finite prefixes of pi > > 3 > 31 > 314 > .. > > This list contains every digit (in order) of the infinite expansion of pi. > > Given the increasing finite prefixes of e > > 2 > 27 > 271 > .. > > This list contains every digit (in order) of the infinite expansion of e. > This one: > Given the increasing finite prefixes of ALL infinite expansions, > that list contains every digit (in order) of every infinite expansion. You provide no justification for that statement. It doesn't follow from what came previously. You just assert it. Sylvia. |