"Tim Little" <tim@little-possums.net> wrote in message news:slrni1tm56.jrj.tim@soprano.little-possums.net... > On 2010-06-21, Peter Webb <webbfamily@DIESPAMDIEoptusnet.com.au> wrote: >> The same algorithm does work for all n. Look at the first n terms to >> n places of accuracy. > > Provide computer code, please. Here's the required function > signature: int getDigit(BigInteger n). You get to fill in the > function body. > > >>> For some fixed M, you could embed M digits of the diagonal into the >>> algorithm, and the algorithm would work for all n < M. But that's not >>> enough: it has to work for all n. >> >> I can in fact compute the nth digit of the anti-diagonal for any n >> by hand. > > "By hand" is irrelevant to the definition of a computable real, > especially when your "hand" is connected to "eyes" that have the list > as input. The list is *not* input to the algorithm. >
Yeah, it is.
Cantor constructs the antidiagonal by taking the list and then computing a missing Real based upon the decimal expansion of every Real on the list.
> >> The same objection could be raised to Cantor's proof; after all, he >> accepts a purportedly infinite list as input to his algorithn as >> well. > > Cantor does not claim to be proving existence of any finite algorithm. > You are. >
Cantor explicitly calculates the real that is missing be examining the decimal expansion of every item in the list. No more or less than I do.
> >> Or are you saying I have to do this without using the infinite list >> itself? > > You may use the infinite list to determine which finite algorithm to > use, but the algorithm must be self-contained and not refer to the > list.
Not in Cantor's proof. He explicitly uses the decimal expansion of every Real on the list to construct the anti-diagonal.
Of course Cantor has know what's on the list to determine an element which is missing. In exactly the same way as I do.
> The *only* input to the algorithm you select will be a natural > number, which denotes n. > > >> I note that Cantor's original proof also requires an infinite list >> of Reals, but you seem to have no problem with that. > > Indeed I don't. Cantor is not claiming that the antidiagonal is a > computable real, and so finite algorithms are irrelevant. >
Yet he explicitly constructs the missing Real to arbitrary accuracy.
> You *are* claiming that the antidiagonal is a computable real, and so > you must prove the existence of a suitable finite algorithm for the > antidiagonal. >
Actually, I only need to be able to specify it to arbitrary accuracy for it to be computable, and I can certainly do that.
Or are you trying to claim that while I can calculate the explicit decimal expansion of the missing computable Real, I can't compute the missing computable Real itself? If I can compute every digit in the missing computable number, I have computed the number itself.
