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.
> 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.
> 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. 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.
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.