"Mike Terry" <email@example.com> wrote in message news:dLGdnUdTkbKpzoPRnZ2dnUVZ8hKdnZ2d@brightview.co.uk... > "Peter Webb" <webbfamily@DIESPAMDIEoptusnet.com.au> wrote in message > news:firstname.lastname@example.org... >> >> Perhaps if you could explain how my proof differs from Cantor's >> >> proof, that would be a start. >> > >> > I have done already, but perhaps I can explain it in different terms >> > for you. >> > >> > >> >> You seem to accept Cantor's proof, but not mine. Yet they are almost >> >> identical, the only difference being I have inserted the word >> >> "computable" in front of "reals". >> >> >> >> I can't see how you can accept one and not the other. >> > >> > Cantor's proof shows that for any mapping L:N->R, antidiag(L) is real >> > and not in range(L). Cantor's proof does include a proof that >> > antidiag(L) is real. It does not show that antidiag(L) is a >> > computable real. >> > >> >> Yes. >> >> > You cannot just drop "computable" in there and suppose that the logic >> > works, just as you cannot drop "rational" in there and suppose that >> > the logic works. >> > >> >> Well, it does work. >> >> > If you want to prove that for any mapping L:N->R, antidiag(L) is >> > computable, you need to include a proof that antidiag(L) satisfies the >> > mathematical definition for a computable real. >> > >> >> OK. >> >> I an specify it to n places for all n using the following simple > algorithm. >> >> Change the nth digit of the nth term from a 1 to a 2 and else to a 1. >> >> Thus if your list contained say: >> >> .1111111 >> .2222222 >> .1111222 >> >> The antidiagonal is 0.212... >> >> Notice that I computed this quite easily? >> >> Every item on the list is computable. So every item is known to at least >> n >> decimal places of accuracy, and Cantor gives us a very easy way to >> compute >> the first n terms on the anti-diagonal given the first n items on the >> list >> to n decimal places. >> >> >> > >> >> It is extremely easy to compute. >> >> >> >> Take the nth decimal place of the nth term. >> > >> > The nth decimal place of the nth term of what? The list L? That's >> > fine if you permit infinite lists of reals as input to the algorithm, >> > but the definition of "computable real" permits no such thing. >> > >> >> No, to calculate the first n terms of the anti-diagonal you require only > te >> first n items on the list to n places of accuracy. > > So you require as input: > - the 1st digit to 1 place of accuracy > - the 2nd digit to 2 places of accuracy > - the 3rd digit to 3 places of accuracy > - ... > > The ... above is the give away! For your algorithm to work its going to > require an INFINITE amount of input data. (Remember, you are trying to > find > a SINGLE stand-alone FINITE algorithm that produces ALL the required > digits > of the antidiagonal.) >
No, I only need a finite algorithm which will calculate it to n decimal places with finite input. The definition of computable only requires me to produce it to arbitrary accuracy.
I'm pretty sure you can't produce an algorithm which will produce *every* digit in sqrt(2) in finite time. Sqrt(2) is computable because we can calculate it to arbitrary accuracy, not because we can explicitly list every digit in its decimal expansion in finite time; we can't.
> So Tim was quite right, and you haven't answered his point correctly > yet... > > Regards, > Mike. > >