"Peter Webb" <webbfamily@DIESPAMDIEoptusnet.com.au> wrote in message news:4c1e3743$0$1028$afc38c87@news.optusnet.com.au... > >> 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.)
So Tim was quite right, and you haven't answered his point correctly yet...
