"Tim Little" <firstname.lastname@example.org> wrote in message news:email@example.com... > On 2010-06-19, Peter Webb <webbfamily@DIESPAMDIEoptusnet.com.au> wrote: >>> The relevant point: the *only* input to the Turing machine in the >>> definition is n. The rest of the tape must is blank. Peter >>> apparently believes that a number is still computable even if the >>> Turing machine must be supplied with an infinite amount of input (the >>> list of reals). >> >> No. > > Oh? Then what leads you to believe that the antidiagonal of a (not > necessarily recursive) list of computable reals is computable? >
Cantor provides an explicit construction for the 1st digit, the second digit, the third digit, etc.
If you give me a list of Computable Reals, its trivially easy to explicitly construct the anti-diagonal to any degree of accuracy.
> >> Do you agree that Cantor's diagonal proof when applied to a >> purported list of all computable Reals will produce a computable >> Real not on the list > > No, for the fifth time now. The antidiagonal of a list of all > computable real is not computable. How many more times would you like > me to repeat this simple and mathematically obvious statement? >
Basically, zero more times, because its crap.
If you don't believe that I can explicitly construct a Real not on any list of computable Reals you give me, try giving me a list of computable Reals and I will produce a missing computable Real to any required degree of accuracy.
The only restriction I place is the restriction that Cantor also placed - that I can go through the list, and identify the nth digit of the nth item for all n.
After all, I am trying to make my proof exactly the same as Cantor's, but with the only difference being the word "computable" in front of the word "Real".