"Tim Little" <email@example.com> wrote in message news:firstname.lastname@example.org... > On 2010-06-18, Peter Webb <webbfamily@DIESPAMDIEoptusnet.com.au> wrote: >> If you can construct a list of all computable numbers (which you >> can't), then Cantor's diagonal proof will construct a number not on >> the list. And that number is definitely computable, because there is >> a simple algorithm for producing it. Take the nth digit of the nth >> item on the list > > That requires having the list be computable or provided as input, > neither of which is assumed or proven. >
Cantor's proof that there can be no list of all Reals also requires the list to be provided first.
The form of the proof is identical - you give me a purported list of all Reals with some property, and I prove the existence of at least one Real which has that property which is not on the list, thus proving the list did not contain all numbers with that property. Cantor used the property "is Real", I used the property "is computable".
> Rest snipped as it is based on your false premise. > > > - Tim
If you believe that you can list all computable numbers, you are welcome to try and do so. Its a pointless exercise, as I can always form a computable Real not on the list, but try anyway.
(This is like arguing with somebody who doesn't understand Cantor's proof that you cannot list all Reals. The proof is very simple, but people don't believe it. If all else fails, I get them to propose a list, and show it has a missing Real. I am happy to explicitly construct a computable Real that isn't on your purported list of all computable Reals).