Re: Cantor Diagonal Proof

On 05/04/2025 08:25, Lawrence D'Oliveiro wrote:

On Sat, 5 Apr 2025 04:26:29 +0100, Mike Terry wrote:
 

You don't understand what a computable number is ...

 “A number which can be computed to any number of digits desired by a
Turing machine.”
<https://mathworld.wolfram.com/ComputableNumber.html>
 

Cutting and pasting text doesn't demonstrate understanding - correct use of the terms involved is how you do that.
Anyway, in your quoted definition the key phrase is "... *a* Turing machine", i.e. there must be *one single TM* which can calculate the n'th digit of the number for any n.
Note just to be doubly clear - not "for any digit there is a TM that calculates that digit", which is what you're effectively working with.
Yes, of course any specific digit is computable.  I used a smiley in my earlier post, but I'm not sure you got that, so I'll replace it here with "DUH! Of course any specific /digit/ is computable - any natural number is "computable" by the same measure, and digits only go up to (let's say) 10, so only 10 different TMs are needed to compute them!"
The problem is that to calculate say the first 100 digits of the Cantor "anti-diagonal" of Cantor's list requires input of 100 digits-worth of information out of the list.  More properly, the TM cannot receive 100 digits-worth on its tape, since by definition the tape contains only the number 100 and is otherwise empty.  Instead the 100 digits-worth of information is coded directly somehow into the TM state table.  But now if you need generate the 200th anti-diagonal digit you don't have enough information to do that - you only have information to generate the first 100.
Your responses keep harping on that you can calculate 200 digits (or any other finite number) with only finite data input, but hopefully you understand now that the only way you could do that is to keep introducing newer and newer (and ever larger) TMs to do the calculation as n increases.
For a computable number what you need to have is ONE SINGLE TM that does the whole lot in one go. You don't have that (as I explained a number of times).  [no point explaining this further I think...]
Mike.