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Topic:
Matheology § 257
Replies:
11
Last Post:
Apr 23, 2013 5:05 PM




Re: Matheology � 257
Posted:
Apr 22, 2013 2:17 PM


"WM" <mueckenh@rz.fhaugsburg.de> wrote in message news:800fc6441b3d4161930b9844d1826854@cd3g2000vbb.googlegroups.com...
Matheology § 257
>The set of all possible computer programs is countable {{if going into >concrete details of computers, we can even say more restrictively: The >number of all possible programs is finite and certainly less than >2^10^100.
that is not proof, you are simply guessing.
>But of course Turing did not refer to real computers and did >not know anything about the memory space of an exploitable >surronding. Therefore let us analyze his approach here.}}, therefore >the set of all computable reals is countable, and diagonalizing over >the computable reals immediately yields an uncomputable real. Q.E.D. >{{Well, it is not always that easy to diagonalize over finite >sequences. The following list has no diagonal: >0 >1 >So we must take some more care, as will be done in the following.}} > Let's do it again more carefully. > Make a list of all possible computer programs. Order the programs >by their size, and within those of the same size, order them >alphabetically. The easiest thing to do is to include all the possible >character strings that can be formed from the finite alphabet of the >programming language, even though most of these will be syntactically >invalid programs.
so now you are *not* using computer programs..... a sequence from some finite alphabet, why not just a binary string (representing (not), wait for it, *psudo machine code*! )
> Here's how we define the uncomputable diagonal number 0 < r < 1. >Consider the kth program in our list. If it is syntactically invalid, >or if the kth program never outputs a kth digit, or if the kth digit >output by the kth program isn't a 3, pick 3 as the kth digit of r. >Otherwise, if the kth digit output by the kth program is a 3, pick 4 >as the kth digit of r.
why 3 or 4 ? who is picking ? There Nose ?
>. After all you cannot use a program that is longer than the longest program that is used.
depends on when it is used, then it can add a byte to make itself longer after you have used it.
>Regards, WM



