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Re: The Diagonal Argument
Posted:
Dec 27, 2012 4:25 PM
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> Try to Visualise an example.
>
> L(x,y)
> +---------------->
> | 0. 2 3 4 5 6 7 ..
> | 0. 9 8 7 6 5 5 ..
> | 0. 1 2 3 1 2 3 ..
> | 0. 9 8 9 8 9 8 ..
> | 0. 6 5 6 5 6 5 ..
> | 0. 5 6 5 6 5 6 ..
> |
> v
>
> Now apply your FLIP(d) function to the whole plane
>
> T(x,y)
> +---------------->
> | 0. 6 6 6 6 5 5 ..
> | 0. 5 5 5 5 6 6 ..
> | 0. 6 6 6 6 6 6 ..
> | 0. 5 5 5 5 5 5 ..
> | 0. 5 6 5 6 5 6 ..
> | 0. 6 5 6 5 6 5 ..
> |
> v
>
> Your claim is that is you take any path from
>
> T(1,?)
> T(2,?)
> T(3,?)
> ...
>
> and repeat that process you must end up with an infinite string absent
> from L?
i.e. ANTIDIAG = T(1,1) T(2,2) T(3,3) T(4,4) ...
But Obviously T(1,1) T(2,99) T(3,10110) T(4,7) ...
is not provably absent from L.
Remember Given a Stack of ESSAYS with every possible sentence written
in every possible order, taking the 1st word of Essay 1, changing it,
then the 2nd word of Essay 2, changing it, never produces a unique
sentence or any original writing at all! Similarly the ANTIDIAG
PROCESS never conjures a Unique Digit Sequence!
In fact, using a Symmetric FLIP(d) Function
L(x,y)
+---------------->
| 0. 2 3 4 5 6 7 ..
| 0. 9 8 7 6 5 5 ..
| 0. 1 2 3 1 2 3 ..
| 0. 9 8 9 8 9 8 ..
| 0. 6 5 6 5 6 5 ..
| 0. 5 6 5 6 5 6 ..
|
v
FLIP(d) = 9-d
Minor Problem with:
0.49999...
<=FLIP=>
0.50000...
T(x,y) = FLIP(L(x,y))
+---------------->
| 0. 7 6 5 4 3 2 ..
| 0. 0 1 2 3 4 4 ..
| 0. 8 7 6 8 7 6 ..
| 0. 0 1 0 1 0 1 ..
| 0. 3 4 3 4 3 4 ..
| 0. 4 3 4 3 4 3 ..
|
v
NOW DIAGONAL(T) is supposedly proven absent from L
0.716133.. NOT COUNTED??
yet if L is the Computable Reals then
T=L
PROOF: For every computable real there is another computable real for
all digit changing functions.
which proves the DIGIT FLIP Operation is a NULL OPERATION
THERFORE ANTIDIAGONAL(L) is no more provably absent from L than
DIAGONAL(L).
QED
Herc
--
www.CAMGIRLS.com
TOTAL: $2834 2012-12-21 Fri
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