Regarding the OP, it isn't quite clear what the question is. However, here is one issue related to this point.
I think that perhaps you might be thinking that since the set ##Q## (rational numbers) must have various enormously complicated re-arrangements, shouldn't it be possible to have an arrangement where the "actual" diagonal reads like [without changing any digits that is]:
0.999999999...
0.1234512345... (repeat forever)
0.000000000...
etc.
In other words, can it happen that the diagonal is a rational number? Actually, with a bit of work we can show that the diagonal must always be necessarily irrational (without changing any digits).
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First write the set ##S=\{0,1,2,3,4,5,6,7,8,9\}##. Now we can consider a digit change function ##f:S \rightarrow S## defined as:
##f(n)=n+2## for ##0\leq n<8##
##f(8)=0##
##f(9)=1##
Now given any complete listing of ##Q##, suppose that the diagonal we obtain (without changing any digits) is equal to ##r_1##. And suppose the number we obtain after changing all the digits of the diagonal [using function ##f## mentioned above] is ##r_2##. The main thing to observe is the following:
##r_1 \in Q## iff ##r_2 \in Q##
However, the after effect of diagonalization that we carried out is that it is impossible for ##r_2 \in Q## to be true. And now, since ##r_2 \notin Q##, we also get ##r_1 \notin Q##. Therefore the original diagonal isn't in the actual list either.