> Given their non-constructive nature "real" numbers are unsurprisingly totally incompatible with computation.
It is funny you say that when Turing defined Turing machines to compute real numbers (like π for example). In its original definition, a number was computable if its Turing machine did not stop. Which makes sense since π does not have a finite decimal expansion.
Today, we usually define Turing machines to decide problems and a problem is decidable if for every input its Turing machine stops with a ``yes'' or ``no'' answer. I guess this is what makes people think what you said in the quote above. Maybe this definition is more intuitive but this conclusion from it could not be more wrong.
Think about it for a second, if the computable numbers were countable there would be no uncomputable problem (Turing actually used the classic cantor diagonal argument to prove that there were uncomputable numbers)
he was a physics and math major and did not know eigenvectors and eigenvalues? i would like to know how is this possible. can someone explain it to me?
did not read the whole document. just the proof of theorem 1 and it has a minor error. it is not obvious that C(n) != x. that is because it is not true. to see that, consider a map C(n) such that
C(n) = 0.10000000...
that is: c_1(n) = 1 and c_i(n) = 0 for all other i.
If you let \overline{c}_1(n) = 0 and c_i(n) = 9 for all other i,
then we have that c_i(n) != \overline{c}(c)_i(n) for all i but C(n) = x.
This is a common mistake people make. To fix it, one needs to be more careful when defining \overline{c}_i.
you can make a mobius strip with paper. then get a pencil and try to orient it in the mobius strip. that is, make it normal to the paper then move it around. you will see that if you go though the strip and go back to the starting point the pencil will be in the other direction. thus, the orientation is not continuous so the surface is not orientable
It is funny you say that when Turing defined Turing machines to compute real numbers (like π for example). In its original definition, a number was computable if its Turing machine did not stop. Which makes sense since π does not have a finite decimal expansion.
Today, we usually define Turing machines to decide problems and a problem is decidable if for every input its Turing machine stops with a ``yes'' or ``no'' answer. I guess this is what makes people think what you said in the quote above. Maybe this definition is more intuitive but this conclusion from it could not be more wrong.
Think about it for a second, if the computable numbers were countable there would be no uncomputable problem (Turing actually used the classic cantor diagonal argument to prove that there were uncomputable numbers)