Comment by drdeca
8 months ago
No individual number is uncomputable. There’s no pair of a number and proof in ZFC that [that number] is the value of BB(748). And, so, there’s no program which ZFC proves to output the value of BB(748). There is a program that outputs BB(748) though, just like for any other number.
Individual numbers can be uncomputable! For example, take your favorite enumeration of Turing machines, (T1, T2...) and write down a real number in binary where the first bit is 0 if T1 halts and 1 otherwise, second bit is 0 if T2 halts... clearly this number is real and between 0 and 1, but it cannot be computed in finite time.
That's a number in R, obviously most of them are uncomputable (there is a countable number of Turing machines).
But for every natural number n there is a trivial Turing machine that just prints n and then halts.
Pardon, I meant natural number. I should have specified.
If it had a finite size it would be computable.
I think your mistake is your claim that BB(748) is a natural number. For you to know that, you would necessarily have to know an upper bound for the number of steps it takes for the BB-748 machine (whichever machine it is) to halt. But you definitely don't know that.
Related: It's incorrect to claim that each machine either halts or doesn't halt. To know that that dichotomy holds would require having a halting problem algorithm.
I don’t know it in a constructive sense, sure.
It’s still true though. I’m not wrong.