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Re:BusyBeaver(6) Is Quite Large by drdeca on Saturday June 28, 2025 @03:37PM (#44407506)
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.
Re:BusyBeaver(6) Is Quite Large by boothby on Saturday June 28, 2025 @06:14PM (#44408600)
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.
Re:BusyBeaver(6) Is Quite Large by drdeca on Saturday June 28, 2025 @07:14PM (#44408950)
Pardon, I meant natural number. I should have specified.
Re:BusyBeaver(6) Is Quite Large by Vecr on Saturday June 28, 2025 @06:39PM (#44408772)
If it had a finite size it would be computable.
Re:BusyBeaver(6) Is Quite Large by Scarblac on Sunday June 29, 2025 @04:40AM (#44411344)
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.

Re:BusyBeaver(6) Is Quite Large by cvoss on Saturday June 28, 2025 @09:34PM (#44409666)
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.

Re:BusyBeaver(6) Is Quite Large by drdeca on Sunday June 29, 2025 @12:02PM (#44414135)
I don’t know it in a constructive sense, sure.

It’s still true though. I’m not wrong.