A computable real number is a real number for which a Turing Machine exists that can compute it to any arbitrary precision.
So yes sqrt(2) is computable.
Every BB(n) is computable since every every natutal number can be computed. It's the BB function itself that is not computable in general, not the specific output of that function for a given input.
That doesn’t sound right to me. What about the machines that don’t halt? You can’t compute whether or not to skip them directly.
> A busy beaver hunter who goes by Racheline has shown that the question of whether Antihydra halts is closely related to a famous unsolved problem in mathematics called the Collatz conjecture. Since then, the team has discovered many other six-rule machines with similar characteristics. Slaying the Antihydra and its brethren will require conceptual breakthroughs in pure mathematics.
The claim is that every bb(n) is computable but I don’t think you can compute bb(6) without knowing which machines won’t halt. That doesn’t seem like a finite calculation?
But given the answer, I suppose you could write a program that just returns it. This seems to hinge on the definition of “computable.” It’s an integer, so that fits the definition of a computable number.
Yes. A valid question for a specific n would be whether you can prove the value of BB(n). If you don't care about provability, you can indeed just produce a number that happens to be the right one.
So as you noticed, it only makes sense to talk about whether a function is computable, we can't meaningfully talk of computable numbers.
The main thing to make it clear is that BB(n) for a specific n isn't a function - it's a number. Just like Mult(10,4) isn't a function, it's a number (40).
So a specific BB(n) is just a number and is computable.
So yes sqrt(2) is computable.
Every BB(n) is computable since every every natutal number can be computed. It's the BB function itself that is not computable in general, not the specific output of that function for a given input.