I think it’s much simpler than that. If you remember your maths well enough to remember the proof that the reals are uncountable and the rationals are countable, consider that the list of programs that produce a number is countable. You don’t need to check if the program halts or not, even if we “count” the “numbers” from
the programs that don’t halt _it’s still countable_. But we’re still left with an uncountable set of numbers for which we can’t even write a program.
Easiest way to construct one is to use the diagonal argument that’s used to prove, amongst other things, the halting problem.
Ah, I wrote the same proof above. I agree this is a much simpler way of proving the existence and size of the uncomputible numbers. The busy beavers problem is just an example of an interesting case
By definition, each program only produces the digits of one number. It's okay for more than one program to produce the same number. Am I misunderstanding your point?
I didn’t realize the initial state of the tape was part of the definition of a Turing machine. I was picturing having the same operations applied to different tapes, but that’s not the definition so I’m wrong.
Easiest way to construct one is to use the diagonal argument that’s used to prove, amongst other things, the halting problem.