| > I'd argue that it's clear that if you know x you can at least tell me what the n^th digit of x is. An arbitrary turing machine fails this, an arbitrary turing machine fails being able to tell me what the first digit is, let alone the n^th one. Sound good (assuming we accept the 999... problem of non-uniqueness). So let's assume the machine makes progress in finite time ie. there is a sequence of natural numbers a(n) such that after time a(n) the machine has emitted at least n digits. There's another possibility you mentioned above about a machine taking an input n and finding a value within 2^(-n) of the number. A machine that keeps emitting numbers on either side of an integer, eg. 2.1, 1.99, 2.00001, etc. fails to tell the first digit. But these numbers are arguably even more physical that programs-emitting-digits. They're roughly the real numbers you'd get from doing actual (classical) physical measurements. > Personally I'd like to ask for equality testing too. You didn't answer the question about irrational numbers but do you think we can have pi? It seems infeasible to mechanically determine that an arbitrary program you are given is a valid pi-digit-calculator though. If you don't think you can ever have irrational numbers, then I think I see where you're coming from now. Having a number could be having a finite representation of the number that can be mechanically tested for equality (in time polynomial in the sizes of representations): a string of digits, a ratio of strings of digits, a string of digits with decimal point and a bar over the repeated ones, etc. IOW a normal form in some finitary data type. > Quickly, means in at most polynomial time. In what? You want the nth digit printed by time O(P(n))? If so, that is strictly stronger than finite time progress so we could dispense with that. But polynomial time doesn't make any sense for a machine that emits a finite string of digits and halts because it doesn't have an input. > or alternatively quote notation is rather cool Hah, I was going to ask about p-adics (do I have -1 if I have the machine that emits an infinite string of 1s?). |
I'm really not satisfied with saying that a machine that gives smaller intervals like that is fully sufficient, on the other hand that's really all we're doing when we specify digits...
> They're roughly the real numbers you'd get from doing actual (classical) physical measurements.
I'd argue that a series of physical measurements don't give you a number so much as a probability distribution, even classically.
> do you think we can have pi
I'm not sure.
The problem I have with denying pi is it doesn't make much more sense than denying 1. Base pi is a perfectly rational numbering system. It's perfectly possible to introduce a special 'pi' symbol (rather like 'i') and define rules of arithmetic so that things work with both rationals and rational multiples of pi. And everything I said just applies to infinitely many other irrationals as well (e.g. 2^(1/n))
> It seems infeasible to mechanically determine that an arbitrary program you are given is a valid pi-digit-calculator though.
Indeed, this is true even if you replace "pi" with "1" though, that's another reason why having a program that calculates the digits isn't sufficient.
> Having a number could be having a finite representation of the number that can be mechanically tested for equality (in time polynomial in the sizes of representations)
Yes.
> In what? You want the nth digit printed by time O(P(n))?
It's a vague definition anyways, polynomial in the sum of everything that is relevant... probably O(P(n + the number represented)).