I was being cheeky! It could’ve been that the set of non-Turing-computible problems had measure zero but still infinite cardinality. However there’s the much stronger result that the set of Turing-computible problems actually has measure zero (for which I used 0% and the integer:reals thing as shorthands because I didn’t want to talk measure theory on Lemmy). This is so weird, I never got downvoted for this stuff on Reddit.
The subset of integers in the set of reals is non-zero. Sure, I guess you could represent it as arbitrarily small small as a ratio, but it has zero as an asymptote, not as an equivalent value.
Infinite seems like it’s low-balling it, then. 0% of problems can be solved by Turing machines (same way 0% of real numbers are integers)
Infinite by definition cannot be “low-balling”.
This is incorrect. Any computable problem can be solved by a Turing machine. You can look at the Church-Turing thesis if you want to learn more.
I was being cheeky! It could’ve been that the set of non-Turing-computible problems had measure zero but still infinite cardinality. However there’s the much stronger result that the set of Turing-computible problems actually has measure zero (for which I used 0% and the integer:reals thing as shorthands because I didn’t want to talk measure theory on Lemmy). This is so weird, I never got downvoted for this stuff on Reddit.
Oh, sorry about that! Your cheekiness went right over my head. 😋
The subset of integers in the set of reals is non-zero. Sure, I guess you could represent it as arbitrarily small small as a ratio, but it has zero as an asymptote, not as an equivalent value.
The cardinality is obviously non-zero but it has measure zero. Probability is about measures.