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The rational numbers make up "zero percent" of the real numbers. It's a little hard to properly explain without assuming a degree in math, since the proper way
by GraphEnthusiast 2y ago
The rational numbers make up "zero percent" of the real numbers. It's a little hard to properly explain without assuming a degree in math, since the proper way to treat this requires measure theoretic probability (formally, the rationals have measure zero in the reals for the "standard" measure).
The short version is that the size of the reals is a "bigger infinity" than the size of the rationals, so they effectively have 'zero weight'.
Reference (very technical):
https://math.stackexchange.com/questions/508217/showing-that-rationals-have-lebesgue-measure-zero#508225 https://math.stackexchange.com/questions/508217/showing-that...
- hn_throwaway_99 2y agoBut then the original implication, "100% of real numbers are normal, so that's pretty strong statistical evidence", still doesn't make any sense, as it's essentially using "100%" to imply "strong statistical evidence" that the rationals don't exist, which obviously doesn't follow.
- staunton 2y ago> still doesn't make any sense Right. I'm pretty sure actually that it was a joke...
- adgjlsfhk1 2y agoit was about half a joke. statistical evidence doesn't really exist for the type of problem since polynomialy computable numbers are countably infinite so you can't define a uniform distribution over then
- mhink 2y agoI got the impression that the comment was a bit tongue-in-cheek. The joke lies in the fact that saying "100% of real numbers" isn't *technically* the same thing as saying "all real numbers", because there's not really a good way to define a meaning for "100%" that lets you exclude rational numbers (or any other countable subset of the reals) and get something other than 100%.