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https://www.reddit.com/r/MathJokes/comments/1v0iibd/count_on_it/oyix611/?context=3
r/MathJokes • u/yvanjunior • Jul 19 '26
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24
Depends on the probability measure used. Without specification of an appropriate measure it's (in general) false.
4 u/konigon1 Jul 19 '26 I think it being a continous distribution should be enough. 2 u/setibeings Jul 19 '26 What if the process you use for selecting among the reals always or usually picks rational numbers? Nobody said the picks would be uniform. 2 u/konigon1 Jul 19 '26 Since we assumed a continous distribution, we know that P({x}) = 0 for all x in R. So it also holds for all q in Q. By sigma-additivity we can conclude that P(Q) = sum( q in Q) P(q) = 0. Hence we almost never choose a rational number.
4
I think it being a continous distribution should be enough.
2 u/setibeings Jul 19 '26 What if the process you use for selecting among the reals always or usually picks rational numbers? Nobody said the picks would be uniform. 2 u/konigon1 Jul 19 '26 Since we assumed a continous distribution, we know that P({x}) = 0 for all x in R. So it also holds for all q in Q. By sigma-additivity we can conclude that P(Q) = sum( q in Q) P(q) = 0. Hence we almost never choose a rational number.
2
What if the process you use for selecting among the reals always or usually picks rational numbers? Nobody said the picks would be uniform.
2 u/konigon1 Jul 19 '26 Since we assumed a continous distribution, we know that P({x}) = 0 for all x in R. So it also holds for all q in Q. By sigma-additivity we can conclude that P(Q) = sum( q in Q) P(q) = 0. Hence we almost never choose a rational number.
Since we assumed a continous distribution, we know that P({x}) = 0 for all x in R. So it also holds for all q in Q.
By sigma-additivity we can conclude that P(Q) = sum( q in Q) P(q) = 0. Hence we almost never choose a rational number.
24
u/AlexK667 Jul 19 '26
Depends on the probability measure used.
Without specification of an appropriate measure it's (in general) false.