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Naw most of these wouldn't work when actually written out as Feynman did. Example: you can easily give an upper bound to the "chance that N is a prime" that goe

by tgb 10y ago

Naw most of these wouldn't work when actually written out as Feynman did. Example: you can easily give an upper bound to the "chance that N is a prime" that goes to zero as N increases. But you would also need to show that it's sum from 0 to infinity over all N also goes to zero. In fact, there's the classic result that this probability is about 1/log(N) [1], which diverges towards infinity. Hence you would probabilistically expect infinite primes and would be correct. [1] https://en.wikipedia.org/wiki/Prime_number_theorem https://en.wikipedia.org/wiki/Prime_number_theorem

Someone 10y ago
That's because you start with a set (the integers) that contains disproportionally many primes. Just pick a larger set of numbers to compare things with. If you look at the reals, the probability of finding a rational already is zero, and there aren't primes that aren't rational numbers. Mathematicians of course, would not do that. They do use the technique of estimating the solution size to get a feeling for the difficulty of a problem, but always try to pick a reference set that is such that the exercise teaches them anything, and starting with all reals doesn't (how do they know that? Intuition, or they may do it anyway, but realize half-way through that it is silly, or even publish it, and, eventually, get corrected) Feynman's solution has more good math, but still makes the fatal mistake of stating that "measure zero implies does not occur" (although he probably knew, since he states it was good enough for him) One can 'prove' the non-existence of any countable infinite set of numbers this way.
- tgb 10y ago
  Your arguments are just fundamentally different from Feynman's. We're trying to estimate whether something exists. If you are using a counting measure on the integers, say, then whether or not something exists is whether or not the measure of that set is zero. If you're dealing with, say, a Lebesgue measure on the reals then the measure being zero tells you nothing about whether the set is empty. Put the counting measure on the reals and then you can work, but then the answers you get won't be zero.

by tgb 10y ago

Someone 10y ago

That's because you start with a set (the integers) that contains disproportionally many primes. Just pick a larger set of numbers to compare things with. If you look at the reals, the probability of finding a rational already is zero, and there aren't primes that aren't rational numbers. Mathematicians of course, would not do that. They do use the technique of estimating the solution size to get a feeling for the difficulty of a problem, but always try to pick a reference set that is such that the exercise teaches them anything, and starting with all reals doesn't (how do they know that? Intuition, or they may do it anyway, but realize half-way through that it is silly, or even publish it, and, eventually, get corrected) Feynman's solution has more good math, but still makes the fatal mistake of stating that "measure zero implies does not occur" (although he probably knew, since he states it was good enough for him) One can 'prove' the non-existence of any countable infinite set of numbers this way.

tgb 10y ago
Your arguments are just fundamentally different from Feynman's. We're trying to estimate whether something exists. If you are using a counting measure on the integers, say, then whether or not something exists is whether or not the measure of that set is zero. If you're dealing with, say, a Lebesgue measure on the reals then the measure being zero tells you nothing about whether the set is empty. Put the counting measure on the reals and then you can work, but then the answers you get won't be zero.