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This proof (or "plausibility argument") bugs me so much. Just because something thins out and becomes rare doesn't mean it doesn't exist. As n gets bigger, the

This proof (or "plausibility argument") bugs me so much. Just because something thins out and becomes rare doesn't mean it doesn't exist. As n gets bigger, the probability of n being a perfect square gets smaller and smaller. In the limit, the probability is zero. Does this mean square numbers don't exist?

foxes 8y ago
This isn't a proof.
- jordigh 8y ago
  I know, but it's being exhibited as an example of how back-of-the-envelope type of approximations by physicists can be just as good as rigid mathematical thinking. And I don't find this to be a convincing example of how loose physicist arguments can work. Schwartz distributions, infintesimals; okay, fine, those turned out to be a weird trick that can be formalised. But sometimes their tricks are just plain wrong and this is one example of a trick that just is wrong and can't be formalised.
  - Retric 8y ago
    That’s not what’s happening. Consider, many useful primality tests are statistical in nature. It’s pure math, and exact answer is possible but it’s still useful to get a quick check to see if something is a waste of time. Really, if a full solution takes 20 years you don’t want to actually spend 20 years without having a very good idea it’s going to work.
    - foldr 8y ago
      I think jordigh is saying that the method is not statistically sound, i.e. that it will not (necessarily) give accurate probability estimates. They're not criticizing the method simply for being statistical.
      - gowld 8y ago
        jordigh hasn't given support for that claim.
        jordigh 8y ago
        Using zero probability on infinite sets to ascertain the inexistence of an object doesn't work. Lots of things have measure zero in an infinite set. For example, out of all the integers, the probability of picking 7 is zero. Any finite set will have density zero in the integers too. So I find his small probability, however tiny, that out of all possible integer tuples none of them are a counterexample to be utterly unconvincing and ultimately misguided. To be clear: my complaint is that there is no way to turn this kind of argument into an actual proof. We could salvage other physicist arguments, but not this one. Probability zero on an infinite set cannot mean inexistence. And the rest of what he's doing, trying to determine that counterexamples must be rare, is "well duh, we knew that, because we've been looking for them."
fanzhang 8y ago
By Feynman's argument, you can prove that square numbers almost certainly keep on existing. Roughly, it goes as such: 1) the probability of N being a perfect square is proportional to 1/sqrt(N). 2) For any N_0 arbitrarily high, if you integrate from N_0 to infinity the expression (1/sqrt(N) dN), you get infinity. 3) The expression in 2) is the "Feynman equivalent" of the expected number of square numbers above N_0. So Feynman's nonproof actually turns out to be true, despite it not being a proof in this case as well.
- jordigh 8y ago
  Okay, let's pick something rarer. Rational numbers. If you integrate the characteristic function of the rational numbers over any interval, you get zero because rational numbers are very rare. So they don't exist either? To be less glib, I don't see Feynmann's argument to be bringing anything new. We already knew that counterexamples, if they existed, would be very rare because we tried looking for them with computers and we couldn't find them. But stuff being rare still doesn't prove anything. Many of us were fooled by Skewe's number: https://en.wikipedia.org/wiki/Skewes%27s_number https://en.wikipedia.org/wiki/Skewes%27s_number There's no way to conclude that this exists via brute calculation. It's just inconceivably large and would have eluded any of Feynmann's methods.
  - jstarks 8y ago
    The integral of the characteristic function is zero, but the probability of finding a rational in any interval is one. In Feynman’s argument, it seems he was relying on this latter probability. So I don’t think you have produced a compelling counterexample yet (though I expect you are right that one exists).
  - sweezyjeezy 8y ago
    Your rational numbers argument does indeed fail, but for a different reason. It's not valid to go from a sum over the rationals to an integral in the reals in that way, however it is perfectly valid to go from a sum over the integers to an integral over the reals (in certain situations), if you only want an estimate (see eg. the Euler Maclaurin formula https://en.wikipedia.org/wiki/Euler–Maclaurin_formula https://en.wikipedia.org/wiki/Euler–Maclaurin_formula ). This was fine in Feynman's argument.
sweezyjeezy 8y ago
No one is claiming that this kind of argument constitutes a proof, but this kind of back of the envelope calculation is very powerful. For example in the theory of prime numbers there is a heuristic that says that primes roughly behave like randomly selected numbers where Prob(N is prime) = 1 / log(N) [this is a simplification, but that's the crux of it]. With this heuristic you can accurately predict whether a large class of statements about primes are true or false, and can get extremely precise estimates about things like "how many twin primes are there less than N", or "how many solutions in primes are there to p1 + p2 = 2*N, for some huge N", which is one way of phrasing two famous prime number conjectures. It doesn't lead you directly proof - but often just knowing what the answer 'should be' can be a real guiding light.
SamReidHughes 8y ago
This argument doesn't calculate the probability of an individual number, it sums the probability over every number. This kind of thing is used a lot in number theory to figure out the plausibility of some theorem. A lot of open number theory problems are of the form "Prove [unlikely event] never happens." Edit: Removed braindead argument.

foxes 8y ago

This isn't a proof.

jordigh 8y ago
I know, but it's being exhibited as an example of how back-of-the-envelope type of approximations by physicists can be just as good as rigid mathematical thinking. And I don't find this to be a convincing example of how loose physicist arguments can work. Schwartz distributions, infintesimals; okay, fine, those turned out to be a weird trick that can be formalised. But sometimes their tricks are just plain wrong and this is one example of a trick that just is wrong and can't be formalised.
- Retric 8y ago
  That’s not what’s happening. Consider, many useful primality tests are statistical in nature. It’s pure math, and exact answer is possible but it’s still useful to get a quick check to see if something is a waste of time. Really, if a full solution takes 20 years you don’t want to actually spend 20 years without having a very good idea it’s going to work.
  - foldr 8y ago
    I think jordigh is saying that the method is not statistically sound, i.e. that it will not (necessarily) give accurate probability estimates. They're not criticizing the method simply for being statistical.
    - gowld 8y ago
      jordigh hasn't given support for that claim.
      - jordigh 8y ago
        Using zero probability on infinite sets to ascertain the inexistence of an object doesn't work. Lots of things have measure zero in an infinite set. For example, out of all the integers, the probability of picking 7 is zero. Any finite set will have density zero in the integers too. So I find his small probability, however tiny, that out of all possible integer tuples none of them are a counterexample to be utterly unconvincing and ultimately misguided. To be clear: my complaint is that there is no way to turn this kind of argument into an actual proof. We could salvage other physicist arguments, but not this one. Probability zero on an infinite set cannot mean inexistence. And the rest of what he's doing, trying to determine that counterexamples must be rare, is "well duh, we knew that, because we've been looking for them."

fanzhang 8y ago

By Feynman's argument, you can prove that square numbers almost certainly keep on existing. Roughly, it goes as such: 1) the probability of N being a perfect square is proportional to 1/sqrt(N). 2) For any N_0 arbitrarily high, if you integrate from N_0 to infinity the expression (1/sqrt(N) dN), you get infinity. 3) The expression in 2) is the "Feynman equivalent" of the expected number of square numbers above N_0. So Feynman's nonproof actually turns out to be true, despite it not being a proof in this case as well.

jordigh 8y ago
Okay, let's pick something rarer. Rational numbers. If you integrate the characteristic function of the rational numbers over any interval, you get zero because rational numbers are very rare. So they don't exist either? To be less glib, I don't see Feynmann's argument to be bringing anything new. We already knew that counterexamples, if they existed, would be very rare because we tried looking for them with computers and we couldn't find them. But stuff being rare still doesn't prove anything. Many of us were fooled by Skewe's number: https://en.wikipedia.org/wiki/Skewes%27s_number https://en.wikipedia.org/wiki/Skewes%27s_number There's no way to conclude that this exists via brute calculation. It's just inconceivably large and would have eluded any of Feynmann's methods.
- jstarks 8y ago
  The integral of the characteristic function is zero, but the probability of finding a rational in any interval is one. In Feynman’s argument, it seems he was relying on this latter probability. So I don’t think you have produced a compelling counterexample yet (though I expect you are right that one exists).
- sweezyjeezy 8y ago
  Your rational numbers argument does indeed fail, but for a different reason. It's not valid to go from a sum over the rationals to an integral in the reals in that way, however it is perfectly valid to go from a sum over the integers to an integral over the reals (in certain situations), if you only want an estimate (see eg. the Euler Maclaurin formula https://en.wikipedia.org/wiki/Euler–Maclaurin_formula https://en.wikipedia.org/wiki/Euler–Maclaurin_formula ). This was fine in Feynman's argument.

sweezyjeezy 8y ago

No one is claiming that this kind of argument constitutes a proof, but this kind of back of the envelope calculation is very powerful. For example in the theory of prime numbers there is a heuristic that says that primes roughly behave like randomly selected numbers where Prob(N is prime) = 1 / log(N) [this is a simplification, but that's the crux of it]. With this heuristic you can accurately predict whether a large class of statements about primes are true or false, and can get extremely precise estimates about things like "how many twin primes are there less than N", or "how many solutions in primes are there to p1 + p2 = 2*N, for some huge N", which is one way of phrasing two famous prime number conjectures. It doesn't lead you directly proof - but often just knowing what the answer 'should be' can be a real guiding light.

SamReidHughes 8y ago

This argument doesn't calculate the probability of an individual number, it sums the probability over every number. This kind of thing is used a lot in number theory to figure out the plausibility of some theorem. A lot of open number theory problems are of the form "Prove [unlikely event] never happens." Edit: Removed braindead argument.