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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

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.

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.