Tell HN: 2^4 == 4^2, extended to rationals
Last week there was a discussion about 2^4 == 4^2 (my search-fu is letting me down - I can't find it again, or I'd link it). It proved that that is the only pair of integers with that relationship.
But what about rationals?
First, remember that (a^b)^c) == a^(bc).
Let a = (3/2)^2 and b = (3/2)^3. Then a^b == b^a. (a and b are rational, but a^b is not.)
More generally, for any natural number n, let a = ((n+1)/n)^n and b = ((n+1)/n)^(n+1). Then a^b == b^a.
Note that the first element of this series has a = 2 and b = 4. Also note that as n increases, the series converges to both a and b equal to e.
But when I said "for any natural number n", that was actually an unnecessary restriction. The math works for any n - it doesn't have to be an integer. We could, for instance, use n = 1/2. Then a = ((3/2)/(1/2))^(1/2) and b = ((3/2)/(1/2))^(3/2). a and b are no longer rational, but a^b still equals b^a.
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[ 4.9 ms ] story [ 56.6 ms ] threadDid you prove that this are all the case? Or this are only the cases you found and the general case is still open?
The integer uniquenes proof relied on the shape of the function between 1 and e non inclusive, and 2 is the only integer in the open interval.
(As n goes from 0 to infinity, a goes from 1 to e, so the formula covers the entire range of possibilities.)
In my formula, however, n may be irrational, even for some rational a. I haven't proven that.
Cool, so the conjecture's a theorem.
https://xn--uni-gttingen-8ib.academia.edu/PredaMihailescu