Tell HN: 2^4 == 4^2, extended to rationals
by AnimalMuppet on Hacker News.
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.