Tell HN: 2^4 == 4^2, extended to rationals
6 comments
Integer version: https://news.ycombinator.com/item?id=35635491 (520 points | 2 days ago | 156 comments) It (implicitly) shows why the e number is important and why it appears as the limit of your example.
Did you prove that this are all the case? Or this are only the cases you found and the general case is still open?
Did you prove that this are all the case? Or this are only the cases you found and the general case is still open?
No, I did not prove that this was all the cases - not even that this was all the rational cases. I tried, years ago (I literally made this post by trying to recall what I had found over 40 years ago), but I did not find a proof that this was all the possibilities.
After some time to think, yes, I'm pretty sure that it's all solutions. The shape of the function in the article means that there is only one b for any a. The formula I gave gives a solution, so it must be the solution. If you can find an n to generate an a, the formula must give you the correct b.
(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.
(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.
There are infinete number of rationals that satisfy the condition.
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.
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.
For general solution you can check https://math.stackexchange.com/questions/3072828/why-didnt-t..., which is related to Lambert W function.
I feel rationals is probably the easiest case to prove, still clever though! Interesting (though not surprising) how there's a countably infinite number of rationals that satisfy this.
Probably this is the article you saw https://link.medium.com/bf81IpGM8yb
I think a more interesting problem to consider after solving n^m = m^n is solving n^m = m^n + 1 in the natural numbers.
If we exclude 0, the only solutions are 2^1 = 1^2 + 1 and 3^2 = 2^3 + 1
That is correct. Why are there no others?
Because Catalan’s conjecture.
> Catalan's conjecture was proven by Preda Mihăilescu in April 2002.
Cool, so the conjecture's a theorem.
https://xn--uni-gttingen-8ib.academia.edu/PredaMihailescu
Cool, so the conjecture's a theorem.
https://xn--uni-gttingen-8ib.academia.edu/PredaMihailescu
Yes, but there is a much simpler proof.
Something related to prime factorization maybe?
Do tell.
Do it for imaginary components :tada:
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.