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Topic: a^b = b^a (Read 626 times) |
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Chronos
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Re: a^b = b^a
« Reply #1 on: Feb 8th, 2003, 4:24pm » |
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Was that meant to be "all real solutions" rather than "all rational solutions"? I don't think that there are any rational solutions.
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NickH
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Re: a^b = b^a
« Reply #2 on: Feb 8th, 2003, 4:30pm » |
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I definitely mean rational, though real would be a challenging extension. (I'm not sure how the solutions would be listed.) The simplest solution is a = 2, b = 4, but there are others...
<edit>(If not clear, I mean a and b to be rational, not necessarily ab.)</edit>
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| « Last Edit: Feb 8th, 2003, 4:32pm by NickH » |
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NickH
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Re: a^b = b^a
« Reply #3 on: Feb 20th, 2003, 1:23pm » |
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I almost have a solution to this one; I'm just short on one part of the proof. See hidden text below...
Let b = ra, where r > 1, and necessarily rational.
Then ara = (ra)a = raaa.
So aa(r-1) = ra, and ar-1 = r.
Hence a = r1/(r-1).
Now, I'm pretty sure, and can almost prove, that r1/(r-1) is irrational unless 1/(r-1) is an integer. So I have a countable infinity of solutions, and an incomplete proof.
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Icarus
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Re: a^b = b^a
« Reply #4 on: Feb 20th, 2003, 4:09pm » |
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To complete Nick's proof:
If x, y, and k are positive integers with k>1, then (x+y)k - xk = kxyk-1+ ... > k. Thus the difference between any two distinct kth powers > k.
Now for any positive integers k,l,m,n, with (k,l)=1, (m,n)=1, (m/n)k/l is rational iff both m and n are perfect lth powers.
Let r = m/n. 1/(r-1) = n/(m-n). So for r1/(r-1) to be rational, both m and n must be perfect m-n powers. But then, their difference would have to be greater than m-n unless m-n = 1 (m > n since r > 1).
So the solution is r = (n+1)/n, a=rn, b=rn+1 for integers n > 0.
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