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 Commutative Exponentiation (Posted on 2017-07-07)
In general exponentiation is not commutative. However most people are aware of the pair 2 and 4 as a pair of unequal numbers for which exponentiation is commutative.

Show that there are an infinite number of pairs of unequal numbers for which exponentiation is commutative.

Go further and show that there are an infinite number of pairs of unequal rational numbers for which exponentiation is commutative.

 See The Solution Submitted by Brian Smith Rating: 4.0000 (1 votes)

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 rational solutions | Comment 3 of 7 |
the rational pairs (9/4, 27/8) and (64/27, 256/81) are examples of solutions that I found with a quick software exploration. That suggests a sequence:

2^1/1^1 and 2^2/1^2 (the one we all know)
3^2/2^2 and 3^3/2^2
4^3/3^3 and 4^4/3^4

So I predict that all pairs of the form
n^n-1/n-1^n-1 and n^n/n-1^n will also be solutions. Testing for n=5 gives the pair: (625/256, 3125/1024) and that too is a solution.

But I're really love to see a closed-form proof...

 Posted by Paul on 2017-07-07 14:53:40

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