02/22/2008, 06:51 PM
bo198214 Wrote:[...]Hmm, interesting. I was going to ask what happens if we consider f(x)=c#x instead, but it doesn't change the conclusion because we assume the same operator #. So we established that a higher operation cannot distribute over exponentiation without becoming trivial.
Another idea would then be to investigate the distribution law for exponentiation, i.e. whether there is an operation # such that
(a^b)#c=(a#c)^(b#c).
As a side node I give a proof that there are only trivial operations # satisfying this property. [...]
Now to the proof, if there would be such an operation then we can consider the function f(x)=x#c (fixed c) which satisfies f(a^b)=f(a)^f(b).
[...]
Summarizing
Proposition: Any operation # defined on the positive real numbers such that (a^b)#c=(a#c)^(b#c) (for each a,b,c) and such that f(x)=x#c is continuous (for each c), is either given by x#c=x or by x#c=1 for all x,c.
I wonder what is the root cause of this... maybe it has something to do with the non-associativity of exponentiation?
