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exponential distributivity
#1
quickfur Wrote:I forgot to note, what I had in mind here is and .
Oh right, misinterpreted that.
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. Note however that on the just discussed hierarchy of operations {n}, there is always distributivity from {n+1} over {n}, but exponentiation does not belong to this hierarchy.

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). Then:
f(a^(b^n))=f((...(a^b)^b...)^b)=f(a)^(f(b)^n) but also
f(a^(b^n))=f(a)^f(b^n)=f(a)^(f(b)^f(n)) hence

f(a)^(f(b)^f(n))=f(a)^(f(b)^n) then for f(a)!=1, f(a)>0
f(b)^f(n)=f(b)^n and again for f(b)!=1, f(b)>0
f(n)=n

f(a)^m=f(a^m)=f((a^(m/n))^n)=(f(a)^f(m/n))^n=f(a)^(f(m/n)n), then for f(a)!=1, f(a)>0:
m=f(m/n)n
f(m/n)=m/n

If we now suppose that f is continuos we extend this law from the fractional to the positive real numbers:
f(x)=x for all positive real x.

This was however under the assumption that there is an with . So the other possibility is:
f(x)=1 for all positive real x.

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.
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Messages In This Thread
exponential distributivity - by bo198214 - 02/22/2008, 12:36 PM
RE: exponential distributivity - by quickfur - 02/22/2008, 06:51 PM
RE: exponential distributivity - by bo198214 - 02/22/2008, 07:17 PM
RE: exponential distributivity - by quickfur - 02/23/2008, 12:29 AM
RE: exponential distributivity - by JmsNxn - 09/22/2011, 03:27 PM

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