04 Jun '04 08:462 edits

I just thought of a set of operations corresponding to the integers which go as follows:

(i) a<0>b = a+b

(ii) a<n+1>b = exp(log(a)<n>log(b))

If you 'reverse' (ii) you get

(iii) a<n-1>b = log(e^a<n>e^b), allowing one to define <-1> etc.

<0> and <1> correspond in some sense to + and x, and <n> looks like it's associative and commutative. Unfortunately, I can't think of a nice set on which I can make <n> well-defined and closed for n other than 0. Does anyone have any ideas?

Edit: I think I can see how to make <2> well-defined over R\{0,1}, so maybe this idea's going somewhere after all. Can anyone tell me whether <n+1> distributes over <n> in general? At the moment I'm getting an identity element for <n> if n is at least 0, but inverses are hard to come by. Can anyone see how I could get inverses (and an identity for n<0), and make <n> a group operation?

(i) a<0>b = a+b

(ii) a<n+1>b = exp(log(a)<n>log(b))

If you 'reverse' (ii) you get

(iii) a<n-1>b = log(e^a<n>e^b), allowing one to define <-1> etc.

<0> and <1> correspond in some sense to + and x, and <n> looks like it's associative and commutative. Unfortunately, I can't think of a nice set on which I can make <n> well-defined and closed for n other than 0. Does anyone have any ideas?

Edit: I think I can see how to make <2> well-defined over R\{0,1}, so maybe this idea's going somewhere after all. Can anyone tell me whether <n+1> distributes over <n> in general? At the moment I'm getting an identity element for <n> if n is at least 0, but inverses are hard to come by. Can anyone see how I could get inverses (and an identity for n<0), and make <n> a group operation?