Abelian Exponentiation

Unlike addition and multiplication, normal exponentiation is not commutative:

a+b=b+a
a*b=b*a
but
a^b≠b^a

However, it is a relatively simple matter to define an operation similar to normal exponentiation that is commutative:

normal exponentiation:

a^n=a*a*a*...a
        1+1+1+...1=n

Abelian exponentiation (where # is the symbol for it):

a#n=a*a*a*...a
        2*2*2*...2=n
or
a#n=2^{(log₂a)*(log₂n)}

The positive real numbers (R⁺) form an Abelian group under Abelian exponentiation (#) and, (R⁺,*,#) forms a field which is isomorphic to the real field (R,+,*).

Proof that Abelian exponentiation forms and Abelian group:

Associative: for each a, b, c in R⁺,

(a#b)#c=a#(b#c)

Proof:
(a#b)#c=[2^{(log₂a)*(log₂b)}]#c=
2^[[log₂[2^{(log₂a)*(log₂b)}]]*(log₂c)]=
2^{(log₂a)*(log₂b)*(log₂c)}
and
a#(b#c)=a#[2^{(log₂b)*(log₂c)}]=
2^[(log₂a)*[log₂[2^{log₂b)*(log₂c)}]]]=
2^{(log₂a)*(log₂b)*(log₂c)}

Identity: there exists a unique elememt, 2, in R⁺ such that for each a in R⁺,

a#2=a

Proof:
a#2 = 2^{(log₂a)*(log₂2)}= 2^(log₂a)=a

Inverse element: corresponding to each elelment a in R⁺ there is an inverse element 2^{1/(log₂a)} (a≠1) in R⁺ such that

[2^{1/(log₂a)}]#a=2

Proof:
[2^{1/(log₂a)}]#a=
2^[[log₂[2^{1/(log₂a)}]]*(log₂a)]=
2^[{1/(log₂a)}*(log₂a)]=2^1=2

Commutative: for each a, b in R⁺,

a#b=b#a

Proof:
a#b=2^{(log₂a)*(log₂b)}
and
b#a=2^{(log₂b)*(log₂a)}=2^{(log₂a)*(log₂b)}

With the following distribution postulate for Abelian exponentiation, the operations of multiplication and Abelian exponentiation over R⁺ form a field:

Distributive (with respect to multiplication): for each a, b, c in R⁺,

a#(b*c)=(a#b)*(a#c)

Proof:
a#(b*c)=2^[(log₂a)*{log₂(b*c)}]=
2^[(log₂a)*{(log₂b)+(log₂c)}]
and
(a#b)*(a#c)=
[2^{(log₂a)*(log₂b)}]*[2^{(log₂a)*(log₂c)}]=
2^[{(log₂a)*(log₂b)}+{(log₂a)*(log₂c)}]=
2^[(log₂a)*{(log₂b)+(log₂c)}]

This field (R⁺,*,#) is isomorphic with the real field (R,+,*). The isomorphism is indicated by the existence of a 1 to 1 map (R --> R⁺ : f(r) = 2^r) such that for all r₁, r₂ in R:

a) 2^(r₁+r₂)=(2^r₁)*(2^r₂);
b) 2^(r₁*r₂)=(2^r₁)#(2^r₂).

Comments

First, it should be noted that the choice of 2 as the identity element of Abelian exponentiation was somewhat arbitrary. Any positive real number (not equal to the multiplicative identity -- 1, in normal multiplication) could have been used. In the base 10 numbering system, another logical choice for the identity element might be 10. Second, while there may be some degree of uncomfortability with the fact that there can be more than 1 possibility for the identity element, it must be pointed out that this is also true of both addition and multiplication, too (see Mathematical Relativity -- http://mysite.verizon.net/reriker/re03001.html).