09/08/2011, 01:37 AM

Well, the proof is really simple, but it works;

Lets assume we have an operator where , and is the super operator of , furthermore, and

Start off by making our only assumption that and are commutative and associative.

start off with the basic formula:

Now, since is commutative and associative, we can rearrange them in the following manner if :

so that:

therefore, for any a,b,c:

given this law, if we set a = S(q) or the identity for operator

we instantly see that

the only assumption we made was that and be commutative and associative. I think maybe this proof is inadequate at proving it cannot be commutative, but I think it'd be on shaky ground to say they are commutative. But forsure, not associative.

Edit: The proof to make it non-commutative is as follows.

Since the ackermann function is defined as:

where the only law it must obey is:

If we want to be analytic over (which we do), we cannot have being commutative over any strip. because if perhaps we say: all operators including and below multiplication are commutative. This would mean:

but if two functions are analytic and they equal each other over a strip then they must be the same function

therefore:

, for all , but this is clearly untrue because exponentiation is not commutative. Therefore is only commutative at addition (0) and multiplication (1).

I guess our rational operators are going to have to behave like exponentiation, I'm really curious about an analytic and integral calculus attack at this problem. Maybe dynamics ain't the right field. I think logarithmic semi operators are as close as it'll get.

Maybe there's a more natural equation that may have some aesthetic properties in terms of relations to trigonometric functions, or other established functions with maybe some fancy constants involved.

Lets assume we have an operator where , and is the super operator of , furthermore, and

Start off by making our only assumption that and are commutative and associative.

start off with the basic formula:

Now, since is commutative and associative, we can rearrange them in the following manner if :

so that:

therefore, for any a,b,c:

given this law, if we set a = S(q) or the identity for operator

we instantly see that

the only assumption we made was that and be commutative and associative. I think maybe this proof is inadequate at proving it cannot be commutative, but I think it'd be on shaky ground to say they are commutative. But forsure, not associative.

Edit: The proof to make it non-commutative is as follows.

Since the ackermann function is defined as:

where the only law it must obey is:

If we want to be analytic over (which we do), we cannot have being commutative over any strip. because if perhaps we say: all operators including and below multiplication are commutative. This would mean:

but if two functions are analytic and they equal each other over a strip then they must be the same function

therefore:

, for all , but this is clearly untrue because exponentiation is not commutative. Therefore is only commutative at addition (0) and multiplication (1).

I guess our rational operators are going to have to behave like exponentiation, I'm really curious about an analytic and integral calculus attack at this problem. Maybe dynamics ain't the right field. I think logarithmic semi operators are as close as it'll get.

Maybe there's a more natural equation that may have some aesthetic properties in terms of relations to trigonometric functions, or other established functions with maybe some fancy constants involved.