02/24/2008, 11:02 AM

KAR Wrote:Further, from the given legitimacies it is possible to install, that Zeration it is commutative. For this purpose we shall consider two functions at various values :

and .

For operations "division" and "subtraction":

and .

From the reduced example, follows, that inverses operations of type "log" at identical values of argument have constant value. Inverse operation of type "radical" has variable values at a modification of identical arguments and is equal to a constant in case of coincidence with operation of type "log", that is commutabilities.

From a limit follows, that inverse operation of type "radical" for Zeration has unique solution for identical arguments, that is Zeration is commutative.

You mean if we have an operation * and we have two inverses / and \, i.e. a=(a*b)/b=(a/b)*b and b=a\(a*b)=a*(a\b), which are equal, i.e. x/y=y\x, then * must be commutative?

Hm, then (b*a)/b = b\(b*a) = a and hence b*a=a*b, so thats correct.

(Though you didnt even show radical equals log, but rather that the radical type inversion function is constant. That does not necessarily imply that also the radical type equals the log type.)

So what you gave is not a derivation, I admit I also love to speculate and think in analogies. However we are here in the mathematical field. Derivations, strict consequences and proofs are the things that count. So after we got some ideas of how things could possibly work, we are not bound to stay at the state of speculation (or to just postulate things without a possibility to verify, as the philosophers do) but we can go to the core and make things manifest: derive things, proving propositions, etc. And that would be what I am really interested in.

Quote:bo198214 Wrote:No, thats not a strict consequence. The following is a strict consequence:You write n> 1 and safely expand "n" up to complex? You can seriously make matching of any complex number with 1?

We start with the axiom

ao(ao(ao......))) = a+n for integer n>1 (and a being real or complex)

then we conclude

ao(a+n)=a+n+1

Why should I extend this condition to complex n?

I just wanted to show that from the by you given conditions, i.e. aoa=a+2 and ao(aoa)=a+3, etc, from which follows ao(a+n)=a+n+1 for n>1, even if we enhance these conditions, by letting n be real, the by you given zeration is not a consequence but rather an example that satisfies these conditions.

However that the condition does not contain a complex but only real n, does not mean that zeration itself can not be defined on the complex numbers (though you did not define it on the complex numbers).

This would be *my* definition for *complex* :

Let us verify the conditions:

because ,

let then because

and for natural the other conditions follow , etc.