(03/11/2015, 07:56 PM)marraco Wrote: For a simple case, let’s take c = Ttr(a,-1), then it should be

By taking logarithm: c . ln(a) = 0, so, if c is a complex number, it can only be zero, for any real a≠1.

Actually,

Still n=0 is one solution, so it requires a new set of numbers to solve for c, but for n≠0:

and with the same reasoning

this is just logarithm of 1 in base a iterated b times

Why the iterated logarithm of 1? because 1 was the implicit base.

More generally:

Is easy to verify that for any pair of integers n,m

For example:

(when taking power with m≠0, m represent m roots or more, and not all the roots are valid solutions to a^^-1, for example y=-1^(2n/(1+2m)), m produces 1+2m roots, and not all those roots give y=1)

This shows that all the rational numbers are solution for 1^^(-1), but we know that all the real numbers r have the property 1^r=1, so:

-Either there are more solutions than given by the logarithm function, or

-Any real number can be expressed as a rational number. Yet, we know that transcendental numbers are not equal to any rational number, but maybe this is saying that since numbers like PI can be approximated as much as desired, then pi is also equal to a rational number, as 1,999999... is equal to 2.

By the way, it is interesting that ln(x)=Ttr(e,-1,x) ,and e(x)=Ttr(e,1,x)