A more consistent definition of tetration of tetration for rational exponents

[UVIR]

It is in the same sense arbitrary as say that I define
\( {}^{1/n}x=1 \). No rule justifies that \( {}^{1/n}x \) should be the inverse of \( {}^nx \).

Let me try to explain the "obvious" rule behind such a definition. Call the inverse operator of multiplication @t. Then the inverse of multiplication must satisfy:

\( (x*n)@t=x \)

consequently it is easily seen that \( @t=*(1/n) \), which is the only operator which fits the bill. Similarly, call the inverse operator of exponentiation @@k. Then the inverse of exponentiation must satisfy:

\( (x^n)@@k=x \)

consequently it is esily seen that \( @@k={\^}(1/n) \), which is the only operator which fits the bill. Again similarly, call the inverse operator of tetration @@@m. Then the inverse of tetration must satisfy:

\( (@@@m)({^n}x)=x \)

from which it follows that \( @@@m= \)tetraroot of order n of x, since the tetraroot is the only operator which satisfies:

\( (tetraroot-n)(^{n}x)=x \)

Now, the very subtle problem which I guess nobody sees (for some strange reason) is that if the operator for tetrating to (1/n) *is NOT* the same operator as that of the tetraroot of order n, then we have an operator discrepancy at a very low level in the hierarchy of operators:

\( {^{1/n}}({^{n}x)=/= x \)

I am not going to argue more about it. Whoever "sees" it, great. Whoever doesn't, great again.

If you not even demand that \( {}^xe \) is continuous, what will then remain?

Sorry, "demanding" and "constructing" are not the same as "existing". Besides, there's no teling whether tetration as defined using tetraroots is or is not continuous.