A fundamental flaw of an operator who's super operator is addition

[Catullus]

The operator which obeys:
\( a\,\oplus\,a = a+ 2 \)
but who's super operator is not addition is, you've guessed it, logarithmic semi operator base root (2)

or

\( a\,\oplus\,b = a\,\bigtriangleup_{\sqrt{2}}^{-1}\,b = \log_{\sqrt{2}}(\sqrt{2}^a + \sqrt{2}^b) \)

\(\forall a\in\Bbb{R}\log_\sqrt2(\sqrt2^a+\sqrt2^a)=\log_\sqrt2(\sqrt2^a*2)=a+2\) It is not true for all a.

yes that works for 2 a's ...

but not for 3 ...

:s

that's my point!

You cannot have an operator that works for all n a's, only at fix points

Voila! this proves you cannot have an associative and commutative operator who's super operator is addition.

reread my proof.

the one that works for three is:

\(\forall a\in\Bbb{R}\log_\sqrt[3]3(\sqrt[3]3^a+\sqrt[3]3^a+\sqrt[3]3^a)=\log_\sqrt[3]3(\sqrt[3]3^a*3)=a+3\) It is not true for all a.