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Well I came across this little paradox when I was fiddling around with logarithmic semi operators. It actually brings to light a second valid argument for the aesthetic nature of working with base root (2) and furthers my belief that this is the "natural" extension to the ackermann function.

Consider an operator who is commutative and associative, but above all else, has a super operator which is addition. this means

$a\, \oplus\, a = a + 2$

or in general
$a_1\, \oplus\, a_2\,\oplus\,a_3...\oplus\,a_n = a + n$

$a\, \oplus\, a = a + 2$
$(a\, \oplus\, a)\,\oplus\,(a\, \oplus\, a) = (a+2) \, \oplus\,(a +2) = a + 4$

right now:

$[(a\, \oplus\, a)\,\oplus\,(a\, \oplus\, a)]\,\oplus\,[(a\, \oplus\, a)\,\oplus\,(a\, \oplus\, a)] = (a+4)\, \oplus\, (a+4) = a + 6$

However, if you count em, there are exactly eight a's, not six.

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

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

$a\,\oplus\,b = a\,\bigtriangleup_{\sqrt{2}}^{-1}\,b = \log_{\sqrt{2}}(\sqrt{2}^a + \sqrt{2}^b)$
yes that works for 2 a's ...

but not for 3 ...

:s
(09/05/2011, 11:50 PM)tommy1729 Wrote: [ -> ]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

Quote:Voila! this proves you cannot have an associative and commutative operator who's super operator is addition.
$a\,\bigtriangleup_{3^{\frac{1}{3}}}^{-1}\,b\,=\log_{3^{\frac{1}{3}}}(3^{\frac{a}{3}} + 3^{\frac{b}{3}})$