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 A fundamental flaw of an operator who's super operator is addition JmsNxn Long Time Fellow Posts: 283 Threads: 67 Joined: Dec 2010 09/04/2011, 05:16 AM 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 $a_1\, \oplus\, a_2\,\oplus\,a_3...\oplus\,a_n = a + n$ The contradiction comes about here: $a\, \oplus\, a = a + 2$ $(a\, \oplus\, a)\,\oplus\,(a\, \oplus\, a) = (a+2) \, \oplus\,(a +2) = a + 4$ 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 $a\,\oplus\,b = a\,\bigtriangleup_{\sqrt{2}}^{-1}\,b = \log_{\sqrt{2}}(\sqrt{2}^a + \sqrt{2}^b)$ tommy1729 Ultimate Fellow Posts: 1,347 Threads: 323 Joined: Feb 2009 09/05/2011, 11:50 PM yes that works for 2 a's ... but not for 3 ... :s JmsNxn Long Time Fellow Posts: 283 Threads: 67 Joined: Dec 2010 09/06/2011, 02:00 AM (This post was last modified: 09/06/2011, 02:05 AM by JmsNxn.) (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. reread my proof. the one that works for three is: $a\,\bigtriangleup_{3^{\frac{1}{3}}}^{-1}\,b\,=\log_{3^{\frac{1}{3}}}(3^{\frac{a}{3}} + 3^{\frac{b}{3}})$ « Next Oldest | Next Newest »

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