A fundamental flaw of an operator who's super operator is addition - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- Thread: A fundamental flaw of an operator who's super operator is addition (/showthread.php?tid=693) A fundamental flaw of an operator who's super operator is addition - JmsNxn - 09/04/2011 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 $a\,\oplus\,b = a\,\bigtriangleup_{\sqrt{2}}^{-1}\,b = \log_{\sqrt{2}}(\sqrt{2}^a + \sqrt{2}^b)$ RE: A fundamental flaw of an operator who's super operator is addition - tommy1729 - 09/05/2011 yes that works for 2 a's ... but not for 3 ... :s RE: A fundamental flaw of an operator who's super operator is addition - JmsNxn - 09/06/2011 (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: $a\,\bigtriangleup_{3^{\frac{1}{3}}}^{-1}\,b\,=\log_{3^{\frac{1}{3}}}(3^{\frac{a}{3}} + 3^{\frac{b}{3}})$ RE: A fundamental flaw of an operator who's super operator is addition - tetration101 - 05/15/2019 pretty ingenious, PS : About the successor, although some consider the successor operator as the one that comes before addition, but the thing is that is unary. Other way could oe consider the Min operator as in "tropical geometry, but not all like that operator since not produce a change about the initial values... https://en.wikipedia.org/wiki/Tropical_geometry http://staffwww.fullcoll.edu/dclahane/colloquium/scottentalk.pdf ...probably would be nice a half-operation between Min operation and Addition operation RE: A fundamental flaw of an operator who's super operator is addition - Chenjesu - 06/23/2019 Shouldn't this follow directly from the fact that addition is axiomatic? If such a hyperoperation existed below addition, then you'd be contradicting the existence of the axioms upon which you've constructed addition in the first place to even define real numbers.