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 A fundamental flaw of an operator who's super operator is addition JmsNxn Ultimate Fellow Posts: 940 Threads: 112 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: 1,676 Threads: 368 Joined: Feb 2009 09/05/2011, 11:50 PM yes that works for 2 a's ... but not for 3 ... :s JmsNxn Ultimate Fellow Posts: 940 Threads: 112 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: $a\,\bigtriangleup_{3^{\frac{1}{3}}}^{-1}\,b\,=\log_{3^{\frac{1}{3}}}(3^{\frac{a}{3}} + 3^{\frac{b}{3}})$ tetration101 Junior Fellow Posts: 6 Threads: 4 Joined: May 2019 05/15/2019, 09:49 PM (This post was last modified: 05/15/2019, 09:54 PM by tetration101.) 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/co...entalk.pdf ...probably would be nice a half-operation between Min operation and Addition operation Chenjesu Junior Fellow Posts: 21 Threads: 4 Joined: May 2016 06/23/2019, 08:19 PM (This post was last modified: 06/25/2019, 07:29 AM by Chenjesu.) 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. Catullus Fellow Posts: 205 Threads: 46 Joined: Jun 2022 06/13/2022, 11:48 PM (This post was last modified: 06/13/2022, 11:54 PM by Catullus.) (09/04/2011, 05:16 AM)JmsNxn Wrote: 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 $a\,\oplus\,b = a\,\bigtriangleup_{\sqrt{2}}^{-1}\,b = \log_{\sqrt{2}}(\sqrt{2}^a + \sqrt{2}^b)$log(sqrt(2),sqrt(2)^a+sqrt(2)^a) = log(sqrt(2),sqrt(2)^a*2) = a+2, if a ∈ $\mathbb{R}$. Not ∀ a. Quote:tommy1729 Wrote: 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: Wrote: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: $a\,\bigtriangleup_{3^{\frac{1}{3}}}^{-1}\,b\,=\log_{3^{\frac{1}{3}}}(3^{\frac{a}{3}} + 3^{\frac{b}{3}})$log(3^1/3,3^a/3*3) = a+3, if a ∈ $\mathbb{R}$. Not ∀ a. ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Please remember to stay hydrated. Sincerely: Catullus MphLee Long Time Fellow Posts: 321 Threads: 25 Joined: May 2013 06/16/2022, 10:33 PM Honestly, I don't see the point in this nitpicking. One could argue for going thru all the forum's posts adding universal quantifiers to every statement... MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ « Next Oldest | Next Newest »

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