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 "associativity degree" of an operation (from MathSE) MphLee Long Time Fellow Posts: 321 Threads: 25 Joined: May 2013 05/19/2014, 07:44 PM User Carucell on MathSE posted this awesome question. http://math.stackexchange.com/questions/...341_788017 He asks if exist a binary operation * over the naturals wich has an "infinite degree of associativity".The associativity degree is the number of "hyperoperators" of * that are associative. The associativity degree is the term I'm going to use to describe better the question. let define for every $*:\mathbb{N} \times \mathbb{N} \rightarrow \mathbb{N}$ its "hyperoperations family" in this way $*_0:=*$ $m*{i+1}(1)=m$ $m*_{i+1}(n+1)=m*_i(m*_{i+1}n)$ Lets call the image of this indexed family $\mathcal H(*)$ or in other word we define a function $\mathcal H:\mathbb{N}^{ (\mathbb{N} \times \mathbb{N})}\rightarrow \mathcal P(\mathbb{N}^{ (\mathbb{N} \times \mathbb{N})})$ that takes a binary operation and gives the image of its "hyperoperations" indexed family. $\mathcal H(*):=\{*_{i}:i \in \mathbb{N} \}$ The associative degree of $*$ is the number of associative binary operations in $\mathcal H(*)$ $Adeg(*):=|\{*' \in \mathcal H(*):l*'(m*'n)=(l*'m)*'n\} |$ The question of the user Carucell is if exist a binary operation with associative degree $Adeg(*)=\omega$. We know that for example $Adeg(+)=2$ $Adeg(\times)=1$ -Another interesting question is if there are operations with associative degree greater than 2 but finite. -Or maybe another question I have in mind is: every associative operation has at least degree one, do the converse hold? Does every operation with Adegree one is associative? We can abtain other kind of associative degree functions replacing the construction of the "hyperoperation family" $\mathcal H$ with another definition of "hyperoperation family" in order to obtain different $\mathcal F$ and define a general concept of $\mathcal{F}{-Adeg}$ MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ MphLee Long Time Fellow Posts: 321 Threads: 25 Joined: May 2013 05/21/2014, 10:23 PM The user r. e. s. gave an interesting proof about the associativity degree. http://math.stackexchange.com/questions/...-operation 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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