# Infinite associativity condition

How associativity condition may be formulated for a function taking an arbitrary ordinal number of arguments?

For a binary operation $\ast$ it is $(a\ast b)\ast c = a\ast (b\ast c)$, but I want an infinite formula, whose special case is the condition of associativity for a binary operation.

-
Let's look at the next case: four args. What would the rules be? The possibilities are abcd = a*(b*(cd)) = a*((bc)*d) = (ab)*(cd) = (a*(bc)*d) = ... Seems to me that the only reasonable way which avoids enumerating all possible parenthesizations (not a word, but ...) would be a recursive one. – marty cohen Mar 18 '12 at 19:21
There isn't really a notion of associativity when starting with a ternary operator, so an operator taking infinitely many variables probably doesn't have a real notion of associativity. – Thomas Andrews Mar 18 '12 at 19:36
@Thomas: Writing $[abc]$ for the result of applying the operation to the ordered triple $\langle a,b,c\rangle$, the obvious notion of associativity is that $[ab[cde]]=[a[bcd]e]=[[abc]de]$ for all $a,b,c,d,e$. A much less obvious generalization of associativity is $[[abc]de]=[a[bde][cde]]$, obtained by thinking of each element $a$ as a two-place function (where the ordinary $(ab)c=a(bc)$ corresponds to thinking of each element as a one-place function). – Brian M. Scott Mar 18 '12 at 22:57
See a related question at MathOverflow: mathoverflow.net/questions/91761/… – porton Mar 20 '12 at 22:37
@BrianM.Scott I think you've provided useful information, but this doesn't contradict what Thomas Andrews says. Your generalization of the associativity, I agree, generalizes our intuitive concept of associativity to a non-binary operation. Associativity (once given a notational system) gets defined as an equality which involves a binary operation, which is not just an intuition. Hence you may claim to have a concept of ternary associativity, but that is not the concept of associativity, or redundantly, binary associativity. – Doug Spoonwood Apr 11 '12 at 23:33