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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.

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    Let's look at the next case: four args. What would the rules be? The possibilities are a*b*c*d = a*(b*(c*d)) = a*((b*c)*d) = (a*b)*(c*d) = (a*(b*c)*d) = ... Seems to me that the only reasonable way which avoids enumerating all possible parenthesizations (not a word, but ...) would be a recursive one.2012-03-18
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    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.2012-03-18
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    @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).2012-03-18
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    See a related question at MathOverflow: http://mathoverflow.net/questions/91761/formula-for-a-product-of-products-over-ordinal-numbers2012-03-20
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    @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.2012-04-11
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    @Doug: I disagree. I would say that there is no generally accepted definition of associativity for ternary operations, not that associativity is necessarily a property of binary operations.2012-04-12
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    @BrianM.Scott If associativity is not a property of a binary operation, what in the world do you mean by associativity? If we have an equation like (a*(b*c))=((a*b)*c), how is * not a binary operation when it behaves exactly like a binary operation?2012-04-12
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    @Doug: Of course in that example $*$ is a binary operation: that’s an illustration of the accepted notion of associativity for binary operations. It also seems to me to be a non sequitur to my previous comment. Does it bother you that the term *ideal* has been extended to settings very different from its original home? My view is that to date there simply hasn’t been any good reason to extend *associativity* to $n$-ary operations for $n\ne 2$ (and indeed there may never be). You seem to want to rule out any such extension *a priori*; not being a seer, I decline to do so, that’s all.2012-04-12
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    @BrianM.Scott I have no problems with extensions of terms. In fact, I *like* them. Using the term "associativity" for extensions of binary associativity isn't necessarily problematic, but you've then used the term "associativity" in a different way for the extended concept than you did for binary associativity. The meaning of the term would not be the same in both cases.2012-04-12
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    @Doug: Nor is it for ideals. And in fact I suggested *two* ways to extend it to ternary operations, one very obviously a straightforward generalization of the binary property. I see no profit in pursuing the disagreement further, however; I’ve said what I have to say on the subject.2012-04-12
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    @BrianM.Scott Yes, you did. Sorry if it seemed like I implied otherwise.2012-04-12
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    http://www.mathematics21.org/binaries/assoc.pdf2012-04-22

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