Binary Operations - Associativity

[spellbinder]

We all know that Associativity in Binary Operations is defined as (a*b)*c = a*(b*c). My question is: What would the arrangement look like if we're dealing with four or more elements. Let's say... A = {a,b,c,d}? I don't where to place the parentheses.

[aswoods]

Associativity is defined in terms of three elements, but you could say that for four elements a,b,c,d:

((ab)c)d = (a(bc))d =(ab)(cd) = a((bc)d) = a(b(cd))

[spellbinder]

Now I get it. Thanks.

[Martin Rattigan]

But was this really the answer you were looking for? The five relations aswood gives for four elements don't necessarily imply associativity.

E.g. if and we define in , then is a groupoid under satisfying each of the five relations given but it is not associative.

It is easy to prove that when all possible ways of parenthesising elements are identical and , then all possible ways of parenthesising elements are identical, but for there are examples similar to the above in which all identities for elements hold, but which are non-associative.

[cknapp]

Martin, do you keep a bag of instructive counter-examples at your desk?

Anyway, since the notion of groupoid has been brought up, I figure I'll continue this descent into the abstract by pointing out that operads are often used to model structures that are "not quite" associative-- for every set of n elements, you have a bunch of n-ary operations, which often are made to represent the different ways of "multiplying" n elements. e.g. you could have an operation f(a,b,c,d) = (ab)(cd) and g(a,b,c,d) = a(b(cd).

Certain n-ary operations can be said to be equal to express some level of associativity. To express "actual" associativity,
we can define an operad with f(a,b,c) =a (bc) and g(a,b,c) = (ab)c, and then say that f=g, that is, a(bc)=(ab)c.