What is your definition of an identity of a ternary function? Say i is the identity of f. Must it hold that f(i, a, b) == f(a, i, b) == f(a, b, i)? For al

What is your definition of an identity of a ternary function? Say i is the identity of f. Must it hold that f(i, a, b) == f(a, i, b) == f(a, b, i)? For all a, b in f's domain? The seems like the obvious choice, and certainly holds for ternary functions "built up" from associative binary ones, but I've just never seen such a thing. Is there a case where it's useful to talk about the identity of a 3-valued function? I think the thing I find weird about this is that a binary operator op2 over some set S is defined as S^2 -> S. So we can sensibly say: op2(a, i) == a == op2(i, a) But the extension of this to a 3-valued function from S^3 -> S cannot have that middle condition, because it has too many "free variables". We can't say: op3(a, b, i) == op2(a, b) Without establishing a relationship between op3 and op2, but must there be a sensible "decomposition" of op3 into some op2? Surely not. I guess I just don't know what it means to be an identity of a 3-valued function. As an example, consider mult3(a, b, c) := a * b * c. Then, mult3(0, a, b) == mult3(a, 0, b) == mult3(a, b, 0) for all a and b. We intuitively know that 0 is not behaving like an identity. Can we write down an identity rule that must hold for a generic op3 that excludes 0 as the identity of that function? Without relying on some oracle to provide us an appropriate decomposed op2? Edit: I would also like to point out this function over R^3: f(a, b, c) := a*b + a*c + b*c Note that every value in R meets the condition: f(a, x, y) == f(x, a, y) == f(x, y, a) == ax + ay + xy For all x and y. Is every number an identity of this function? In fact, this is trivially true for any completely associative function. I guess I just need to see the definition of an identity of a ternary operator.