4 ms·
I've yet to encounter the second definition, and the first only when f is associative (otherwise f(x,x) = x doesn't seem strong enough). All the others are jus
by contravariant 2y ago
I've yet to encounter the second definition, and the first only when f is associative (otherwise f(x,x) = x doesn't seem strong enough).
All the others are just the many ways x^n can equal x (for all n>0) for various associative operators. Reusing 'f' for both the binary functions and the unary ones is needlessly misleading by the way, the unary f behaves more like x in the other examples.
And yes if you want R^2 = R then the condition only needs to hold for some y. It's related to transitivity.
- GrantMoyer 2y agoYou don't need associativity: (x ⋅ x) ⋅ x (x) ⋅ x x ⋅ (x) x ⋅ (x ⋅ x)
- a1369209993 2y agoYou do need (some subset of) associativity: (y0 * x) * x ?= y0 * (x * x) y1 * x ?= y0 * x y2 ?= y1 y2 is not in general constrained to be the same as y1 otherwise.
- o11c 2y agoGood point about reusing variables, I've renamed all except the ones introduced locally. I would only say that these are "isomorphic to" others; in the real world people do think of them differently, so it is important to enumerate them. And in particular people do think "the function", even though different the functions act differently. (the real question is, did I ever see "left-idempotent" and "right-idempotent" somewhere?)