TITLE: Is there an intuitive reason for a certain operation to be associative?
QUESTION [70 upvotes]: When I read Pinter's A Book of Abstract Algebra, Exercise 7 on page 25 asks whether the operation
$$x*y=\frac{xy}{x+y+1}$$
(defined on the positive real numbers) is associative. At first I considered this to be false, because the expression is so complicated. But when I worked out $(x*y)*z$ and $x*(y*z)$, I found both to be $$\frac{xyz}{xy+yz+zx+x+y+z+1}!$$ Commutativity is easy to see. But associativity can be so counter-intuitive! Can you see this operation is associative without working it out? Are there tricks to do this?

REPLY [7 votes]: Another way to prove associativity quite easily is when your operation is already known to be commutative. Then it is enough (and in fact necessary) to find an expression
$$ (x\ast y) \ast z = \varphi(x,y,z) $$
where $\varphi$ is invariant under (circular) permutation, meaning that $\varphi(t_1,t_2,t_3) = \varphi(t_{\sigma(1)},t_{\sigma(2)},t_{\sigma(3)})$ for all (circular) permutations $\sigma \in \mathfrak S_3$.
Indeed, you then get
$$ (x\ast y) \ast z = \varphi(x,y,z) = \varphi(y,z,x) = (y\ast z)\ast x = x \ast (y \ast z) $$
where the last equality holds by commutativity of $\ast$.
(Taking a good look at what happen above, showing that $\varphi$ is invariant under the transposition $(1\,3)$ is actually enough.)
In you case, commutativity of $\ast$ is directly given by commutativity of product and sum of reals. And the expression you found for $(x \ast y)\ast z$ is clearly invariant by permutation of $x,y,z$ (again by commutativity of products and sums of reals). So you can conclude that $\ast$ is associative.