Is there an intuitive reason for a certain operation to be associative?

Question

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?

Answer 1

A common way to build weird-looking associative operations is to start from a known one, such as multiplication, say on the real numbers or some subset of them, and then to transform it through some bijection $\alpha$, by defining $$x\ast y=\alpha^{-1}(\alpha(x)\cdot\alpha(y)).$$ Indeed this is equivalent to $\alpha(x\ast y)=\alpha(x)\cdot \alpha(y)$ (so that $\alpha$ is actually an isomorphism), and it is then easy to check associativity by noticing that \begin{align*}\alpha(x\ast (y\ast z)) & =\alpha(x)\cdot \alpha(y\ast z) = \alpha(x)\cdot(\alpha(y)\cdot \alpha(z))\\ & =(\alpha(x)\cdot \alpha(y))\cdot \alpha(z) = \alpha(x\ast y)\cdot \alpha(z)\\ & =\alpha((x\ast y)\ast z),\end{align*} which implies that $x\ast (y\ast z)=(x\ast y)\ast z$ since $\alpha$ is bijective. Other properties, such as commutativity or existence of neutral or inverses, can be done in the same way, depending on the cases.

In this case, we can see that $$\frac{1}{x\ast y}=\frac{1}{x}+\frac{1}{y}+\frac{1}{xy}$$so that $$1+\frac{1}{x\ast y}=1+\frac{1}{x}+\frac{1}{y}+\frac{1}{xy}=\left(1+\frac{1}{x}\right)\cdot\left(1+\frac{1}{y}\right),$$ so if you define $\alpha(x)=1+\frac{1}{x}$, you can check that it defines a bijection $(0,+\infty)\to (1,+\infty)$, and $\ast$ is just a transformation of the multiplication on $(1,+\infty)$, which explain why it is associative. In fact you can also see right away that it must also be commutative, but that it can't have a neutral element (otherwise $(1,+\infty)$ would have one).

Answer 2

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.