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**

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).

**Attribution***Source : Link , Question Author : Zirui Wang , Answer Author : Arnaud D.*