TITLE: Associative law is not self evident
QUESTION [6 upvotes]: The statement:
"It is important to understand that the associative law is not
self-evident; indeed, if $a*b=a/b$ for positive numbers $a$ and $b$
then, in general, $(a*b)*c\ne a*(b*c)$." - p. 3, A. Beardon, Algebra
and Geometry.
I am unclear as to how to take the assumption $a*b=a/b$. Does he mean
that we suppose, as an alternative (thought experiment),
that $a*b :=
a/b$ and substitute
$(a/b)$ for each $(a*b)$ and prove the inequality of the resulting
statement? Or, should one notice that
$a*b=a/b$ is true when $b=1$, but then note that the domain was
assumed to be
 $\mathbb{R}_+$, therefore $(a*b)*c \ne a*(b*c)$ in general?

REPLY [7 votes]: Here, the $*$ represents an arbitrary binary operation, not multiplication as you think. The definition of $*$ is a map $\mathbb{R}_+ \times \mathbb{R}_+ \rightarrow \mathbb{R}_+$ that is defined by $*(a,b)=\frac{a}{b}$. Now, $a*b$ is just an alternate way of representing $*(a,b)$.
Now why isn't $*$ associative? Well, we have $(a*b)*c=(a/b)*c=(a/b)/c=\frac{a}{bc}$, and $a*(b*c)=a*(b/c)=a/(b/c)=\frac{ac}{b}$, hence $(a*b)*c \not= a*(b*c)$.