Associative law is not self evident

Question

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?

Answer 1

You are basically right in your guesses. What is being emphasized is that not all binary operations (i.e. things that take two inputs from your set and return a third element of the set) that one can define are necessarily associative just because they are binary operations. In this example they are saying that if you define the particular example of a binary operation, to which they give the name $*$, on the set $\Bbb R^+$, as in this example, i.e. $a*b={a\over b}$, then

$$(a*b)*c={{a\over b}\over c}={a\over bc}$$

and

$$a*(b*c)={a\over {b\over c}}={ac\over b}$$

by direct computation.

At this point you might object and say "but doesn't $*$ mean multiplication?" If so, you can replace the symbol $*$ by another one if it helps you, $\oplus$ is a nice one as well, the point is it's the binary operation we've defined, no matter what it looks like on paper as a symbol.

Now, by definition two real numbers, $x,y\in\Bbb R$, the definition of equality is

$$x=y\iff x-y=0$$

and given two fractions, ${p\over q}, {r\over s}\in\Bbb R^+$ the definition of equality is

$${p\over q}={r\over s}\iff ps-qr=0.$$

In your case:

$${a\over bc}-{ac\over b}=0$$ $$\iff {a\over bc}={ac\over b}$$ $$\iff ab-abc^2=0$$ $$\iff ab=abc^2$$

for every $a,b,c\in\Bbb R^+$, however this condition clearly requires $c^2=1$ which is only true for $c=1$ over the positive real numbers, so the operation is not associative, because the relation only holds for a single positive real number, $c=1$, rather than all positive reals.

Answer 2

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 $(a*b)*c \not= a*(b*c)$.

Answer 3

I think you're assuming that * means multiplication. But in this situation it represents a general binary operation, and in particular, division.

Answer 4

In general, without making any assumptions, $\ast$ stands just for a group operation. In your cited example, it is defined to be division. Therefore, in this case, $(a \ast b) \ast c = (a/b)/c \neq a/(b/c) = a \ast (b \ast c)$, except for trivial cases.