Basic algebra: associativity vs. commutativity? While I was reviewing someone's homework for them, I came across the following:
$$(2)(x/3)(5) = 11$$
The workbook claims that rearranging the equation to:
$$(2)(5)(x/3) = 11$$
is an example of the associative property.
To me this looks much more like the commutative property. So, which is it?
 A: Actually you have to use both the associative and the commutative property.
$$2 \cdot\frac{x}{3} \cdot 5  \tag{1}$$
is an abbreviation for 
$$(2 \cdot\frac{x}{3}) \cdot 5\tag{2}$$
because multiplication is evaluated from left to right.
Now using associativity you can transform $(2)$ to
$$2 \cdot (\frac{x}{3} \cdot 5) \tag{3}$$
and now you use the commutative property to get
$$2 \cdot (5 \cdot \frac{x}{3} )\tag{4} $$
Now you use again associativity and get 
$$(2 \cdot 5 ) \cdot \frac{x}{3} \tag{5} $$
which can be abbreviated to
$$2 \cdot 5  \cdot \frac{x}{3} \tag{6} $$

So if an operation $\cdot$ has the associative and the commutative property you have
$$(a \cdot b) \cdot c = (a \cdot c) \cdot b, \forall\;  a,b,c \tag{7}$$
But   commutative property  is not sufficient that $(7)$ holds for the  operation $\cdot$. The following operation $\cdot$ will show  this :
. 0 1 2
-+-----
0|0 1 1
1|1 0 2
2|1 2 2

This operation is defined on the set $\{0,1,2\}$ and it is commutative because the Cayley table is symmetric. But you have
$$(0 \cdot  1) \cdot 2= 1 \cdot 2=2 \tag{8}$$
but
$$(0 \cdot 2) \cdot 1= 1 \cdot 1= 0  \tag{8}$$
So
$$(a \cdot b) \cdot c \ne (a \cdot c) \cdot b \tag{10}$$
for some $a,b,c$ even if $\cdot$ is commutative.
