Basic algebra: associativity vs. commutativity?

Question

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?

Answer 1

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 $$2 \cdot 5 \cdot \frac{x}{3} \tag{6} $$

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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.