For the binary operation $^*$ defined on the set $R - \{-1\}$ by $a ^* b = \frac{a}{b+1}$,determine whether $^*$ is commutative or associative.

The binary operation is defined as $a ^* b = \frac{a}{b+1}$ for $a, b \in R - \{-1\}$.
$1$. Commutativity:
To check if $^*$ is commutative,we compare $a ^* b$ and $b ^* a$.
$a ^* b = \frac{a}{b+1}$
$b ^* a = \frac{b}{a+1}$
Since $\frac{a}{b+1} \neq \frac{b}{a+1}$ in general (e.g.,let $a=1, b=2$,then $1 ^* 2 = \frac{1}{3}$ and $2 ^* 1 = \frac{2}{2} = 1$),the operation $^*$ is not commutative.
$2$. Associativity:
To check if $^*$ is associative,we compare $(a ^* b) ^* c$ and $a ^* (b ^* c)$.
$(a ^* b) ^* c = (\frac{a}{b+1}) ^* c = \frac{\frac{a}{b+1}}{c+1} = \frac{a}{(b+1)(c+1)}$
$a ^* (b ^* c) = a ^* (\frac{b}{c+1}) = \frac{a}{\frac{b}{c+1} + 1} = \frac{a}{\frac{b+c+1}{c+1}} = \frac{a(c+1)}{b+c+1}$
Since $\frac{a}{(b+1)(c+1)} \neq \frac{a(c+1)}{b+c+1}$ in general (e.g.,let $a=1, b=2, c=3$,then $(1 ^* 2) ^* 3 = \frac{1}{(3)(4)} = \frac{1}{12}$ and $1 ^* (2 ^* 3) = \frac{1(4)}{2+3+1} = \frac{4}{6} = \frac{2}{3}$),the operation $^*$ is not associative.