The binary operation * on R - {-1} defined by | vedclass

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(C) Given the binary operation $a * b = \frac{a}{b+1}$.Check for commutativity:$a * b = \frac{a}{b+1}$$b * a = \frac{b}{a+1}$Since $\frac{a}{b+1} 
eq

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$*$ is neither associative nor commutative

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(C) Given the binary operation $a * b = \frac{a}{b+1}$.Check for commutativity:$a * b = \frac{a}{b+1}$$b * a = \frac{b}{a+1}$Since $\frac{a}{b+1} 
eq \frac{b}{a+1}$ for all $a, b \in R - \{-1\}$,the operation is not commutative.Check for associativity:$(a * b) * c = \left(\frac{a}{b+1}ight) * c = \frac{\frac{a}{b+1}}{c+1} = \frac{a}{(b+1)(c+1)}$$a * (b * c) = a * \left(\frac{b}{c+1}ight) = \frac{a}{\frac{b}{c+1} + 1} = \frac{a(c+1)}{b+c+1}$Since $\frac{a}{(b+1)(c+1)} 
eq \frac{a(c+1)}{b+c+1}$,the operation is not associative.Therefore,the operation $*$ is neither commutative nor associative.

$^*$	$1$	$2$	$3$	$4$	$5$
$1$	$1$	$1$	$1$	$1$	$1$
$2$	$1$	$2$	$2$	$2$	$2$
$3$	$1$	$2$	$3$	$3$	$3$
$4$	$1$	$2$	$3$	$4$	$4$
$5$	$1$	$2$	$3$	$4$	$5$

The binary operation $$ on $R - \{-1\}$ defined by $a b = \frac{a}{b+1}$ is:

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The binary operation $*$ on $R - \{-1\}$ defined by $a * b = \frac{a}{b+1}$ is: