Let * be a binary operation defined on R by a | vedclass

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(B) Given that,$a * b = \frac{a+b}{4}$.For commutativity,we check $b * a = \frac{b+a}{4} = \frac{a+b}{4} = a * b$. Thus,the operation is commutative.F

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Commutative but not Associative

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(B) Given that,$a * b = \frac{a+b}{4}$.For commutativity,we check $b * a = \frac{b+a}{4} = \frac{a+b}{4} = a * b$. Thus,the operation is commutative.For associativity,we check $a * (b * c)$ and $(a * b) * c$.$a * (b * c) = a * \left(\frac{b+c}{4}ight) = \frac{a + \frac{b+c}{4}}{4} = \frac{4a + b + c}{16}$.$(a * b) * c = \left(\frac{a+b}{4}ight) * c = \frac{\frac{a+b}{4} + c}{4} = \frac{a + b + 4c}{16}$.Since $\frac{4a + b + c}{16} 
eq \frac{a + b + 4c}{16}$,the operation is not associative.Therefore,the operation is commutative but not associative.

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

Let $$ be a binary operation defined on $R$ by $a b = \frac{a+b}{4}$ for all $a, b \in R$. Then the operation $*$ is:

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Let $*$ be a binary operation defined on $R$ by $a * b = \frac{a+b}{4}$ for all $a, b \in R$. Then the operation $*$ is: