Question

Let $\ast$ be a binary operation defined on $R$ by $a\ast b$ = $\frac{a+b}{4}\forall a,b\in R$ then the operation $\ast$ is

• A. Commutative and Associative
• B. Commutative but not Associative
• C. Associative but not Commutative
• D. Neither Associative nor Commutative

Answer 1

Answer: (B)

Solution: We have, $a\ast b=\frac{a+b}{4}$
For commutative, $a\ast b=b\ast a$
$\therefore \frac{a+b}{4}=\frac{b+a}{4}$
Hence, $\ast$ is commutative.
For associative, $a\ast (b\ast c)=(a\ast b)\ast c$
LHS $=a\ast \left(\frac{b+c}{4}\right)=\frac{a+\frac{b+c}{4}}{4}$
$=\frac{4a+b+c}{16}$
$\text{ RHS }=(a\ast b)\ast c=\left(\frac{a+b}{4}\right)\ast c$
$=\frac{\frac{a+b}{4}+c}{4}=\frac{a+b+4c}{16}$
Since, $a\ast (b\ast c)\ne (a\ast b)\ast c$
So, ${}^{\ast }$ is not associative.