Question

Let * be a binary operation on $\mathrm{N}$ defined by $\mathrm{a}^* \mathrm{~b}=2^{5 \mathrm{ab}} ; \mathrm{a}, \mathrm{b} \in \mathrm{N}$. then the binary operation * is

(A) Only commutative

(C) Both commutative and associative

(B) Only associative

(D) Neither commutative nor associative

Answer 1

For $a, b \in N$, we have

$a * b=2^{5 a b}$

Commutativity. Let $\mathrm{a}, \mathrm{b} \in \mathrm{N}$

$\therefore a{ }^* b=2^{5 a b}=2^{5 b a}=b^* a \\$

$\therefore a{ }^* b=b^* a$

$\therefore{ }^*$ is commutative.

Associativity. Let $a=1, b=2, c=3$

$\therefore\left(a^* b\right) * c=(1 * 2) * 3=2^{5(A)(B) *} 3=2^{10 *} 3=2^{5\left(2^{10}\right) 3}=2^{15.2^{10}}$

Also $a^*\left(b^* c\right)=1^*\left(2^* 3\right)=1^* 2^{5(B)(C)}=1^* 2^{30}=2^{5(1)\left(2^{30}\right)}=2^{5.2^{30}}$

$\therefore\left(a{ }^* b\right)^* c \neq a{ }^*\left(b{ }^* c\right) \text { for } a=1, b=2, c=3 \\$

$\therefore \text { In general, }\left(a{ }^* b\right){ }^* c \neq a{ }^*\left(b{ }^* c\right) \\$

$\therefore \text { * is not associative. }$