# Mark the tick against the correct answer in the following:

Question:

Mark the tick against the correct answer in the following:

Let $Z^{+}$be the set of all positive integers. Then, the operation $*$ on $Z^{+}$defined bya $* b=a^{b}$ is

A. commutative but not associative

B. associative but not commutative

C. neither commutative nor associative

D. both commutative and associative

Solution:

According to the question,

$\mathrm{Q}=\{$ All integers $\}$

$\mathrm{R}=\left\{(\mathrm{a}, \mathrm{b}): \mathrm{a}^{*} \mathrm{~b}=\mathrm{a}^{\mathrm{b}}\right\}$

Formula

$*$ is commutative if $a * b=b^{*} a$

$*$ is associative if $(a * b) * c=a *(b * c)$

Check for commutative

Consider, $a * b=a^{b}$

And, $b * a=b^{a}$

Both equations are not the same and will not always be true .

Therefore, $*$ is not commutative (1)

Check for associative

Consider, $(a * b) * c=\left(a^{b}\right) * c=\left(a^{b}\right)^{c}$

And, $a^{*}\left(b^{*} c\right)=a^{*}\left(b^{c}\right)=a^{\left(b^{c}\right)}$

$E x a=2 b=3 c=4$

$\left(a^{*} b\right) * c=\left(2^{3}\right) * c=(8)^{4}$

$a^{*}\left(b^{*} c\right)=2 *\left(3^{4}\right)=2^{(81)}$

Both the equation are not the same and therefore will not always be true.

Therefore, $*$ is not associative (2)

Now, according to the equations (1), (2)

Correct option will be (C)