Let * be a binary operation on N, defined

Solution:

To prove: $*$ is neither commutative nor associative

Let us assume that $*$ is commutative

$\Rightarrow a^{b}=b^{a}$ for $a l l a, b \in N$

This is valid only for $\mathrm{a}=\mathrm{b}$

For example take $a=1, b=2$

$1^{2}=1$ and $2^{1}=2$

So * is not commutative

Let us assume that * is associative

$\Rightarrow\left(a^{b}\right)^{c}=a^{b^{c}}$ for $a l l a, b, c \in N$

$\Rightarrow a^{b c}=a^{b^{c}}$ for all $a, b, c \in N$

This is valid in the following cases:

(i) $a=1$

(ii) $b=0$

(iii) $\mathrm{bc}=\mathrm{b}^{\mathrm{C}}$

For example, let $\mathrm{a}=2, \mathrm{~b}=1, \mathrm{c}=3$

$a^{b c}=2^{(1 \times 3)}=2^{3}=8$

$a^{b^{c}}=2^{1^{3}}=2$

So * is not associative