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(N/A) binary operation $\ast$ on a set $S$ is commutative if $a \ast b = b \ast a$ for all $a, b \in S$. For addition $(+)$ on $R$: Since $a + b = b +

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(N/A) binary operation $\ast$ on a set $S$ is commutative if $a \ast b = b \ast a$ for all $a, b \in S$. For addition $(+)$ on $R$: Since $a + b = b + a$ for all $a, b \in R$,the operation $+$ is commutative. For multiplication $(	imes)$ on $R$: Since $a 	imes b = b 	imes a$ for all $a, b \in R$,the operation $	imes$ is commutative. For subtraction $(-)$ on $R$: Since $a - b 
eq b - a$ in general (e.g.,$3 - 4 = -1$ and $4 - 3 = 1$),the operation $-$ is not commutative. For division $(\div)$ on $R_*$: Since $a \div b 
eq b \div a$ in general (e.g.,$3 \div 4 = 0.75$ and $4 \div 3 = 1.33$),the operation $\div$ is not commutative.

Show that $+: R \times R \rightarrow R$ and $\times: R \times R \rightarrow R$ are commutative binary operations,but $-: R \times R \rightarrow R$ and $\div: R_* \times R_* \rightarrow R_*$ are not commutative.

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Show that $+: R \times R \rightarrow R$ and $\times: R \times R \rightarrow R$ are commutative binary operations,but $-: R \times R \rightarrow R$ and $\div: R_* \times R_* \rightarrow R_*$ are not commutative.

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Let $^*$ be a binary operation on the set $Q$ of rational numbers defined as $a \,^*\, b = a b^{2}$. Determine whether the operation is commutative and associative.

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