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On $Q,$ the operation $^*$ is defined as $a ^* b = a - b.$$1$. Commutativity:For the operation to be commutative,$a ^* b = b ^* a$ must hold for all $

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On $Q,$ the operation $^*$ is defined as $a ^* b = a - b.$$1$. Commutativity:For the operation to be commutative,$a ^* b = b ^* a$ must hold for all $a, b \in Q.$Consider $a = \frac{1}{2}$ and $b = \frac{1}{3}.$$\frac{1}{2} ^* \frac{1}{3} = \frac{1}{2} - \frac{1}{3} = \frac{3 - 2}{6} = \frac{1}{6}$$\frac{1}{3} ^* \frac{1}{2} = \frac{1}{3} - \frac{1}{2} = \frac{2 - 3}{6} = -\frac{1}{6}$Since $\frac{1}{6} 
eq -\frac{1}{6},$ the operation $^*$ is not commutative.$2$. Associativity:For the operation to be associative,$(a ^* b) ^* c = a ^* (b ^* c)$ must hold for all $a, b, c \in Q.$Consider $a = \frac{1}{2}, b = \frac{1}{3}, c = \frac{1}{4}.$$(a ^* b) ^* c = (\frac{1}{2} - \frac{1}{3}) ^* \frac{1}{4} = \frac{1}{6} ^* \frac{1}{4} = \frac{1}{6} - \frac{1}{4} = \frac{2 - 3}{12} = -\frac{1}{12}$$a ^* (b ^* c) = \frac{1}{2} ^* (\frac{1}{3} - \frac{1}{4}) = \frac{1}{2} ^* \frac{1}{12} = \frac{1}{2} - \frac{1}{12} = \frac{6 - 1}{12} = \frac{5}{12}$Since $-\frac{1}{12} 
eq \frac{5}{12},$ the operation $^*$ is not associative.

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

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

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State whether the following statement is true or false and justify your answer: If $^*$ is a commutative binary operation on $N$,then $a ^* (b ^* c) = (c ^* b) ^* a$.

Show that the operation $*: R \times R \rightarrow R$ defined by $a * b = a + 2b$ is not associative.

Let $*$ be a binary operation defined on the set of rational numbers $Q$. Determine whether the binary operation defined by $a * b = (a - b)^{2}$ for all $a, b \in Q$ is commutative.

Determine whether or not each of the definitions of $*$ given below gives a binary operation. In the event that $*$ is not a binary operation,give justification for this. On $Z^+$,define $*$ by $a * b = |a - b|$.

Let $^*$ be the binary operation on $N$ given by $a \,^*\, b = \text{L.C.M. of } a \text{ and } b$. Is $^*$ associative?