Show that $*: R \times R \rightarrow R$ given by $(a, b) \rightarrow a+4 b^{2}$ is a binary operation.

Which of the following is not a group with respect to the given operation?

Let $A = \{0, 1, 2, 3, 4, 5, 6\}$. If $a, b \in A$,and $a * b$ is defined as the remainder when $ab$ is divided by $7$,then find the inverse of $2$ under the operation $*$.

Consider the binary operations $^*: R \times R \rightarrow R$ and $o: R \times R \rightarrow R$ defined as $a \,^*\, b = |a-b|$ and $a \,o\, b = a$,$\forall \, a, b \in R$. Show that $^*$ is commutative but not associative,and $o$ is associative but not commutative. Further,show that $\forall \, a, b, c \in R, a \,^*\, (b \,o\, c) = (a \,^*\, b) \,o\, (a \,^*\, c)$. [If it is so,we say that the operation $^*$ distributes over the operation $o$]. Does $o$ distribute over $^*$? Justify your answer.

Is $^*$ defined on the set $\{1, 2, 3, 4, 5\}$ by $a \,^*\, b = \text{L.C.M. of } a \text{ and } b$ a binary operation? Justify your answer.

Let $*$ be a binary operation defined on the set of rational numbers $Q$. Determine whether the binary operation defined by $a * b = a + ab$ for all $a, b \in Q$ is commutative.