Let $*$ be a binary operation defined on $R$ by $a * b = \frac{a+b}{4}$ for all $a, b \in R$. Then the operation $*$ is:

On the set $Z$ of all integers,the operation $*$ is defined by $a * b = a + b - 5$. If $2 * (x * 3) = 5$,then $x$ is equal to:

Show that the $\vee: R \times R \rightarrow R$ given by $(a, b) \rightarrow \max\{a, b\}$ and the $\wedge: R \times R \rightarrow R$ given by $(a, b) \rightarrow \min\{a, b\}$ are binary operations.

Show that $-a$ is not the inverse of $a \in N$ for the addition operation $+$ on $N$ and $\frac{1}{a}$ is not the inverse of $a \in N$ for the multiplication operation $\times$ on $N$,for $a \neq 1$.

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.

$A$ group $(G, *)$ has $10$ elements. The minimum number of elements of $G$,which are their own inverses is