Let $R$ be a relation from $Q$ to $Q$ defined by $R = \{(a, b) : a, b \in Q \text{ and } a - b \in Z \}$. Show that $(a, b) \in R$ and $(b, c) \in R$ implies that $(a, c) \in R$.

Give an example of a relation on a set $A = \{4, 6, 8\}$ which is reflexive and symmetric but not transitive.

If $n(A) = m$,then the total number of reflexive relations that can be defined on $A$ is-

Let $R = \{(a, a)\}$ be a relation on a set $A$. Then $R$ is

Let $A = \{2, 3, 4\}$ and $B = \{8, 9, 12\}$. Then the number of elements in the relation $R = \{((a_1, b_1), (a_2, b_2)) \in (A \times B) \times (A \times B) : a_1 \text{ divides } b_2 \text{ and } a_2 \text{ divides } b_1\}$ is:

The relation $R$ defined on a set $A$ is antisymmetric if $(a, b) \in R$ and $(b, a) \in R$ implies $a = b$ for all $a, b \in A$. Based on this definition,the relation $R$ is antisymmetric if $(a, b) \in R$ and $(b, a) \in R$ implies $a = b$,which is equivalent to saying that if $a \neq b$,then it is not possible for both $(a, b) \in R$ and $(b, a) \in R$ to be true. Therefore,the condition is that for $a \neq b$,we cannot have both $(a, b) \in R$ and $(b, a) \in R$.