Let $A=\{1,2,3,4\}$ and $R$ be a relation on the set $A \times A$ defined by $R=\{((a, b),(c, d)): 2 a+3 b=4 c+5 d\}$. Then the number of elements in $R$ is:
$6$
$5$
$4$
$3$
Among the relations $S =\left\{( a , b ): a , b \in R -\{0\}, 2+\frac{ a }{ b } > 0\right\}$ And $T =\left\{( a , b ): a , b \in R , a ^2- b ^2 \in Z \right\}$,
In the set $A = \{1, 2, 3, 4, 5\}$, a relation $R$ is defined by $R = \{(x, y)| x, y$ $ \in A$ and $x < y\}$. Then $R$ is
Let $A=\{1,2,3\} .$ Then number of relations containing $(1,2)$ and $(1,3)$ which are reflexive and symmetric but not transitive is
Show that each of the relation $R$ in the set $A =\{x \in Z : 0 \leq x \leq 12\},$ given by $R =\{(a, b):|a-b| $ is a multiple of $4\}$
Consider set $A = \{1,2,3\}$ . Number of symmetric relations that can be defined on $A$ containing the ordered pair $(1,2)$ & $(2,1)$ is