Browse Questions

# Let $A = \{1, 2, 3\}.$ Then number of relations containing $(1, 2)\;$ and $\;(1, 3)$ which are reflexive and symmetric but not transitive is

$$(A) \quad 1\qquad(B) \quad 2 \qquad(C) \quad 3 \qquad(D)\quad 4\qquad$$

Toolbox:
• A relation R in a set A is called reflexive. if $(a,a) \in R\;for\; all\; a\in A$
• A relation R in a set A is called symmetric. if $(a_1,a_2) \in R\;\Rightarrow \; (a_2,a_1)\in R \;$ for $\;a_1,a_2 \in A$
• A relation R in a set A is called transitive. if $(a_1,a_2) \in\; R$ and $(a_2,a_3)\in R \Rightarrow \;(a_1,a_3)\in R\;$for all $\; a_1,a_2,a_3 \in A$
Consider the relation $R$ in $A=\{1,2,3\}\;$ where $R=\{(1,1),(2,2),(3,3)(1,2)(1,3)(2,1)(3,1)(3,2)(2,3)\}$
We need to work with the relations that contains $(1,2), (3,1)$
Relation R is reflexive since $(1,1)(2,2)(3,3) \in R$
Relation R is symmetric since $(1,2),(2,1) \in R (1,3)(3,1) \in R$
Relation R is not transitive since $(3,1)(1,2) \in R\;but \;(3,2) \not \in R$
Therefore the total number of relation containing (1,2)(1,3) which are reflexive ,symmetric but not transitive in 1
However if we add the pair (3,2) and (2,3) to relation R then it will become transitive.
Therefore, the correct answer is 1 (A).
edited Mar 20, 2013