Question
Let $$R = \left\{ {\left( {x,\,y} \right):x,\,y\, \in \,N{\text{ and }}{x^2} - 4xy + 3{y^2} = 0} \right\},$$ where $$N$$ is the set of all natural numbers. Then the relation $$R$$ is :
A.
reflexive but neither symmetric nor transitive
B.
symmetric and transitive
C.
reflexive and symmetric
D.
reflexive and transitive
Answer :
reflexive and transitive
Solution :
$$\eqalign{
& R = \left\{ {\left( {x,\,y} \right):x,\,y\, \in \,N{\text{ and }}{x^2} - 4xy + 3{y^2} = 0} \right\}, \cr
& {\text{Now, }}{x^2} - 4xy + 3{y^2} = 0\,\,\, \Rightarrow \left( {x - y} \right)\left( {x - 3y} \right) = 0 \cr
& \therefore \,x = y{\text{ or }}x = 3y \cr
& \therefore \,R = \left\{ {\left( {1,\,1} \right),\,\left( {3,\,1} \right),\,\left( {2,\,2} \right),\,\left( {6,\,2} \right),\,\left( {3,\,3} \right),\,\left( {9,\,3} \right),.....} \right\} \cr} $$
Since $$\left( {1,\,1} \right),\,\left( {2,\,2} \right),\,\left( {3,\,3} \right),.....$$ are present in the relation, therefore $$R$$ is reflexive.
Since $$\left( {3,\,1} \right)$$ is an element of $$R$$ but $$\left( {1,\,3} \right)$$ is not the element of $$R$$ is not symmetric.
$$\eqalign{
& {\text{Here }}\left( {3,\,1} \right)\, \in \,R{\text{ and }}\left( {1,\,1} \right)\, \in \,R\, \Rightarrow \left( {3,\,1} \right)\, \in \,R \cr
& \left( {6,\,2} \right)\, \in \,R{\text{ and }}\left( {2,\,2} \right)\, \in \,R\, \Rightarrow \left( {6,\,2} \right)\, \in \,R \cr
& {\text{For all such }}\left( {a,\,b} \right)\, \in \,R{\text{ and }}\left( {b,\,c} \right)\, \in \,R \Rightarrow \left( {a,\,c} \right)\, \in \,R \cr
& {\text{Hence }}R{\text{ is transitive}}{\text{.}} \cr} $$