R ⇒ T is congruent to T ⇒ T is congruent to T ⇒ (T , T ) ∈ R. Hence, R is symmetric. Moreover, (T , T ), (T , T ) ∈ R ⇒ T is congruent to T and T is congruent to T ⇒ T is congruent to T ⇒ (T , T ) ∈ R. Therefore, R is an equivalence relation.
Example Let L be the set of all lines in a plane and R be the relation in L defined as R = {(L , L ) : L is perpendicular to L }. Show that R is symmetric but neither reflexive nor transitive. Solution R is not reflexive, as a line L can not be perpendicular to itself, i.e., (L , L ) ∉ R. R is symmetric as (L , L ) ∈ R ⇒ L is perpendicular to L ⇒ L is perpendicular to L ⇒ (L , L ) ∈ R.
R is not transitive. Indeed, if L is perpendicular to L and L is perpendicular to L , then L can never be perpendicular to L . In fact, L is parallel to L , i.e., (L , L ) ∈ R, (L , L ) ∈ R but (L , L ) ∉ R. Example Show that the relation R in the set { , , } given by R = {( , ), ( , ), ( , ), ( , ), ( , )} is reflexive but neither symmetric nor transitive.
Solution R is reflexive, since ( , ), ( , ) and ( , ) lie in R. Also, R is not symmetric, as ( , ) ∈ R but ( , ) ∉ R. Similarly, R is not transitive, as ( , ) ∈ R and ( , ) ∈ R but ( , ) ∉ R. Example