shows that gof = I N and fog = I Y , which implies that f is invertible and g is the inverse of f . Miscellaneous Examples Example If R and R are equivalence relations in a set A, show that R ∩ R is also an equivalence relation. Solution Since R and R are equivalence relations, ( a , a ) ∈ R , and ( a , a ) ∈ R ∀ a ∈ A. This implies that ( a , a ) ∈ R ∩ R , ∀ a , showing R ∩ R is reflexive.
Further, ( a , b ) ∈ R ∩ R ⇒ ( a , b ) ∈ R and ( a , b ) ∈ R ⇒ ( b , a ) ∈ R and ( b , a ) ∈ R ⇒ ( b , a ) ∈ R ∩ R , hence, R ∩ R is symmetric. Similarly, ( a , b ) ∈ R ∩ R and ( b , c ) ∈ R ∩ R ⇒ ( a , c ) ∈ R and ( a , c ) ∈ R ⇒ ( a , c ) ∈ R ∩ R . This shows that R ∩ R is transitive. Thus, R ∩ R is an equivalence relation.
Example Let R be a relation on the set A of ordered pairs of positive integers defined by ( x , y ) R ( u , v ) if and only if xv = yu . Show that R is an equivalence relation. Solution Clearly, ( x , y ) R ( x , y ), ∀ ( x , y ) ∈ A, since xy = yx . This shows that R is reflexive.
Further, ( x , y ) R ( u , v ) ⇒ xv = yu