Let N be the set of natural numbers and R be the relation on N × N defined by ( a , b ) R ( c , d ) iff a d = b c for all a , b , c , d ∈ N . Show that R is an equivalence relation.

Given that, N is the set of natural numbers and R be the relation on N × N defined by ( a , b ) R ( c , d ) iff a d = b c for all a , b , c , d ∈ N We know that, a relation R that is reflexive, symmetric and transitive is called an equivalence relation. Reflexive: For any ( a , b ) ∈ N × N a · b = b · a ∴ ( a , b ) R ( a , b ) Thus, R is reflexive Symmetric: For any ( a , b ) , ( c , d ) ∈ N × N ( a , b ) R ( c , d ) ⇒ a · d = b · c ⇒ c · b = d · a ⇒ ( c , d ) R ( a , b ) Thus, R is symmetric Transitive: For any ( a , b ) , ( c , d ) , ( e , f ) , ∈ N × N ( a , b ) R ( c , d ) and ( c , d ) R ( e , f ) ⇒ a · d = b · c and c · f = d · e ⇒ a · d · c · f = b · c · d · e ⇒ a · f = b · e ∴ ( a , b ) R ( e , f ) Thus, R is transitive Therefore, R is an equivalence Relation.

Check whether the relation R defined in the set 1 , 2 , 3 , 4 , 5 , 6 as R = a , b : b = a + 1 is reflexive, symmetric or transitive.

Suppose A = 1 , 2 , 3 , 4 , 5 , 6 . A relation R is defined on set A as: R = { ( a , b ) : b = a + 1 } ∴ R = { ( 1, 2 ) , 2 , 3 , 3 , 4 , 4 , 5 , 5 , 6 } Reflexive Property : A relation R in a set A is said to be reflexive if x , x ∈ R , for every x ∈ A Now, 1 , 1 , 2 , 2 , 3 , 3 , 4 , 4 , 5 , 5 , 6 , 6 ∉ R ∴ R is not reflexive. Symmetric Property : A relation R in a set A is said to be symmetric if x , y ∈ R implies that y , x ∈ R for all x , y ∈ A . Now it can be observed that 1 , 2 ∈ R , but 2 , 1 ∉ R . Hence , R is not symmetric. Transitive Property : A relation R in a set A is said to be transitive if x , y ∈ R and y , z ∈ R implies that x , z ∈ R for all x , y , z ∈ A . Now, 1 , 2 , 2 , 3 ∈ R but, ( 1 , 3 ) ∉ R ∴ R is not transitive. Hence, R is neither reflexive, nor symmetric, nor transitive.

Show that the relation R on R is defined as R = { ( a , b ) : a ≤ b } , is reflexive and transitive but not symmetric.

Given, R = a , b : a ≤ b , a , b ∈ R For the relation R to be reflexive: If the relation R is reflexive, then a , a ∈ R i.e. a ≤ a , which is always true for all the values of a , a ∈ R So, the given relation R is reflexive. For the relation R to be symmetric: If a , b ∈ R then b , a ∈ R i.e. if a ≤ b then b ≤ a , which is not possible for all the values of a , b ∈ R . For example: if a = 2 and b = 5 Here, a < b but b > a So, the given relation R is not symmetric. For the relation R to be transitive: If a , b ∈ R and b , c ∈ R then a , c ∈ R i.e.,if a ≤ b and b ≤ c then a ≤ c , which is true for all the values of a , b , c ∈ R . For example: if a = 2 , b = 3 and c = 4 Here, a < b < c So, the given relation is transitive. Hence, the given relation R is reflexive and transitive but not symmetric.