For the relation R 1 defined on R by the rule a , b ∈ R 1 ⇔ 1 + a b > 0 . Prove that: a , b ∈ R 1 and b , c ∈ R 1 ⇒ a , c ∈ R 1 is not true for all a , b , c ∈ R .

Given, for the relation R 1 defined on R by the rule a , b ∈ R 1 ⇔ 1 + a b > 0 and a , b ∈ R Let 2 , - 1 4 ∈ R 1 ⇒ 1 + 2 × - 1 4 = 1 2 > 0 - 1 4 , - 3 ∈ R 1 ⇒ 1 + - 1 4 - 3 > 0 But 2 , - 3 ∉ R 1 ∵ 1 + 2 - 3 < 0 Hence proved.

For the relation R 1 defined on R by the rule a , b ∈ R 1 ⇔ 1 + a b > 0 . Prove that: a , b ∈ R 1 and b , c ∈ R 1 ⇒ a , c ∈ R 1 is not true for all a , b , c ∈ R .

Given, for the relation R 1 defined on R by the rule a , b ∈ R 1 ⇔ 1 + a b > 0 and a , b ∈ R Let 2 , - 1 4 ∈ R 1 ⇒ 1 + 2 × - 1 4 = 1 2 > 0 - 1 4 , - 3 ∈ R 1 ⇒ 1 + - 1 4 - 3 > 0 But 2 , - 3 ∉ R 1 ∵ 1 + 2 - 3 < 0 Hence proved.

Let R be a relation on N × N defined by a , b R c , d ⇔ a + d = b + c for all a , b , c , d ∈ N × N Show that: a , b R a , b for all a , b ∈ N × N

Given R be a relation on N × N defined by a , b R c , d ⇔ a + d = b + c for all a , b , c , d ∈ N × N Consider a + b = a + b for all a , b ∈ N is true ∴ a , b R a , b for all a , b ∈ N × N

Let R be a relation on N × N defined by a , b R c , d ⇔ a + d = b + c for all a , b , c , d ∈ N × N Show that: a , b R c , d ⇒ c , d R a , b for all a , b , c , d ∈ N × N

Given a , b R c , d ⇒ a + d = b + c for all ( a , b ) , ( c , d ) ∈ N × N ⇒ c + b = d + a ⇒ c , d R a , b for all ( c , d ) , ( a , b ) ∈ N × N Hence, proved

Let R be a relation on N × N defined by a , b R c , d ⇔ a + d = b + c for all a , b , c , d ∈ N × N Show that: ( a , b ) R ( c , d ) and c , d R e , f ⇒ ( a , b ) R ( e , f ) for all a , b , c , d , e , f ∈ N × N

a , b R c , d ⇒ a + d = b + c … . (i) c , d R e , f ⇒ c + f = d + e … . (ii) Adding (i) and (ii), a + d + c + f = b + c + d + e ⇒ a + f = b + e ⇒ a , b R e , f ⇒ a , b R e , f ∀ a , b , c , d , e , f ∈ N × N Hence proved

If A = { 1 , 2 , 4 } , B = { 2 , 4 , 5 } and C = { 2 , 5 } write ( A - C ) × ( B - C )

Given, A = { 1 , 2 , 4 } , B = { 2 , 4 , 5 } and C = { 2 , 5 } From the definition of set difference, we know that A - B = x | x ∈ A ∧ x ∉ B Therefore, A - C = { 1 , 4 } and B - C = { 4 } From the definition of cross product of two sets, we know that A × B = x , y | x ∈ A , y ∈ B Thus, we have, ( A - C ) × ( B - C ) = { ( 1 , 4 ) , ( 4 , 4 ) }

If R is a relation defined on the set Z of integers by the rule ( x , y ) ∈ R ⇔ x 2 + y 2 = 9 then write domain of R .

Here, Relation R is defined on the set Z . And by the given definition of relation, R = x , y : x , y ∈ Z , x 2 + y 2 = 9 ⇒ R = - 3 , 0 , 0 , - 3 , 3 , 0 , 0 , 3 Now we know, Domain is the set which consist all first elements of ordered pairs in relation R . So, Domain R = - 3 , 0 , 3