If A = { ( 0 , 1 , 2 , … , 9 } and the relation R on A is defined by R = { ( x , y ) : x , y ∈ A , y = 2 x + 1 } , then determine whether the relation R is reflexive.

Given that, A = { 0 , 1 , 2 , … , 9 } and relation R = { ( x , y ) ; x , y ∈ A , y = 2 x + 1 } . We know that, The reflexive property states that any real number a , is equal to itself. That is, a = a . Reflexive: Here, 1 ≠ 2 × 1 + 1 = 3 ∴ ( 1 , 1 ) ∉ R ∴ R is not reflexive on A .

If A = { ( 0 , 1 , 2 , … , 9 } and the relation R on A is defined by R = { ( x , y ) : x , y ∈ A , y = 2 x + 1 } , then determine whether the relation R is symmetric.

Given that, A = { 0 , 1 , 2 , … , 9 } and relation R = { ( x , y ) ; x , y ∈ A , y = 2 x + 1 } . We know that, the symmetric property states that for any real numbers, a and b , if a = b then b = a Symmetric: Here, ( 1 , 3 ) ∈ R Since, 3 = 2 · 1 + 1 But, ( 3 - 1 ) ∉ R ∵ 1 ≠ 2 × 3 + 1 = 7 ∴ R is not symmetric.

If A = { ( 0 , 1 , 2 , … , 9 } and the relation R on A is defined by R = { ( x , y ) : x , y ∈ A , y = 2 x + 1 } , then determine whether the relation R is transitive.

Given that, A = { 0 , 1 , 2 , … , 9 } and relation R = { ( x , y ) ; x , y ∈ A , y = 2 x + 1 } . Transitive: Here, ( 1 , 3 ) ∈ R ∵ 3 = 2 · 1 + 1 = 3 And ( 3 , 7 ) ∈ R ∵ 7 = 2 × 3 + 1 = 7 But ( 1 , 7 ) ∉ R ∵ 7 ≠ 2 × 1 + 1 = 3 ∴ For any 1 , 3 , 7 ∈ A , ( 1 , 3 ) , ( 3 , 7 ) ∈ R But ( 1 , 7 ) ∉ R Thus R is not transitive on A .

Is the relation R on the set R of real numbers defined by R = { ( a , b ) : a , b ∈ R , 1 + a b ≥ 0 } transitive? Justify your answer.

Given relation R on the set of real numbers is given by R = { ( a , b ) : a , b ∈ R , 1 + a b ≥ 0 } Now, ( 1 , - 1 ) ∈ R ∵ 1 + 1 × ( - 1 ) ≥ 0 ⇒ 0 ≥ 0 which is true. And ( - 1 , - 2 ) ∈ R ∵ 1 + ( - 1 ) ( - 2 ) = 1 + 2 = 3 ≥ 0 But ( 1 , - 2 ) ∉ R ∵ 1 + 1 × ( - 2 ) = 1 - 2 = - 1 < 0 ∴ R is not transitive on set of real numbers.

Is the relation R on the set Q of rational numbers are defined by R = ( x , y ) : x , y ∈ Q , x < y 2 , symmetric? Justify your answer.

Given that, relation R = ( x , y ) ; x , y ∈ Q , x < y 2 We know that, the symmetric property states that for any real numbers, a and b , if a = b then b = a . ∀ x , y ∈ Q , ( x , y ) ∈ R ⇒ x < y 2 y 2 < x ( y , x ) ∉ R For example: ( 1 , 2 ) ∈ R , ∵ 1 < 2 2 = 4 But, ( 2 , 1 ) ∈ R ∵ 2 2 = 4 > 1 , 4 > 1 Thus R is not symmetric on Q .