let * be a binary operation on R defined by a * b = a 4 + b 7 , a , b ∈ R .Show that * is not commutative and associative.

Given a * b = a 4 + b 7 Case - 1 Checking for commutative a * b = a 4 + b 7 .....equation 1 and b * a = b 4 + a 7 ......equation 2 From equation 1 and 2 we can conclude that they are not commutative Case 2 Checking for Associative a * b * c = a 4 + b 7 * c = a 4 + b 7 4 + c 7 .......equation 3 a * b * c = a 4 + b * c 7 = a 4 + b 4 + c 7 7 .........equation 4 From equation 3 and 4 we conclude that they are not associative .

let * be a binary operation on R defined by a * b = a 4 + b 7 , a , b ∈ R .Show that * is not commutative and associative.

Given a * b = a 4 + b 7 Case - 1 Checking for commutative a * b = a 4 + b 7 .....equation 1 and b * a = b 4 + a 7 ......equation 2 From equation 1 and 2 we can conclude that they are not commutative Case 2 Checking for Associative a * b * c = a 4 + b 7 * c = a 4 + b 7 4 + c 7 .......equation 3 a * b * c = a 4 + b * c 7 = a 4 + b 4 + c 7 7 .........equation 4 From equation 3 and 4 we conclude that they are not associative .

Let N denote the set of all natural numbers. Define two binary relations on N as R 1 = x , y ∈ N × N : 2 x + y = 10 and R 2 = x , y ∈ N × N : x + 2 y = 10 . Then

both R 1 and R 2 are transitive relations

both R 1 and R 2 are symmetric relations

Show that : R × R → R defined by a * b = a + 2 b is neither commutative nor associative.

let a = 1 , b = 0 ∈ R a * b = 1 + 2 0 = 1 And b * a = b + 2 a = 0 + 2 = 2 Here, a * b ≠ b * a Hence * is not commutative. Let c = 3 . a * b * c = 1 * c = 1 + 2 3 = 6 a * b * c = a * 2 = 1 * 2 = 1 + 2 2 = 5 Here, a * b * c ≠ a * b * c Hence * is not associative.

Let * be a binary operation on Q , defined by a * b = a b 2 ; a , b ∈ Q . Determine whether * is associative or not.

Given that, * is a binary operation on Q , defined by a * b = a b 2 ; a , b ∈ Q A binary operation * on a set A is associative if a * b * c = a * b * c , for all a , b , c ∈ A [ A is Non-empty set] Now, a * b * c = a b 2 * c = a b 2 × c 2 = a b c 4 Similarly, a * b * c = a * b c 2 = a × b c 2 2 = a b c 4 Since, ( a * b ) * c = a * ( b * c ) , ∀ a , b , c ∈ Q Hence, * is an associative binary operation.