On the set Z of integers a binary operation * is defined by a * b = a b + 1 for all a , b ∈ Z . Prove that * is not associative on Z .

Given binary operation on Z is a * b = a b + 1 for all a , b ∈ Z . * is associative if a * b * c = a * b * c Let a , b , c ∈ Z a * ( b * c ) = a * ( b c + 1 ) = a ( b c + 1 ) + 1 = a b c + a + 1 ( a * b ) * c = ( a b + 1 ) * c = ( a b + 1 ) c + 1 = a b c + c + 1 Thus, a * ( b * c ) ≠ ( a * b ) * c Thus, * is not associative on Z .

On the set Z of integers a binary operation * is defined by a * b = a b + 1 for all a , b ∈ Z . Prove that * is not associative on Z .

Given binary operation on Z is a * b = a b + 1 for all a , b ∈ Z . * is associative if a * b * c = a * b * c Let a , b , c ∈ Z a * ( b * c ) = a * ( b c + 1 ) = a ( b c + 1 ) + 1 = a b c + a + 1 ( a * b ) * c = ( a b + 1 ) * c = ( a b + 1 ) c + 1 = a b c + c + 1 Thus, a * ( b * c ) ≠ ( a * b ) * c Thus, * is not associative on Z .

Let S be the set of all real numbers except − 1 and let ' * ' be an operation defined by a * b = a + b + a b for all a , b ∈ S . Determine whether ' * ' is a binary operation on S . If yes, check its commutativity and associativity. Also, solve the equation ( 2 * x ) * 3 = 7 .

Given a * b = a + b + a b for all a , b ∈ S . S be the set of all real numbers except - 1 . Checking for binary operation: Let a , b ∈ S . Then a , b ∈ R and a ≠ - 1 , b ≠ - 1 a * b = a + b + a b We need to prove that a + b + a b ∈ S . For this we have to prove that a + b + a b ∈ R and a + b + a b ≠ - 1 Since a , b ∈ R , a + b + a b ∈ R , let us assume that a + b + a b = - 1 a + b + a b + 1 = 0 ⇒ a + a b + b + 1 = 0 ⇒ a 1 + b + 1 1 + b = 0 ⇒ a + 1 b + 1 = 0 a = - 1 , b = - 1 , which is not possible. a + b + a b ≠ - 1 Therefore, a + b + a b ∈ S Thus, * is binary operation on S . * is commutative if a * b = b * a Commutativity: Let a , b ∈ S . Then, a * b = a + b + a b = b + a + b a = b * a Therefore, a * b = b * a , ∀ a , b ∈ S Thus, * is commutative on S . * is associative if a * b * c = a * b * c Associativity: Let a , b , c ∈ S a * b * c = a * b + c + b c = a + b + c + b c + a b + c + b c = a + b + c + b c + a b + a c + a b c a * b * c = a + b + a b * c = a + b + a b + c + a + b + a b c = a + b + a b + c + a c + b c + a b c Therefore, a * b * c = a * b * c , ∀ a , b , c ∈ S Thus, * is associative on S . Now, Given: 2 * x * 3 = 7 ⇒ 2 + x + 2 x * 3 = 7 ⇒ 2 + 3 x * 3 = 7 ⇒ 2 + 3 x + 3 + 2 + 3 x 3 = 7 ⇒ 5 + 3 x + 6 + 9 x = 7 ⇒ 12 x + 11 = 7 ⇒ 12 x = - 4 ⇒ x = - 4 12 ⇒ x = - 1 3

On Q , the set of all rational numbers, * is defined by a * b = a - b 2 , shown that * is no associative.

Given binary operation on Q is a * b = a - b 2 , for all a , b ∈ Q . * is associative if a * b * c = a * b * c Let a , b , c ∈ Q . Then, a * b * c = a * b - c 2 = a - b - c 2 2 ⇒ a * b * c = 2 a - b + c 4 a * b * c = a - b 2 * c = a - b 2 - c 2 ⇒ a * b * c = a - b - 2 c 4 Thus, a * b * c ≠ a * b * c If a = 1 , b = 2 , c = 3 1 * 2 * 3 = 1 * 2 - 3 2 = 1 * - 1 2 = 1 + 1 2 2 = 3 4 1 * 2 * 3 = 1 - 2 2 * 3 = - 1 2 * 3 = - 1 2 - 3 2 = - 7 4 Therefore, a = 1 , b = 2 , c = 3 ∈ R such that a * b * c ≠ a * b * c Thus, * is not associative on Q .

On the set Q of all rational numbers if a binary operation * is defined by a * b = a b 5 , prove that * is associative on Q .

Given binary operation on Q is a * b = a b 5 , for all a , b ∈ Q * is associative if a * b * c = a * b * c Let a , b , c ∈ Q . Then, a * b * c = a * b c 5 = a b c 5 5 = a b c 25 a * b * c = a b 5 * c = a b 5 c 5 = a b c 25 Therefore, a * b * c = a * b * c , ∀ a , b , c ∈ Q . Thus, * is associative on Q .

The binary operation * is defined by a * b = a b 7 on the set Q of all rational numbers. Show that * is associative.

Given binary operation on Q is a * b = a b 7 , for all a , b ∈ Q * is associative if a * b * c = a * b * c Let a , b , c ∈ Q . Then, a * b * c = a * b c 7 = a b c 7 7 = a b c 49 a * b * c = a b 7 * c = a b 7 c 7 = a b c 49 Therefore, a * b * c = a * b * c , ∀ a , b , c ∈ Q . Thus, * is associative on Q .

On Q , the set of all rational numbers a binary operation * is defined by a * b = a + b 2 . Show that * is not associative on Q .

Given binary operation on Q is a * b = a + b 2 for all a , b ∈ Q . * is associative if a * b * c = a * b * c Let a , b , c in Q . Then a * b * c = a * b + c 2 = a + b + c 2 2 ⇒ a * b * c = 2 a + b + c 4 a * b * c = a + b 2 * c = a + b 2 + c 2 = a + b + 2 c 4 Thus, a * b * c ≠ a * b * c Thus, * is not associative on Q .

Let S be the set of all rational numbers except 1 and * be defined on S by a * b = a + b - a b , for all a , b ∈ S . Prove that * is commutative as well as associative.

We have, S = Q - { 1 } and * is defined on S as a * b = a + b - a b , for all a , b ∈ S * is commutative if a * b = b * a Commutativity : Let a , b ∈ S . Then, a * b = a + b - a b = b + a - b a = b * a Therefore a * b = b * a , ∀ a , b ∈ S Thus, * is commutative on S . * is associative if a * b * c = a * b * c Associativity : Let a , b , c ∈ S . Then, a * b * c = a * b + c - b c = a + b + c - b c - a b + c - b c = a + b + c - b c - a b - a c + a b c a * b * c = a + b - a b * c = a + b - a b + c - a + b - a b c = a + b + c - a b - a c - b c + a b c Therefore, a * b * c = a * b * c , ∀ a , b , c ∈ S Thus , * is associative on S . So, * is commutative as well as associative.