Question

Check the commutativity and associativity of each of the following binary operations:

(i) ‘*’ on Z defined by a * b = a + b + ab for all a, b ∈ Z 

(ii) ‘*’ on N defined by a * b = 2^ab for all a, b ∈ N

(iii) ‘*’ on Q defined by a * b = a – b for all a, b ∈ Q

(iv) ‘⊙’ on Q defined by a ⊙ b = a² + b² for all a, b ∈ Q

(v) ‘o’ on Q defined by a o b = (ab/2) for all a, b ∈ Q

Answer 1

(i) First Let us check commutativity of *

Let a, b ∈ Z

Then a * b = a + b + ab

= b + a + ba

= b * a

So,

a * b = b * a, ∀ a, b ∈ Z

Let us prove associativity of *

Let a, b, c ∈ Z, Then,

a * (b * c) = a * (b + c + bc)

= a + (b + c + bc) + a (b + c + bc)

= a + b + c + bc + ab + ac + abc

(a * b) * c = (a + b + ab) * c

= a + b + a b + c + (a + b + a b) c

= a + b + a b + c + a c + b c + a b c

So,

a * (b * c) = (a * b) * c, ∀ a, b, c ∈ Z

Thus, * is associative on Z.

(ii) Let us check commutativity of *

Let a, b ∈ N

a * b = 2^ab

= 2^ba

= b * a

So, a * b = b * a, ∀ a, b ∈ N

Thus, * is commutative on N

Let us check associativity of *

Let a, b, c ∈ N

Then, a * (b * c) = a * (2^bc)

=2a ∗ 2bc

(a * b) * c = (2^ab) * c

=2ab ∗ 2c

So, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on N

(iii) Let us check commutativity of *

Let a, b ∈ Q, then

a * b = a – b

b * a = b – a

So, a * b ≠ b * a

Thus, * is not commutative on Q

Let us check associativity of *

Let a, b, c ∈ Q, then

a * (b * c) = a * (b – c)

= a – (b – c)

= a – b + c

(a * b) * c = (a – b) * c

= a – b – c

Therefore,

a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Q

(iv) Let us check commutativity of ⊙

Let a, b ∈ Q, then

a ⊙ b = a² + b²

= b² + a²

= b ⊙ a

So, a ⊙ b = b ⊙ a, ∀ a, b ∈ Q

Thus, ⊙ on Q

Let us check associativity of ⊙

Let a, b, c ∈ Q, then

a ⊙ (b ⊙ c) = a ⊙ (b² + c²)

= a² + (b² + c²)²

= a² + b⁴ + c⁴ + 2b²c²

(a ⊙ b) ⊙ c = (a² + b²) ⊙ c

= (a² + b²)² + c²

= a⁴ + b⁴ + 2a²b² + c²

So,

(a ⊙ b) ⊙ c ≠ a ⊙ (b ⊙ c)

Thus, ⊙ is not associative on Q.

(v) Let us check commutativity of o

Let a, b ∈ Q, then

a o b = (ab/2)

= (ba/2)

= b o a

So, a o b = b o a, ∀ a, b ∈ Q

Thus, o is commutative on Q

Let us check associativity of o

Let a, b, c ∈ Q, then

a o (b o c) = a o (b c/2)

= [a (bc/2)]/2

= [a (bc/2)]/2

= (a b c)/4

(a o b) o c = (ab/2) o c

= [(ab/2)c]/2

= (a b c)/4

So, a o (b o c) = (a o b) o c, ∀ a, b, c ∈ Q

Hence, o is associative on Q.