Question

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

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

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

(iii) ‘*’ on Q defined by a * b = (ab/4) for all a, b ∈ Q

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

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

Answer 1

(i) Let us check the commutativity of *

Let a, b ∈ N, then

a * b = a^b

b * a = b^a

So, a * b ≠ b * a

Thus, * is not commutative on N.

Let us check the associativity of *

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

= 

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

= (a^b)^c

= a^bc

So, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on N

(ii) Let us check the commutativity of *

Let a, b ∈ Z, then

a * b = a – b

b * a = b – a

So,

a * b ≠ b * a

Thus, * is not commutative on Z.

Let us check the associativity of *

Let a, b, c ∈ Z, then

a * (b * c) = a * (b – c)

= a – (b – c)

= a – (b + c)

(a * b) * c = (a – b) – c

= a – b – c

So, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Z

(iii) Let us check the commutativity of *

Let a, b ∈ Q, then

a * b = (ab/4)

= (ba/4)

= b * a

So, a * b = b * a, for all a, b ∈ Q

Thus, * is commutative on Q

Let us check the associativity of *

Let a, b, c ∈ Q, then

a * (b * c) = a * (b c/4)

= [a (b c/4)]/4

= (a b c/16)

(a * b) * c = (ab/4) * c

= [(ab/4) c]/4

= a b c/16

So,

a * (b * c) = (a * b) * c for all a, b, c ∈ Q

Thus, * is associative on Q.

(iv) Let us check the commutativity of *

Let a, b ∈ Z, then

a * b = a + b – ab

= b + a – ba

= b * a

Therefore, a * b = b * a, for all a, b ∈ Z

Thus, * is commutative on Z.

Let us check the associativity of *

Let a, b, c ∈ Z

a * (b * c) = a * (b + c – b c)

= a + b + c- b c – ab – ac + a b c

(a * b) * c = (a + b – a b) c

= a + b – ab + c – (a + b – ab) c

= a + b + c – ab – ac – bc + a b c

So,

a * (b * c) = (a * b) * c, for all a, b, c ∈ Z

Thus, * is associative on Z.

(v) Let us check the commutativity of *

Let a, b ∈ N, then

a * b = gcd (a, b)

= gcd (b, a)

= b * a

So, a * b = b * a, for all a, b ∈ N

Thus, * is commutative on N.

Let us check the associativity of *

Let a, b, c ∈ N

a * (b * c) = a * [gcd (a, b)]

= gcd (a, b, c)

(a * b) * c = [gcd (a, b)] * c

= gcd (a, b, c)

So,

a * (b * c) = (a * b) * c, for all a, b, c ∈ N

Hence, * is associative on N.