1.

Check the commutativity and associativity of each of the following binary operations:(i) ‘*’ on Q defined by a * b = ab2 for all a, b ∈ Q(ii) ‘*’ on Q defined by a * b = a + ab for all a, b ∈ Q(iii) ‘*’ on R defined by a * b = a + b - 7 for all a, b ∈ R(iv) ‘*’ on Q defined by a * b = (a – b)2 for all a, b ∈ Q(v) ‘*’ on Q defined by a * b = ab + 1 for all a, b ∈ Q

Answer»

(i) Let us check commutativity of *

Let a, b ∈ Q, then

a * b = ab2

b * a = ba2

So,

a * b ≠ b * a

Thus, * is not commutative on Q

Let us check the associativity of *

Let a, b, c ∈ Q, then

a * (b * c) = a * (bc2)

= a(bc2)2

= ab2c4

(a * b) * c = (ab2) * c

= ab2c2

So, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Q.

(ii) We have to check commutativity of *

Let a, b ∈ Q, then

a * b = a + ab

b * a = b + ba

= b + ab

So, a * b ≠ b * a

Thus, * is not commutative on Q.

Let us prove associativity on Q.

Let a, b, c ∈ Q, then

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

= a + a (b + b c)

= a + ab + a b c

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

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

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

So, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Q.

(iii) Let us check the commutativity of *

Let a, b ∈ R, then

a * b = a + b – 7

= b + a – 7

= b * a

Therefore,

a * b = b * a, for all a, b ∈ R

Thus, * is commutative on R

Let us prove associativity of * on R.

Let a, b, c ∈ R, then

a * (b * c) = a * (b + c – 7)

= a + b + c - 7 - 7

= a + b + c – 14

(a * b) * c = (a + b – 7) * c

= a + b – 7 + c – 7

= a + b + c – 14

So,

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

Thus, * is associative on R.

(iv) Let us check commutativity of *

Let a, b ∈ Q, then

a * b = (a – b)2

= (b – a)2

= b * a

Therefore,

a * b = b * a, for all a, b ∈ Q

Thus, * is commutative on Q

Let us prove the associativity of * on Q

Let a, b, c ∈ Q, then

a * (b * c) = a * (b – c)2

= a * (b2 + c2 – 2 b c)

= (a – b2 – c2 + 2bc)2

(a * b) * c = (a – b)2 * c

= (a2 + b2 – 2ab) * c

= (a2 + b2 – 2ab – c)2

Therefore, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Q.

(v) Let us check the commutativity of *

Let a, b ∈ Q, then

a * b = ab + 1

= ba + 1

= b * a

Therefore

a * b = b * a, for all a, b ∈ Q

Thus, * is commutative on Q

Let us prove the associativity of * on Q

Let a, b, c ∈ Q, then

a * (b * c) = a * (bc + 1)

= a (b c + 1) + 1

= a b c + a + 1

(a * b) * c = (ab + 1) * c

= (ab + 1) c + 1

= a b c + c + 1

So, a * (b * c) ≠ (a * b) * c

Hence, * is not associative on Q.


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