1.

Determine which of the following binary operation is associative and which is commutative:(i) * on N defined by a * b = 1 for all a, b ∈ N(ii) * on Q defined by a * b = (a + b)/2 for all a, b ∈ Q

Answer»

(i) Let us prove commutativity of *

Let a, b ∈ N

a * b = 1

b * a = 1

So,

a * b = b * a, for all a, b ∈ N

Thus * is commutative on N.

Let us prove associativity of *

Let a, b, c ∈ N

Then a * (b * c) = a * (1)

= 1

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

= 1

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

Hence, * is associative on N.

(ii) Let us commutativity of *

Let a, b ∈ N

a * b = (a + b)/2

= (b + a)/2

= b * a

So, a * b = b * a, ∀ a, b ∈ N

Thus * is commutative on N.

Let us prove associativity of *

Let a, b, c ∈ N

a * (b * c) = a * (b + c)/2

= [a + (b + c)]/2

= (2a + b + c)/4

Now, (a * b) * c = (a + b)/2 * c

= [(a + b)/2 + c]/2

= (a + b + 2c)/4

Thus, a * (b * c) ≠ (a * b) * c

If a = 1, b = 2, c = 3

1 * (2 * 3) = 1 * (2 + 3)/2

= 1 * (5/2)

= [1 + (5/2)]/2

= 7/4

(1 * 2) * 3 = (1 + 2)/2 * 3

= 3/2 * 3

= [(3/2) + 3]/2

= 4/9

So, there exist a = 1, b = 2, c = 3 ∈ N such that a * (b * c) ≠ (a * b) * c

Hence, * is not associative on N.


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