(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.