(i) Given that ‘×₆‘ on S = {1, 2, 3, 4, 5} defined by a × ₆b = Remainder when ab is divided by 6.

Consider the table,

Here all elements of table are not in S.

⇒ For a = 2 and b = 3,

a × ₆b = 2 × ₆3 = remainder when 6 divided by 6 = 0 ≠ S

So, ×₆ is not a binary operation on S.

(ii) Given ‘+₆’ on S = {0, 1, 2, 3, 4, 5} defined by a + ₆b

= {(a + b, if a + b < 6), (a + b - 6, if a + b ≥ 6)

Consider the table

Here all elements of table are not in S.

⇒ For a = 2 and b = 3,

a × ₆b = 2 × ₆3 = remainder when 6 divided by 6 = 0 ≠ Thus, ×₆ is not a binary operation on S.

(iii) Given that ‘⊙’ on N defined by a ⊙ b = a^b + b^a for all a, b ∈ N

Let a, b ∈ N. Then,

a^b, b^a ∈ N

⇒ a^b + b^a ∈ N [∵Add in binary operation on N]

⇒ a ⊙ b ∈ N

So, ⊙ is a binary operation on N.

(iv) Given ‘*’ on Q defined by a * b = (a – 1)/(b + 1) for all a, b ∈ Q

If a = 2 and b = -1 in Q,

a * b = (a – 1)/(b + 1)

= (2 – 1)/(- 1 + 1)

= 1/0 [which is not defined]

For a = 2 and b = -1

a * b does not belongs to Q

Therefore, * is not a binary operation in Q.