Determine whether the following operation define a binary operation on the given set or not: (iii) ‘*’ on N defined by a * b = a + b – 2 for all a, b ∈ N (iv) ‘×6‘ on S = {1, 2, 3, 4, 5} defined by a ×6 b = Remainder when a b is divided by 6.
Determine whether the following operation define a binary operation on the given set or not: (iii) ‘*’ on N defined by a * b = a + b – 2 for all a, b ∈ N (iv) ‘×6‘ on S = {1, 2, 3, 4, 5} defined by a ×6 b = Remainder when a b is divided by 6.

(iii)  Given ‘*’ on N defined by a * b = a + b – 2 for all a, b ∈ N

    \[\begin{array}{*{35}{l}} If~a~=\text{ }1\text{ }and~b\text{ }=\text{ }1,  \\ a\text{ }*\text{ }b\text{ }=\text{ }a\text{ }+\text{ }b\text{ }~-2  \\ =\text{ }1\text{ }+\text{ }1\text{ }~-2  \\ =\text{ }0~\notin \text{ }N  \\ ~  \\ \end{array}\]

Thus, there exist a = 1 and b = 1 such that a * b ∉ N

So, * is not a binary operation on N.

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

Consider the composition table,

X6 1 2 3 4 5
1 1 2 3 4 5
2 2 4 0 2 4
3 3 0 3 0 3
4 4 2 0 4 2
5 5 4 3 2 1

Here all the elements of the table are not in S.

⇒ For a = 2 and b = 3,

a ×6 b = 2 ×6 3 = remainder when 6 divided by 6 = 0 ≠ S

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