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