Question -
Answer -
(i) Given ‘*’ on Ndefined by a * b = ab for all a, b ∈ N.
Let a, b ∈ N. Then,
ab ∈ N [∵ ab≠0 and a,b is positive integer]
⇒ a * b ∈ N
Therefore,
a * b ∈ N, ∀ a, b ∈ N
Thus, * is a binaryoperation on N.
(ii) Given ‘O’ on Zdefined by a O b = ab for all a, b ∈ Z.
Both a =3 and b = -1 belong to Z.
⇒ a * b = 3-1
= 1/3 ∉ Z
Thus, * is not abinary operation on Z.
(iii) Given ‘*’on N defined by a * b = a + b – 2 for all a, b ∈ N
If a = 1and b = 1,
a * b = a + b – 2
= 1 + 1 – 2
= 0 ∉ N
Thus, there exist a =1 and b = 1 such that a * b ∉ N
So, * is not a binaryoperation 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 thecomposition 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 elementsof 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.
(v) Given ‘+6’on S = {0, 1, 2, 3, 4, 5} defined by a +6 b
Consider thecomposition table,
| +6 | 0 | 1 | 2 | 3 | 4 | 5 |
| 0 | 0 | 1 | 2 | 3 | 4 | 5 |
| 1 | 1 | 2 | 3 | 4 | 5 | 0 |
| 2 | 2 | 3 | 4 | 5 | 0 | 1 |
| 3 | 3 | 4 | 5 | 0 | 1 | 2 |
| 4 | 4 | 5 | 0 | 1 | 2 | 3 |
| 5 | 5 | 0 | 1 | 2 | 3 | 4 |
Here all the elementsof 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 ≠ Thus, ×6 is not a binary operation on S.
(vi) Given ‘⊙’ on N defined by a ⊙ b= ab + ba forall a, b ∈ N
Let a, b ∈ N. Then,
ab, ba ∈ N
⇒ ab +ba ∈N [∵Addition is binary operation on N]
⇒ a ⊙ b ∈ N
Thus, ⊙ is a binary operation on N.
(vii) Given ‘*’ on Qdefined by a * b = (a – 1)/ (b + 1) for all a, b ∈ Q
If a = 2 and b = -1 inQ,
a * b = (a – 1)/ (b +1)
= (2 – 1)/ (- 1 + 1)
= 1/0 [which is notdefined]
For a = 2 and b = -1
a * b does not belongsto Q
So, * is not a binaryoperation in Q.