Question
1
:
Determine whether the following operation define a binary operation onthe given set or not:
(i) ‘*’ on N defined by a * b = ab for all a, b ∈ N.
(ii) ‘O’ on Z defined by a O b = ab for all a, b ∈ Z.
(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.
(v) ‘+6’ on S = {0, 1, 2, 3, 4, 5} defined by a +6 b
(vi) ‘⊙’ on N defined by a ⊙ b= ab + ba for all a, b ∈ N
(vii) ‘*’ on Q defined by a * b = (a – 1)/ (b + 1) for all a, b ∈ Q
Answer
1 :
(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.
Question
2
:
Determine whether or not the definition of * given below gives a binaryoperation. In the event that * is not a binary operation give justification ofthis.
(i) On Z+, defined *by a * b = a – b
(ii) On Z+, define * by a*b = ab
(iii) On R, define * by a*b = ab2
(iv) On Z+ define * by a * b =|a − b|
(v) On Z+ define * by a * b = a
(vi) On R, define * by a * b = a + 4b2
Here, Z+ denotes the set of all non-negativeintegers.
Answer
2 :
(i) Given On Z+,defined * by a * b = a – b
If a = 1 and b = 2 inZ+, then
a * b = a – b
= 1 – 2
= -1 ∉ Z+ [because Z+ isthe set of non-negative integers]
For a = 1 and b = 2,
a * b ∉ Z+
Thus, * is not abinary operation on Z+.
(ii) Given Z+,define * by a*b = a b
Let a, b ∈ Z+
⇒ a, b ∈ Z+
⇒ a * b ∈ Z+
Thus, * is a binaryoperation on R.
(iii) Given on R,define by a*b = ab2
Let a, b ∈ R
⇒ a, b2 ∈ R
⇒ ab2 ∈ R
⇒ a * b ∈ R
Thus, * is a binaryoperation on R.
(iv) Given on Z+ define* by a * b = |a − b|
Let a, b ∈ Z+
⇒ | a – b | ∈ Z+
⇒ a * b ∈ Z+
Therefore,
a * b ∈ Z+, ∀ a, b ∈ Z+
Thus, * is a binaryoperation on Z+.
(v) Given on Z+ define* by a * b = a
Let a, b ∈ Z+
⇒ a ∈ Z+
⇒ a * b ∈ Z+
Therefore, a * b ∈ Z+ ∀ a, b ∈ Z+
Thus, * is a binaryoperation on Z+.
(vi) Given On R,define * by a * b = a + 4b2
Let a, b ∈ R
⇒ a, 4b2 ∈ R
⇒ a + 4b2 ∈ R
⇒ a * b ∈ R
Therefore, a *b ∈ R, ∀ a, b ∈ R
Thus, * is a binaryoperation on R.
Question
3
:
Let * be a binary operation on the set I ofintegers, defined by a * b = 2a + b − 3.Find the value of 3 * 4.
Answer
3 :
Given a * b =2a + b – 3
3 * 4 = 2 (3) + 4 – 3
= 6 + 4 – 3
= 7
Question
4
:
Is * defined on the set {1, 2, 3, 4, 5} by a * b =LCM of a and b a binary operation? Justify your answer.
Answer
4 :
| LCM | 1 | 2 | 3 | 4 | 5 |
| 1 | 1 | 2 | 3 | 4 | 5 |
| 2 | 2 | 2 | 6 | 4 | 10 |
| 3 | 3 | 5 | 3 | 12 | 15 |
| 4 | 4 | 4 | 12 | 4 | 20 |
| 5 | 5 | 10 | 15 | 20 | 5 |
In the givencomposition table, all the elements are not in the set {1, 2, 3, 4, 5}.
If weconsider a = 2 and b = 3, a * b = LCMof a and b = 6 ∉ {1, 2, 3, 4, 5}.
Thus, * is not abinary operation on {1, 2, 3, 4, 5}.
Question
5
:
Let S = {a, b, c}. Find the total number of binaryoperations on S.
Answer
5 :
Number of binaryoperations on a set with n elements is nn2
Here, S ={a, b, c}
Number of elementsin S = 3
Number of binaryoperations on a set with 3 elements is 332
Question
6
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6 :
Question
7
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7 :
Question
8
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8 :
Question
9
:
The binary operation& : R × R → R is defined as a * b = 2a + b.
Find (2*3)*4.
Answer
9 :
Question
10
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10 :