Class 12 RD Sharma Solutions β Chapter 3 Binary Operations β Exercise 3.1
Question 1. Determine whether the following operation define a binary operation on the given set or not:
(i) β*β on N defined by a * b = ab for all a, b β N.
(ii) βOβ on Z defined by a O b = ab for all a, b β Z.
(iii) β*β on N defined by a * b = a + b β 2 for all a, b β N
(iv) βΓ6β on S = {1, 2, 3, 4, 5} defined by a Γ 6 b = Remainder when a b is divided by 6.
(v) β+6β on S = {0, 1, 2, 3, 4, 5} defined by a +6 b
(vi) βββ on N defined by a β b= ab + ba for all a, b β N
(vii) β*β on Q defined by a * b = (a β 1)/ (b + 1) for all a, b β Q
Solution:
(i) Given β*β on N defined by a * b = ab for all a, b β N.
Let a, b β N. Then,
ab β N [β΅ abβ 0 and a, b is positive integer]
β a * b β N
Therefore,
a * b β N, β a, b β N
Thus, * is a binary operation on N.
(ii) Given βOβ on Z defined by a O b = ab for all a, b β Z.
Both a = 3 and b = -1 belong to Z.
β a * b = 3-1
= β Z
Thus, * is not a binary operation on Z.
(iii) Given β*β on N defined by a * b = a + b β 2 for all a, b β N
If a = 1 and b = 1,
a * b = a + b β 2
= 1 + 1 β 2
= 0 β N
Thus, there exist a = 1 and b = 1 such that a * b β N
So, * is not a binary operation on N.
(iv) Given βΓ6β on S = {1, 2, 3, 4, 5} defined by a Γ6 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.
β For a = 2 and b = 3,
a Γ6 b = 2 Γ6 3 = remainder when 6 divided by 6 = 0 β S
Thus, Γ6 is not a binary operation on S.
(v) Given β+6β on S = {0, 1, 2, 3, 4, 5} defined by a +6 b
Consider the composition 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 elements of the table are not in S.
β For a = 2 and b = 3,
a Γ6 b = 2 Γ6 3 = remainder when 6 divided by 6 = 0 β Thus, Γ6 is not a binary operation on S.
(vi) Given βββ on N defined by a β b= ab + ba for all a, b β N
Let a, b β N. Then,
ab, ba β N
β ab + ba β N [β΅Addition is binary operation on N]
β a β b β N
Thus, β is a binary operation on N.
(vii) Given β*β on Q defined by a * b = (a β 1)/ (b + 1) for all a, b β Q
If a = 2 and b = -1 in Q,
a * b =
=
= [which is not defined]
For a = 2 and b = -1
a * b does not belongs to Q
So, * is not a binary operation in Q.
Question 2. Determine whether or not the definition of * given below gives a binary operation. In the event that * is not a binary operation give justification of this.
(i) On Z+, defined * by a * b = a β 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 β 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-negative integers.
Solution:
(i) Given On Z+, defined * by a * b = a β b
If a = 1 and b = 2 in Z+, then
a * b = a β b
= 1 β 2
= -1 β Z+ [because Z+ is the set of non-negative integers]
For a = 1 and b = 2,
a * b β Z+
Thus, * is not a binary operation on Z+.
(ii) Given Z+, define * by a*b = a b
Let a, b β Z+
β a, b β Z+
β a * b β Z+
Thus, * is a binary operation on R.
(iii) Given on R, define by a*b = ab2
Let a, b β R
β a, b2 β R
β ab2 β R
β a * b β R
Thus, * is a binary operation on R.
(iv) Given on Z+ define * by a * b = |a β b|
Let a, b β Z+
β | a β b | β Z+
β a * b β Z+
Therefore,
a * b β Z+, β a, b β Z+
Thus, * is a binary operation on Z+.
(v) Given on Z+ define * by a * b = a
Let a, b β Z+
β a β Z+
β a * b β Z+
Therefore, a * b β Z+ β a, b β Z+
Thus, * is a binary operation on Z+.
(vi) Given On R, define * by a * b = a + 4b2
Let a, b β R
β a, 4b2 β R
β a + 4b2 β R
β a * b β R
Therefore, a *b β R, β a, b β R
Thus, * is a binary operation on R.
Question 3. Let * be a binary operation on the set I of integers, defined by a * b = 2a + b β 3. Find the value of 3 * 4.
Solution:
Given:
a * b = 2a + b β 3
3 * 4 = 2 (3) + 4 β 3
= 6 + 4 β 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.
Solution:
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 given composition table, all the elements are not in the set {1, 2, 3, 4, 5}.
If we consider a = 2 and b = 3, a * b = LCM of a and b = 6 β {1, 2, 3, 4, 5}.
Thus, * is not a binary operation on {1, 2, 3, 4, 5}.
Question 5. Let S = {a, b, c}. Find the total number of binary operations on S.
Solution:
Number of binary operations on a set with n elements is
Here, S = {a, b, c}
Number of elements in S = 3
Number of binary operations on a set with 3 elements is
Question 6. Find the total number of binary operations on {a, b}.
Solution:
We have,
S = {a, b}
The total number of binary operation on S = {a, b} in
Question 7. Prove that the operation * on the set
M= defined by A + B = AB is a binary operation.
Solution:
We have,
and
A + B = AB for all A, B β M
Let A =\ and B =
Now, AB =
Therefore, a β R, b β R, c β R and d β R
β ac β R and bd β R
β
β A * B β M
Hence, the operator * defines a binary operation on M
Question 8. Let S be the set of all rational numbers of the form where m β Z and n = 1, 2, 3. Prove that * on S defined by a * b = ab is not a binary operation
Solution:
S = set of rational numbers of the form where m β Z and n = 1, 2, 3
Also, a * b = ab
Let a β S and b β S
β ab =
Therefore, a * b β S
Hence, the operator * does not defines a binary operation on S
Question 9. The binary operation & : R Γ R β R is defined as a*b = 2a + b
Solution:
It is given that, a*b = 2a + b
Now,
(2*3) = 2 Γ 2 + 3
= 4 + 3
(2*3)*4 = 7*4 = 2 Γ 7 + 4
= 14 + 4
= 18
Question 10. Let * be a binary operation on N given by a*b = LCM(a, b) for all a, b β N. Find 5*7.
Solution:
It is given that a*b = LCM (a, b)
Now,
5*7 = LCM (5, 7)
= 35