Class 12 RD Sharma Solutions Chapter 3 Binary Operations Exercise 3.1 (original) (raw)
Last Updated : 23 Jul, 2025
**Question 1. Determine whether the following operation define a binary operation on the 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**
=\begin{cases}a+b,\ if\ a+b<6\\a+b-6,\ if\ a+b\ge 6\end{cases}
****(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**
**Solution:
****(i)** Given ‘*’ on N defined 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 binary operation on N.
****(ii)** Given ‘O’ on Z defined by a O b = ab for all a, b ∈ Z.
Both a = 3 and b = -1 belong to Z.
⇒ a * b = 3-1
= \frac{1}{3} ∉ Z
Thus, * is not a binary operation on Z.
****(iii)** Given ‘*’ on N defined by a * b = a + b – 2 for all a, b ∈ N
If a = 1 and 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 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.
****(v)** Given ‘+6’ on S = {0, 1, 2, 3, 4, 5} defined by a +6 b
=\begin{cases}a+b,\ if\ a+b<6\\a+b-6,\ if\ a+b\ge 6\end{cases}
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.
⇒ 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 for all 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 Q defined by a * b = (a – 1)/ (b + 1) for all a, b ∈ Q
If a = 2 and b = -1 in Q,
a * b = \frac{(a – 1)}{ (b + 1)}
= \frac{(2 – 1)}{ (- 1 + 1)}
= \frac{1}{0} [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 – b
****(ii) On Z** + , define * by a*b = ab
****(iii) On R, define * by a*b = ab** 2
****(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 + 4b** 2
**Here, Z + denotes the set of all non-negative integers.
**Solution:
****(i)** Given On Z+, defined * by a * b = a – b
If a = 1 and b = 2 in Z+, then
a * b = a – b
= 1 – 2
= -1 ∉ Z+ [because Z+ is the set of non-negative integers]
For a = 1 and b = 2,
a * b ∉ Z+
Thus, * is not a binary 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 binary operation 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 binary operation 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 binary operation 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 binary operation 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 binary operation on R.
**Question 3. Let * be a binary operation on the set I of integers, defined by a * b = 2a + b − 3. Find the value of 3 * 4.
**Solution:
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.
**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 ∉ {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 n^{n^2}
Here, S = {a, b, c}
Number of elements in S = 3
Number of binary operations on a set with 3 elements is 3^{3^2}=3^9
**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 2^{2^2}=2^4=16
**Question 7. Prove that the operation * on the set
****M=**\left\{\begin{bmatrix}a&0\\0&b\end{bmatrix} : a, b ∈ R=\{0\}\right\} defined by A + B = AB is a binary operation.
**Solution:
We have,
\left\{\begin{bmatrix}a&0\\0&b\end{bmatrix} : a, b ∈ R=\{0\}\right\} and
A + B = AB for all A, B ∈ M
Let A =\\begin{bmatrix}a&0\\0&b\end{bmatrix}∈M and B = \begin{bmatrix}c&0\\0&d\end{bmatrix}∈M
Now, AB = \begin{bmatrix}a&0\\0&b\end{bmatrix}\begin{bmatrix}c&0\\0&d\end{bmatrix}=\begin{bmatrix}ac&0\\0&bd\end{bmatrix}
Therefore, a ∈ R, b ∈ R, c ∈ R and d ∈ R
⇒ ac ∈ R and bd ∈ R
⇒ \begin{bmatrix}ac&0\\0&bd\end{bmatrix}∈M
⇒ A * B ∈ M
Hence, the operator * defines a binary operation on M
**Question 8. Let S be the set of all rational numbers of the form \frac{m}{n} where m ∈ 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\frac{m}{n} where m ∈ Z and n = 1, 2, 3
Also, a * b = ab
Let a ∈ S and b ∈ S
⇒ ab = \frac{35}{6}∈S
Therefore, a * b ∉ S
Hence, the operator * does not defines a binary operation on S
**Question 9. The binary operation & : R × R → R is defined as a*b = 2a + b
**Solution:
It is given that, a*b = 2a + b
Now,
(2*3) = 2 × 2 + 3
= 4 + 3
(2*3)*4 = 7*4 = 2 × 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 ∈ N. Find 5*7.
**Solution:
It is given that a*b = LCM (a, b)
Now,
5*7 = LCM (5, 7)
= 35
Summary
Exercise 3.1 in Chapter 3 of RD Sharma's Class 12 mathematics textbook likely focuses on binary operations. This exercise typically covers the definition and properties of binary operations, including closure, associativity, commutativity, and the existence of identity and inverse elements. It may also explore examples of binary operations in various mathematical structures such as real numbers, matrices, and sets, and examine how to determine whether a given operation is a binary operation on a specific set.
Practice Questions
1. Determine if the operation * defined by a * b = a² + b² is a binary operation on the set of real numbers.
2. Let A = {1, 2, 3, 4}. Define the operation ⊕ on A by a ⊕ b = (a + b) mod 5. Construct the operation table for ⊕.
3. Prove that the operation * defined by a * b = max(a, b) is commutative but not associative on the set of real numbers.
4. For the set of 2x2 matrices, is matrix multiplication a binary operation? Justify your answer.
5. Let S be the set of all non-zero real numbers. Is the operation ÷ (division) a binary operation on S? Explain.
6. Define a * b = a - b on the set of integers. Find the identity element for this operation, if it exists.
7. For the operation * defined by a * b = ab + a + b on the set of real numbers, find a * 0 and 0 * a.
8. Prove that the operation ⊕ defined by a ⊕ b = a + b + 1 on the set of integers is associative.
9. Let A = {0, 1, 2, 3, 4}. Define a * b = (a + b) mod 5. Find all elements x in A such that 2 * x = 4.
10. For the binary operation * defined on the set of real numbers by a * b = ab/2, find the inverse of 4, if it exists.