The RD Sharma Solutions for Class 12 Maths Chapter 3 – Binary Operations

The RD Sharma Solutions for Class 12 Maths Chapter 3 – Binary Operations are available here in PDF format. Students who face difficulties in solving the exercise-wise problems of Chapter 3 can easily download the solutions PDF of RD Sharma Solutions. These solutions have been carefully crafted by expert subject tutors as per the latest CBSE guidelines and updated marking schemes. Moreover, the solutions are presented in a step-by-step manner to help students learn concepts more effectively.

The RD Sharma Solutions for Class 12 are designed in a simple and clear format according to the CBSE board syllabus. Students who aspire to score high in their Class 12 exams can refer to these PDFs as an essential study resource to strengthen their exam preparation. For an in-depth understanding of the concepts covered in this chapter, students can access the RD Sharma Solutions for Class 12 Chapter 3 Binary Operations PDF from the download links provided here.

RD Sharma Solutions for Class 12 Chapter Binary Operations

A binary operation is a mathematical operation that combines two elements from a set to produce another element from the same set. Binary operations are fundamental in various areas of mathematics, including algebra and set theory. The RD Sharma Solutions for Class 12 introduces students to the concept of binary operations, the properties of binary operations such as closure, associativity, identity, and invertibility, and how to verify these properties with step-by-step examples.

Download RD Sharma Class 12 Chapter Binary Operations PDF Solutions

Students can easily download the RD Sharma Class 12 Maths Chapter Binary Operations Solutions PDF, which covers all exercises, solved examples, and additional practice questions. These PDF solutions are ideal for quick revision and help build strong conceptual understanding for binary operation-based questions in board and competitive exams.

RD Sharma Solutions for Binary Operations: Solved Questions

Question 1 Determine whether the operation * defined by a * b = ab for all a, b ∈ ℕ is a binary operation on ℕ.

Step 1: For an operation to be a binary operation on a set S, it must satisfy the closure property: if a, b ∈ S, then a * b ∈ S.

Step 2: Given operation: a * b = ab where a, b ∈ ℕ

Step 3: Check closure property:

• Let a, b ∈ ℕ (natural numbers)
• Then a * b = ab = a × b
• Since the product of two natural numbers is always a natural number, ab ∈ ℕ

Answer: Yes, * is a binary operation on ℕ because it satisfies the closure property.

Question 2 Given O on ℤ defined by a O b = ab for all a, b ∈ ℤ, is O a binary operation on ℤ?

Step 1: Check closure property for the operation a O b = ab on ℤ (integers).

Step 2: Let a, b ∈ ℤ

Step 3: Then a O b = ab = a × b

Step 4: The product of two integers is always an integer, so ab ∈ ℤ

Answer: Yes, O is a binary operation on ℤ because it satisfies the closure property.

Problem: On ℕ, * is defined by a * b = a + b - 2. Does * define a binary operation on ℕ?

Step 1: Check closure property: if a, b ∈ ℕ, then a * b must be in ℕ.

Step 2: Given: a * b = a + b - 2 where a, b ∈ ℕ

Step 3: Test with specific values:

• Let a = 1, b = 1 (both in ℕ)
• a * b = 1 + 1 - 2 = 0
• But 0 ∉ ℕ (natural numbers start from 1)

Step 4: Since we found a case where a * b ∉ ℕ, the closure property is violated.

Answer: No, * does not define a binary operation on ℕ because it fails the closure property.

Problem: Let S = {1, 2, 3, 4, 5}, define a ×₆ b as the remainder when ab is divided by 6. Is ×₆ a binary operation on S?

Step 1: Check if a ×₆ b ∈ S for all a, b ∈ S

Step 2: The operation a ×₆ b gives the remainder when ab is divided by 6, so the result is in {0, 1, 2, 3, 4, 5}

Step 3: Test some cases:

• 1 ×₆ 1 = remainder of (1×1)/6 = 1 ∈ S ✓
• 2 ×₆ 3 = remainder of (2×3)/6 = remainder of 6/6 = 0 ∉ S ✗

Step 4: Since 0 ∉ S but can be obtained as a result, closure property fails.

Answer: No, ×₆ is not a binary operation on S because the result can be 0, which is not in S.

Problem: For S = {0,1,2,3,4,5}, define a ⊕₆ b = a+b if a+b<6, and a+b-6 if a+b≥6. Is ⊕₆ a binary operation on S?

Step 1: Check closure property for all a, b ∈ S

Step 2: Case 1: If a + b < 6, then a ⊕₆ b = a + b

• Since a, b ∈ {0,1,2,3,4,5} and a + b < 6, we have a + b ∈ {0,1,2,3,4,5} = S

Step 3: Case 2: If a + b ≥ 6, then a ⊕₆ b = a + b - 6

• Maximum value: a = 5, b = 5 ⟹ a + b - 6 = 10 - 6 = 4 ∈ S
• Minimum value when a + b ≥ 6: a + b = 6 ⟹ a + b - 6 = 0 ∈ S
• So a + b - 6 ∈ {0,1,2,3,4,5} = S

Answer: Yes, ⊕₆ is a binary operation on S because it satisfies the closure property in both cases.

Problem: Define a ⊛ b = ab + ba for a, b ∈ ℕ. Check if ⊛ is a binary operation on ℕ.

Step 1: Note that ab + ba = ab + ab = 2ab for a, b ∈ ℕ

Step 2: So a ⊛ b = 2ab

Step 3: Check closure property:

• Let a, b ∈ ℕ
• Then ab ∈ ℕ (product of natural numbers)
• Therefore 2ab ∈ ℕ (2 times a natural number)

Answer: Yes, ⊛ is a binary operation on ℕ because 2ab ∈ ℕ for all a, b ∈ ℕ.

Problem: Let * on ℝ be a * b = a² + b². Does * have an identity element in ℝ?

Step 1: An identity element e must satisfy: a * e = e * a = a for all a ∈ ℝ

Step 2: For our operation: a * e = a² + e² and e * a = e² + a²

Step 3: For e to be identity: a² + e² = a for all a ∈ ℝ

Step 4: This gives us e² = a - a² for all a ∈ ℝ

Step 5: This is impossible because:

• For a = 0: e² = 0 - 0 = 0, so e = 0
• For a = 1: e² = 1 - 1 = 0, so e = 0
• For a = 2: e² = 2 - 4 = -2 < 0, which is impossible for real e

Answer: No, the operation * does not have an identity element in ℝ.

Problem: On ℝ, a * b = (a + b)/2. Is the operation commutative and/or associative?

Step 1: Check commutativity: Does a * b = b * a?

• a * b = (a + b)/2
• b * a = (b + a)/2 = (a + b)/2
• Therefore a * b = b * a

Step 2: Check associativity: Does (a * b) * c = a * (b * c)?

• LHS: (a * b) * c = ((a + b)/2) * c = ((a + b)/2 + c)/2 = (a + b + 2c)/4
• RHS: a * (b * c) = a * ((b + c)/2) = (a + (b + c)/2)/2 = (2a + b + c)/4
• LHS ≠ RHS in general (e.g., when a = 1, b = 0, c = 0: LHS = 1/2, RHS = 1/2, but when a = 2, b = 0, c = 0: LHS = 1, RHS = 1)

Step 3: Let's check with specific values: a = 1, b = 2, c = 3

• LHS = (1 + 2 + 2×3)/4 = 9/4
• RHS = (2×1 + 2 + 3)/4 = 7/4
• Since 9/4 ≠ 7/4, the operation is not associative

Answer: The operation is commutative but not associative.

Problem: Let * on ℤ be a * b = a - b. Is * commutative? Is it associative?

Step 1: Check commutativity: Does a * b = b * a?

• a * b = a - b
• b * a = b - a
• In general, a - b ≠ b - a
• Counter-example: 3 * 2 = 3 - 2 = 1, but 2 * 3 = 2 - 3 = -1

Step 2: Check associativity: Does (a * b) * c = a * (b * c)?

• LHS: (a * b) * c = (a - b) * c = (a - b) - c = a - b - c
• RHS: a * (b * c) = a * (b - c) = a - (b - c) = a - b + c
• LHS = a - b - c, RHS = a - b + c
• LHS ≠ RHS unless c = 0

Answer: The operation * is neither commutative nor associative.

Problem: Find the identity element (if it exists) for the operation a * b = a + b on ℤ.

Step 1: An identity element e must satisfy: a * e = e * a = a for all a ∈ ℤ

Step 2: For our operation a * b = a + b:

• a * e = a + e = a
• This gives us e = 0

Step 3: Verify: e * a = e + a = 0 + a = a ✓

Step 4: Check for all integers:

• a * 0 = a + 0 = a ✓
• 0 * a = 0 + a = a ✓

Answer: The identity element is 0.

Problem: Let a * b = ab mod 5 on {0,1,2,3,4}. Make a composition table for this binary operation.

Step 1: Calculate a * b = (ab) mod 5 for all pairs (a,b)

Step 2: Composition Table:

*	0	1	2	3	4
0	0	0	0	0	0
1	0	1	2	3	4
2	0	2	4	1	3
3	0	3	1	4	2
4	0	4	3	2	1

Step 3: Verification of some entries:

• 2 * 3 = (2×3) mod 5 = 6 mod 5 = 1
• 4 * 4 = (4×4) mod 5 = 16 mod 5 = 1
• 3 * 4 = (3×4) mod 5 = 12 mod 5 = 2

Problem: On S = {a, b}, list all possible binary operations that can be defined on set S.

Step 1: For a binary operation on S = {a, b}, we need to define the result of each of the four possible operations: a*a, a*b, b*a, b*b

Step 2: Each result must be in S, so each can be either 'a' or 'b'

Step 3: This gives us 2⁴ = 16 possible binary operations

Step 4: List all operations by their composition tables:

Operation 1:

*	a	b
a	a	a
b	a	a

Operation 2:

*	a	b
a	a	a
b	a	b

... (continuing similarly for all 16 operations)

Answer: There are 16 possible binary operations on S = {a, b}.

Problem: Find the total number of possible binary operations on a set with three elements S = {a, b, c}.

Step 1: For a set with n elements, a binary operation must assign to each ordered pair (x, y) an element from the set.

Step 2: For S = {a, b, c}, we have 3 elements.

Step 3: Number of ordered pairs = 3² = 9

These are: (a,a), (a,b), (a,c), (b,a), (b,b), (b,c), (c,a), (c,b), (c,c)

Step 4: For each ordered pair, we can choose any of the 3 elements as the result.

Step 5: Total number of binary operations = 3⁹ = 19,683

Answer: There are 3⁹ = 19,683 possible binary operations on a set with three elements.

Problem: If * on ℚ⁺ is defined as a * b = ab/2, find the identity element and the inverse of 4 under this operation.

Step 1: Find the identity element e such that a * e = e * a = a for all a ∈ ℚ⁺

Step 2: For our operation: a * e = ae/2 = a

• This gives us ae/2 = a
• Therefore e/2 = 1
• So e = 2

Step 3: Verify: e * a = 2a/2 = a ✓

Step 4: Find the inverse of 4:

The inverse of 4, call it 4⁻¹, must satisfy: 4 * 4⁻¹ = e = 2

Step 5: Calculate:

• 4 * 4⁻¹ = (4 × 4⁻¹)/2 = 2
• This gives us (4 × 4⁻¹)/2 = 2
• So 4 × 4⁻¹ = 4
• Therefore 4⁻¹ = 1

Step 6: Verify: 4 * 1 = (4 × 1)/2 = 2 = e ✓

Answer: Identity element is 2, and the inverse of 4 is 1.

Problem: Define a ◊ b = |a-b| on ℤ. Is ◊ a binary operation on ℤ? Check for the existence of an identity element.

Step 1: Check if ◊ is a binary operation (closure property)

• For any a, b ∈ ℤ, |a - b| is the absolute value of their difference
• Since a - b ∈ ℤ, and |x| ∈ ℤ for any x ∈ ℤ, we have |a - b| ∈ ℤ
• Therefore, closure property is satisfied

Step 2: Check for identity element

An identity element e must satisfy: a ◊ e = e ◊ a = a for all a ∈ ℤ

Step 3: For our operation: a ◊ e = |a - e| = a

Step 4: This means |a - e| = a for all a ∈ ℤ

• For a = 1: |1 - e| = 1
• This gives us either 1 - e = 1 or 1 - e = -1
• So either e = 0 or e = 2

Step 5: Test e = 0:

• For a = 1: 1 ◊ 0 = |1 - 0| = 1 ✓
• For a = -1: (-1) ◊ 0 = |-1 - 0| = 1 ≠ -1 ✗

Step 6: Test e = 2:

• For a = 1: 1 ◊ 2 = |1 - 2| = 1 ✓
• For a = 3: 3 ◊ 2 = |3 - 2| = 1 ≠ 3 ✗

Answer: Yes, ◊ is a binary operation on ℤ. However, there is no identity element because |a - e| = a cannot be satisfied for all integers a with any fixed value of e.