RS Aggarwal Solutions for Class 12 Chapter 03 Binary Operations

 

Question 1. Let * be a binary operation on the set I of all integers, defined by a * b = 3a + 4b - 2. Find the value of 4 * 5.
Answer: We need to calculate 4*5 using the given definition. Substituting a = 4 and b = 5 into the formula: 4*5 = 3(4) + 4(5) - 2 = 12 + 20 - 2 = 30.
In simple words: Replace a with 4 and b with 5 in the formula, then work through the arithmetic to get 30.

Exam Tip: Always substitute the values carefully into the given operation formula and follow the order of operations to avoid calculation errors.

 

Question 2. The binary operation * on R is defined by a * b = 2a + b. Find (2 * 3) * 4.
Answer: First, calculate 2*3: 2*3 = 2(2) + 3 = 4 + 3 = 7. Then apply the operation to 7 and 4: 7*4 = 2(7) + 4 = 14 + 4 = 18. Therefore, (2*3)*4 = 18.
In simple words: Work from the inside out - first find 2*3, which gives 7, then combine that result with 4 using the same rule.

Exam Tip: When dealing with nested operations, always evaluate the innermost operation first, then use that result for the next step.

 

Question 3. Let * be a binary operation on the set of all nonzero real numbers, defined by a * b = ab/5. Find the value of x given that 2 * (x * 5) = 10.
Answer: Using the definition a*b = ab/5, first calculate x*5: x*5 = (x·5)/5 = x. Next, apply the operation to 2 and x: 2*x = (2x)/5. Setting this equal to 10: (2x)/5 = 10, which gives 2x = 50, so x = 25.
In simple words: Simplify x*5 to get x, then set up an equation with 2*x equal to 10, and solve for x.

Exam Tip: Simplify intermediate results before substituting them into further operations - this makes the final equation much easier to solve.

 

Question 4. Let *: R × R → R be a binary operation given by a * b = a + 4b². Then, compute( - 5) * (2 * 0).
Answer: First, calculate 2*0 using a*b = a + 4b²: 2*0 = 2 + 4(0)² = 2 + 0 = 2. Then apply the operation to - 5 and 2: ( - 5)*2 = - 5 + 4(2)² = - 5 + 4(4) = - 5 + 16 = 11.
In simple words: Find the inner result 2*0 first, which equals 2, then use that to compute ( - 5)*2.

Exam Tip: Pay close attention to the order of operations and bracket notation - evaluate expressions within brackets before applying further operations.

 

Question 5. Let be a binary operation on the set Q of all rational numbers given as a * b = (2a - b)² for all a, b ∈ Q. Find 3 * 5 and 5 * 3. Is 3 * 5 = 5 * 3?
Answer: Calculate 3*5: 3*5 = (2(3) - 5)² = (6 - 5)² = 1² = 1. Calculate 5*3: 5*3 = (2(5) - 3)² = (10 - 3)² = 7² = 49. Since 1 ≠ 49, we have 3*5 ≠ 5*3.
In simple words: The operation gives different results depending on the order of the numbers, showing it is not commutative.

Exam Tip: Always test commutativity by computing the operation both ways - if the results differ, the operation is not commutative.

 

Question 6. Let * be a binary operation on N given by a * b = LCM of a and b. Find the value of 20 * 16. Is * (i) commutative, (ii) associative?
Answer: To find LCM of 20 and 16, first find the prime factorizations: 20 = 2² × 5 and 16 = 2⁴. The LCM takes the highest power of each prime factor: LCM(20, 16) = 2⁴ × 5 = 16 × 5 = 80.
(i) For commutativity, note that finding the LCM involves taking the highest power of each prime factor from both numbers, regardless of their order. The prime factorizations remain identical whether we compute LCM(a,b) or LCM(b,a). Therefore, LCM(a,b) = LCM(b,a), making the operation commutative.
(ii) For associativity, we verify that LCM[LCM(a,b),c] = LCM[a,LCM(b,c)]. When three numbers have prime factors p₁, p₂, p₃, and p₄, the LCM operation selects the highest power of each factor. Applying this rule sequentially in either order yields the same result: p₁^q₁ × p₂^q₂ × p₃^q₃ × p₄^q⁴. Thus, the operation is associative.
In simple words: The LCM operation is commutative because the order doesn't matter when finding the highest power of each prime factor. It's associative because grouping the numbers differently doesn't change the final result.

Exam Tip: For LCM-based operations, recall that commutativity and associativity follow naturally from the definition involving prime factorizations, which are independent of order.

 

Question 7. If * be the binary operation on the set Z of all integers defined by a * b = (a + 3b²), find 2 * 4.
Answer: Using a*b = a + 3b², substitute a = 2 and b = 4: 2*4 = 2 + 3(4)² = 2 + 3(16) = 2 + 48 = 50.
In simple words: Plug in the values into the formula, calculate 3 times 16, and add 2 to get 50.

Exam Tip: Be careful with the order of operations - square the value first, multiply by the coefficient, then add the other term.

 

Question 8. Show that * on Z⁺ defined by a * b = |a - b| is not a binary operation.
Answer: For a binary operation on Z⁺, the result of combining any two positive integers must also be a positive integer. When a = b, we have a*b = |a - b| = |0| = 0. However, 0 is neither positive nor negative, so it does not belong to Z⁺. Since there exist elements a and b in Z⁺ (specifically when a = b) such that a*b ∉ Z⁺, the set is not closed under this operation. Therefore, * is not a binary operation on Z⁺.
In simple words: A binary operation must always produce a result that stays in the original set. Here, when equal numbers are combined, we get 0, which isn't a positive integer, so the operation fails.

Exam Tip: To show an operation is not binary, find a counterexample where the result falls outside the given set - the simplest counterexample often involves equal elements or boundary cases.

 

Question 9. Let * be a binary operation on N, defined by a * b = aᵇ for all a, b ∈ N. Show that * is neither commutative nor associative.
Answer: To show the operation is not commutative, we demonstrate that aᵇ ≠ bᵃ in general. For example, take a = 1 and b = 2: 1² = 1, but 2¹ = 2. Since 1 ≠ 2, the operation fails commutativity except when a = b. To show it is not associative, we check whether (aᵇ)ᶜ = a^(bᶜ). Let a = 2, b = 1, and c = 3: (2¹)³ = 2³ = 8, but 2^(1×3) = 2³ = 8. However, with a = 2, b = 1, c = 3, we compute (aᵇ)ᶜ = (2¹)³ = 2³ = 8 and a^(bᶜ) = 2^(1³) = 2¹ = 2. Since 8 ≠ 2, associativity does not hold.
In simple words: With exponentiation, swapping the numbers gives different results (1² ≠ 2¹), and grouping matters because (2¹)³ ≠ 2^(1×3).

Exam Tip: For exponential operations, always remember that the order is critical and parentheses affect which operation happens first - these are classic sources of non-commutativity and non-associativity.

 

Question 10. Let a * b = LCM(a, b) for all values of a, b ∈ N.
(i) Find (12 * 16).
(ii) Show that * is commutative on N.
(iii) Find the identity element in N.
(iv) Find all invertible elements in N.
Answer:
(i) To find the LCM of 12 and 16, use prime factorizations: 12 = 2² × 3 and 16 = 2⁴. The LCM takes the highest power of each prime: LCM(12, 16) = 2⁴ × 3 = 16 × 3 = 48.
(ii) The LCM is computed by taking the highest power of each prime factor from the numbers involved. Since prime factorizations are independent of order, LCM(a,b) = LCM(b,a) for any positive integers a and b. Therefore, the operation is commutative.
(iii) For an identity element e, we need a*e = a for all a ∈ N. This means LCM(a, e) = a. This holds when e = 1, because LCM(a, 1) = a. Thus, the identity element is 1.
(iv) For an element to be invertible, there must exist another element y such that a*y = e (the identity). Here, LCM(a, y) = 1. The LCM of two positive integers equals 1 only when both integers are 1. So a = y = 1, making 1 the only invertible element in N with respect to this operation.
In simple words: The LCM of 12 and 16 is 48. The operation is commutative because order doesn't affect the highest prime powers. The number 1 is the identity because any number's LCM with 1 is that number itself. Only 1 is invertible because it's the only number whose LCM with another number can be 1.

Exam Tip: When finding identity and inverse elements, always use the definition directly: substitute the candidate into the operation and check if it yields the required result.

 

Question 11. Let Q be the set of all positive rational numbers.
(i) Show that the operation * on Q⁺ defined by a * b = (1/2)(a + b) is a binary operation.
(ii) Show that * is commutative.
(iii) Show that * is not associative.
Answer:
(i) For * to be a binary operation on Q⁺, the result must always lie in Q⁺. For any positive rationals a = 1 and b = 2, we have a*b = (1/2)(1 + 2) = (1/2)(3) = 1.5, which is a positive rational number. Since the sum of positive numbers is positive and one-half of a positive number remains positive, the operation is closed on Q⁺. Therefore, * is a binary operation.
(ii) For commutativity, verify a*b = b*a: a*b = (1/2)(a + b) and b*a = (1/2)(b + a). Since addition is commutative, a + b = b + a, so a*b = b*a. The operation is commutative.
(iii) To show non-associativity, compute (a*b)*c and a*(b*c) with specific values. Let a = 1, b = 2, c = 3. Then (a*b)*c = ((1/2)(1 + 2))*3 = 1.5*3 = (1/2)(1.5 + 3) = (1/2)(4.5) = 2.25. And a*(b*c) = 1*(1/2)(2 + 3) = 1*2.5 = (1/2)(1 + 2.5) = (1/2)(3.5) = 1.75. Since 2.25 ≠ 1.75, the operation is not associative.
In simple words: The operation takes the average of two positive rationals, which stays positive. Averaging is commutative because a + b = b + a. But grouping affects the result: averaging (1 and 2) with 3 gives a different answer than averaging 1 with (2 and 3).

Exam Tip: For testing associativity, always use concrete numbers and compare (a*b)*c with a*(b*c) side-by-side to see the difference clearly.

 

Question 12. Show that the set A = {-1, 0, 1} is not closed for addition.
Answer: A set is closed under an operation if combining any two elements from the set produces a result that remains in the set. Consider a = 1 and b = 1, both in A. Their sum is 1 + 1 = 2, which is not a member of the set A = {-1, 0, 1}. Since adding two elements from A produces a result outside A, the set is not closed for addition.
In simple words: When we add 1 + 1, we get 2, which isn't in the set. This breaks the closure property, so the set is not closed under addition.

Exam Tip: To prove closure fails, find just one counterexample - two elements whose operation produces a result outside the set.

 

Question 13. Show that * on R - {-1}, defined by a * b = a/(b + 1) is neither commutative nor associative.
Answer: To show non-commutativity, take a = 1 and b = 0, both in R - {-1}. Then a*b = 1/(0 + 1) = 1, while b*a = 0/(1 + 1) = 0. Since 1 ≠ 0, we have a*b ≠ b*a, so the operation is not commutative. To show non-associativity, let a = 1, b = 0, c = 3. Then (a*b)*c = 1*3 = 1/(3 + 1) = 1/4. And a*(b*c) = a*0 = 1/(0 + 1) = 1*0 = 1/(0 + 1) = 1. Wait, let me recalculate: b*c = 0/(3 + 1) = 0, so a*(b*c) = a*0 = 1*0 (where the result of b*c is 0). Actually, a*(b*c) = a*(0/(c+1)) requires careful substitution. Let's use: (a*b)*c = (1/(0+1))*c = 1*c = 1/(c+1) = 1/4. And a*(b*c) = a*(0/(3+1)) means we compute a*(0) = 1/(0+1) = 1. Since 1/4 ≠ 1, the operation is not associative.
In simple words: The order matters: a*b gives 1 but b*a gives 0. Also, grouping differently changes the result, so neither property holds.

Exam Tip: When dealing with fractional operations, substitute carefully and double-check your arithmetic, especially with nested operations.

 

Question 14. For all a, b ∈ R, we define a * b = |a - b|. Show that * is commutative but not associative.
Answer: For commutativity, examine a*b and b*a. By definition, a*b = |a - b|. Using the property that |x| = |-x|, we have |a - b| = |-(a - b)| = |b - a| = b*a. Therefore, a*b = b*a, making the operation commutative. For associativity, we need (a*b)*c = a*(b*c), or ||a - b| - c| = |a - |b - c||. Using a = 1, b = -1, c = 5: LHS = ||1 - (-1)| - 5| = ||2| - 5| = |2 - 5| = |-3| = 3. RHS = |1 - |-1 - 5|| = |1 - |-6|| = |1 - 6| = |-5| = 5. Since 3 ≠ 5, the operation is not associative.
In simple words: The absolute value of a difference is the same regardless of order (commutative). But when we nest the operations, the grouping matters and gives different results (not associative).

Exam Tip: Use the absolute value property |a - b| = |b - a| to quickly confirm commutativity, then pick numbers with a mix of signs to test associativity.

 

Question 15. For all a, b ∈ N, we define a * b = a³ + b³. Show that * is commutative but not associative.
Answer: For commutativity, compute a*b and b*a. By definition, a*b = a³ + b³ and b*a = b³ + a³. Since addition is commutative, a³ + b³ = b³ + a³, so a*b = b*a. The operation is commutative. For associativity, check whether (a*b)*c = a*(b*c). Let a = 1, b = 2, c = 3. Then a*b = 1³ + 2³ = 1 + 8 = 9. So (a*b)*c = 9*3 = 9³ + 3³ = 729 + 27 = 756. And b*c = 2³ + 3³ = 8 + 27 = 35, so a*(b*c) = 1*35 = 1³ + 35³ = 1 + 42875 = 42876. Since 756 ≠ 42876, the operation is not associative.
In simple words: Adding cubes is commutative because a³ + b³ = b³ + a³. But the grouping matters: the cubes combine differently depending on the order of operations, producing different sums.

Exam Tip: For sum-based operations, commutativity is usually immediate from properties of addition, but associativity often fails when the operation applies the same formula repeatedly.

 

Question 16. Let X be a nonempty set and * be a binary operation on P(X), the power set of X, defined by A * B = A ∩ B for all A, B ∈ P(X).
(i) Find the identity element in P(X).
(ii) Show that X is the only invertible element in P(X).
Answer:
(i) For e to be an identity element, we need A*e = A for any set A in P(X). Using the operation A*e = A ∩ e, we require A ∩ e = A. This is true when e ⊇ A for all A ∈ P(X). The only set that contains all subsets of X is X itself. Therefore, A ∩ X = A for every A ∈ P(X), making X the identity element.
(ii) For an element to be invertible, there must exist an inverse B such that A*B = X and B*A = X. Since A*B = A ∩ B = X, we need A ∩ B = X. For the intersection of two sets to equal X, both A and B must contain all elements of X. Since both are subsets of X, this means A = X and B = X. Thus, X is the only element with an inverse (which is itself), making it the only invertible element in P(X).
In simple words: The identity is X because intersecting any set with X gives that set back. Only X is invertible because to get X from an intersection, both sets must already be X.

Exam Tip: For set operations, visualize with a Venn diagram - the identity must "preserve" every set, and invertibility is rare because most set operations are restrictive (intersection removes elements, union combines them).

 

Question 17. A binary operation * on the set (0, 1, 2, 3, 4, 5) is defined as:
\( a * b = \begin{cases} a + b; & \text{if } a + b < 6 \\ a + b - 6; & \text{if } a + b \geq 6 \end{cases} \)
Show that 0 is the identity for this operation and each element a has an inverse (6 - a).
Answer: To verify 0 is the identity, check that e*b = b and b*e = b. For any b in the set, if we take e = 0, then 0 + b = b, which is always less than 6. So 0*b = 0 + b = b by the first condition. Similarly, b*0 = b + 0 = b < 6, so b*0 = b. Thus, 0 serves as both left and right identity. To verify that 6 - a is the inverse of a, we need a*(6-a) = 0. Note that a + (6-a) = 6, which satisfies a + (6-a) ≥ 6. By the second condition, a*(6-a) = a + (6-a) - 6 = 6 - 6 = 0. Similarly, (6-a)*a = (6-a) + a - 6 = 0. Since 0 ≤ a ≤ 5, we have 1 ≤ 6-a ≤ 6, but we must check: when a = 0, 6-a = 6, which is not in the set {0,1,2,3,4,5}. However, a = 0 is its own inverse since 0*0 = 0. For a ≠ 0, the value 6-a ranges from 1 to 5, all in the set. So c = 6-a is the inverse for each non-zero element a, and 0 is its own inverse.
In simple words: Zero is the identity because a + 0 always equals a in this operation. Each element a has an inverse 6-a because adding them gives 6, which by the operation's second rule produces 0.

Exam Tip: For operations with conditional cases, check both conditions separately - identify which condition applies based on whether the sum is below or above the threshold.

 

Question 1. Define * on N by m * n = LCM(m, n). Show that * is a binary operation which is commutative as well as associative.
Answer: For * to be a binary operation on N, the result must be a natural number. For any natural numbers m and n, their LCM is also a natural number, so the operation is closed. Example: m = 2, n = 3 gives m*n = LCM(2, 3) = 6 ∈ N, confirming the operation is binary. For commutativity, the LCM operation depends on finding the highest power of each prime factor in both numbers. Since prime factorizations are the same regardless of order, LCM(m, n) = LCM(n, m). Thus m*n = n*m, making the operation commutative. For associativity, we verify m*(n*p) = (m*n)*p. Taking m = 2, n = 3, p = 4: m*(n*p) = 2*LCM(3,4) = 2*12 = LCM(2, 12) = 12. And (m*n)*p = LCM(2,3)*4 = 6*4 = LCM(6, 4) = 12. Since the results match, and this holds generally because LCM involves selecting the maximum power of each prime from all numbers involved, the operation is associative.
In simple words: LCM is always a natural number, so it's binary. The order doesn't matter for LCM (commutative). Grouping doesn't matter either because we're just picking the highest prime powers from all the numbers (associative).

Exam Tip: For LCM-based operations on N, commutativity and associativity follow almost immediately from the definition - no special cases to worry about.

 

Question 2. Define * on Z by a * b = a - b + ab. Show that * is a binary operation on Z which is neither commutative nor associative.
Answer: For * to be a binary operation, the result of a*b must be an integer whenever a and b are integers. Since a - b + ab is a combination of integer operations (subtraction and multiplication), the result is always an integer. Thus * is a binary operation. For commutativity, compute a*b and b*a: a*b = a - b + ab while b*a = b - a + ba = b - a + ab (since multiplication is commutative). Notice a - b ≠ b - a unless a = b. For instance, with a = 1 and b = 2: a*b = 1 - 2 + 1(2) = 1 - 2 + 2 = 1, while b*a = 2 - 1 + 2(1) = 2 - 1 + 2 = 3. Since 1 ≠ 3, the operation is not commutative. For associativity, compute (a*b)*c and a*(b*c). With a = 1, b = 2, c = 1: (a*b)*c = 1*1 = 1 - 1 + 1(1) = 1. And a*(b*c) = a*(2 - 1 + 2(1)) = a*3 = 1 - 3 + 1(3) = 1. This example doesn't show non-associativity, so let's try a = 2, b = 1, c = 3: (a*b)*c = (2 - 1 + 2)*3 = 3*3 = 3 - 3 + 9 = 9. And a*(b*c) = a*(1 - 3 + 3) = a*1 = 2 - 1 + 2 = 3. Since 9 ≠ 3, the operation is not associative.
In simple words: Integer subtraction and multiplication always give integers, so it's binary. But since subtraction is not commutative (a - b ≠ b - a), the whole operation fails commutativity. The non-commutativity also ruins associativity.

Exam Tip: When an operation includes subtraction, expect non-commutativity - use simple numerical examples like a = 1, b = 2 to expose it quickly.

 

Question 3. Define * on Z by a * b = a + b - ab. Show that * is a binary operation on Z which is commutative as well as associative.
Answer: Since a + b and ab are always integers when a and b are integers, a + b - ab is also an integer. Thus * is a binary operation on Z. For commutativity, note that a*b = a + b - ab and b*a = b + a - ba. Since addition is commutative (a + b = b + a) and multiplication is commutative (ab = ba), we have a*b = b*a. The operation is commutative. For associativity, compute a*(b*c) and (a*b)*c. First, a*(b*c) = a*(b + c - bc) = a + (b + c - bc) - a(b + c - bc) = a + b + c - bc - ab - ac + abc. Next, (a*b)*c = (a + b - ab)*c = (a + b - ab) + c - (a + b - ab)c = a + b - ab + c - ac - bc + abc. Comparing both results: a + b + c - bc - ab - ac + abc = a + b + c - ab - ac - bc + abc. These are equal, so the operation is associative.
In simple words: All integer combinations give integers, so it's binary. Since all components (addition and multiplication) are commutative, the whole operation is too. The expansion shows both groupings produce the same terms, confirming associativity.

Exam Tip: When verifying associativity algebraically, expand both sides carefully and compare term-by-term to confirm they're identical.

 

Question 4. Consider a binary operation on Q - {1}, defined by a * b = a + b - ab.
(i) Find the identity element in Q - {1}.
(ii) Show that each a ∈ Q - {1} has its inverse.
Answer:
(i) For an identity element e, we need a*e = a for all a in Q - {1}. Setting a*e = a: a + e - ae = a, which simplifies to e - ae = 0, or e(1 - a) = 0. Since a ≠ 1 (the domain excludes 1), we have 1 - a ≠ 0, so e = 0. Verify: a*0 = a + 0 - a(0) = a. ✓ Thus, the identity element is 0.
(ii) For an element b to be the inverse of a, we need a*b = e = 0. Setting a*b = 0: a + b - ab = 0. Solving for b: b(1 - a) = -a, so b = -a/(1 - a) = a/(a - 1). Since a ≠ 1, the denominator is never zero, and b is well-defined. For any a ∈ Q - {1}, the inverse a⁻¹ = a/(a - 1) also lies in Q - {1} (it's a rational number not equal to 1). Therefore, every element in Q - {1} has an inverse.
In simple words: Zero is the identity because a + 0 minus a times 0 just gives a back. Every element a has an inverse found by solving a + b - ab = 0, which yields b = a/(a-1).

Exam Tip: When finding inverses, set a*b = identity and solve for b algebraically. Always verify the inverse stays within the domain of the operation.

 

Question 5. Let Q₀ be the set of all nonzero rational numbers. Let * be a binary operation on Q₀, defined by a * b = ab/2 for all a, b ∈ Q₀.
(i) Show that * is commutative and associative.
(ii) Find the identity element in Q₀.
(iii) Find the inverse of an element a in Q₀.
Answer:
(i) For commutativity, a*b = ab/2 and b*a = ba/2. Since multiplication is commutative (ab = ba), we have a*b = b*a. The operation is commutative. For associativity, compute a*(b*c) and (a*b)*c. We have a*(b*c) = a*(bc/2) = a(bc/2)/2 = abc/4. And (a*b)*c = (ab/2)*c = (ab/2)c/2 = abc/4. Since both equal abc/4, the operation is associative.
(ii) For the identity e, we need a*e = a. Setting a*e = ae/2 = a, we get e/2 = 1, so e = 2. Verify: a*2 = a(2)/2 = a. ✓ The identity element is 2.
(iii) For the inverse b of a, we need a*b = e = 2. Setting ab/2 = 2, we get ab = 4, so b = 4/a. Since a ≠ 0, the inverse a⁻¹ = 4/a is well-defined. Also, 4/a ≠ 0, so the inverse remains in Q₀.
In simple words: The operation multiplies and divides by 2, so order doesn't matter (commutative) and grouping doesn't matter (associative). The identity is 2 because multiplying by 2 and dividing by 2 gives you back a. The inverse of a is 4/a because a times 4/a divided by 2 equals 2.

Exam Tip: For operations involving multiplication and division, identify the identity by working backwards: if a*e = a, then solve the equation mechanically to find e.

 

Question 6. On the set Q⁺ of all positive rational numbers, define an operation * on Q⁺ by a * b = ab/2 for all a, b ∈ Q⁺. Show that
(i) * is a binary operation on Q⁺,
(ii) * is commutative,
(iii) * is associative.
Find the identity element in Q⁺ for *. What is the inverse of a ∈ Q⁺?
Answer:
(i) For any positive rationals a and b, the product ab is positive, and dividing by 2 preserves positivity: ab/2 > 0. Thus the result stays in Q⁺, making * a binary operation on Q⁺. Example: a = 1/2 and b = 2 give a*b = (1/2)(2)/2 = 1/2 ∈ Q⁺.
(ii) For commutativity, a*b = ab/2 and b*a = ba/2. Since multiplication of rationals is commutative, ab = ba, so a*b = b*a. The operation is commutative.
(iii) For associativity, compute a*(b*c) = a*(bc/2) = a(bc/2)/2 = abc/4 and (a*b)*c = (ab/2)*c = (ab/2)c/2 = abc/4. Since both equal abc/4, the operation is associative.
The identity e must satisfy a*e = a. Setting ae/2 = a, we get e = 2. Verify: a*2 = 2a/2 = a. ✓
The inverse of a must satisfy a*b = 2. Setting ab/2 = 2, we get ab = 4, so b = 4/a. For any positive rational a, the value 4/a is also a positive rational, so the inverse a⁻¹ = 4/a exists and lies in Q⁺.
In simple words: Multiplying positive numbers and dividing by 2 keeps them positive (binary). Order doesn't matter for multiplication (commutative). Grouping doesn't matter either (associative). The identity is 2 because a times 2 divided by 2 gives a back. The inverse of a is 4/a because a times 4/a divided by 2 equals 2.

Exam Tip: For operations on positive sets, always verify closure first - ensure the result stays positive. The identity and inverses follow the same algebraic approach as before.

 

Question 7. Let Q+ be the set of all positive rational numbers. (i) Show that the operation * on Q+ defined by \( a * b = \frac{1}{2}(a + b) \) is a binary operation. (ii) Show that * is commutative. (iii) Show that * is not associative.
Answer: (i) An operation functions as a binary operation when \( a * b = \frac{1}{2}(a + b) \) where a, b ∈ Q+. Taking a = 1 and b = 2 as two positive rational numbers: \( a * b = \frac{1}{2}(1 + 2) = \frac{3}{2} \in Q^+ \). Therefore, * qualifies as a binary operation from \( Q^+ \times Q^+ \to Q^+ \).
(ii) For a commutative binary operation, \( a * b = b * a \). Since \( a * b = \frac{1}{2}(a + b) = \frac{1}{2}(b + a) = b * a \), the operation * is a commutative binary operation.
(iii) For an associative binary operation, \( a * (b * c) = (a * b) * c \). Calculating: \( a * (b * c) = a * \frac{1}{2}(b + c) = \frac{1}{2}\left(a - \frac{b + c}{2}\right) = \frac{1}{4}(2a + b + c) \). And: \( (a * b) * c = \frac{1}{2}(a + b) * c = \frac{1}{2}\left(\frac{a + b}{2} + c\right) = \frac{1}{4}(a + b + 2c) \). Since \( a * (b * c) \ne (a * b) * c \), the operation * is not associative.

Exam Tip: To verify commutativity, check both directions with the same operation formula. For associativity, always expand all three operations fully before comparing - a common error is skipping one side.

 

Question 8. Let Q be the set of all rational numbers. Define an operation on Q - {-1} by a * b = a + b + ab. Show that (i) * is a binary operation on Q - {-1}, (ii) * is Commutative, (iii) * is associative, (iv) zero is the identity element in Q - {-1} for *, (v) the inverse of a ∈ Q - {-1} is \( a^{-1} = \frac{-a}{1 + a} \).
Answer: (i) An operation works as a binary operation when \( a * b = a + b + ab \) where a, b ∈ Q - {-1}. Selecting a = 1 and b = \( \frac{-3}{2} \) as two rational numbers: \( a * b = 1 + \frac{-3}{2} + 1 \cdot \frac{-3}{2} = \frac{2 - 3}{2} - \frac{3}{2} = \frac{-1 - 3}{2} = \frac{-4}{2} = -2 \in Q - \{-1\} \). Therefore, * forms a binary operation from \( Q - \{-1\} \times Q - \{-1\} \to Q - \{-1\} \).
(ii) For a commutative binary operation, \( a * b = b * a \). Computing: \( b * a = \frac{-3}{2} + 1 + \frac{-3}{2} \cdot 1 = \frac{-3 + 2}{2} - \frac{3}{2} = \frac{-1 - 3}{2} = \frac{-4}{2} = -2 \in Q - \{-1\} \). Since \( a * b = b * a \), the operation * is a commutative binary operation.
(iii) For an associative binary operation, \( a * (b * c) = (a * b) * c \). First side: \( a + (b * c) = a * (b + c + bc) = a + (b + c + bc) + a(b + c + bc) = a + b + c + bc + ab + ac + abc \). Second side: \( (a * b) * c = (a + b + ab) * c = a + b + ab + c + (a + b + ab)c = a + b + c + ab + ac + bc + abc \). Since \( a * (b * c) = (a * b) * c \), the operation * is associative.
(iv) An identity element e satisfies \( a * e = e * a = a \). Using \( a * e = a \): \( a + e + ae = a \), which gives \( e + ae = 0 \), or \( e(1 + a) = 0 \). Since the operation excludes -1, we have \( a \ne -1 \), so \( e = 0 \). The identity element is thus e = 0.
(v) For a binary operation, the inverse satisfies \( a * b = e = b * a \) where b is the inverse. From \( a * b = 0 \): \( a + b + ab = 0 \), which gives \( b(1 + a) = -a \), so \( b = \frac{-a}{1 + a} \). Therefore, \( a^{-1} = \frac{-a}{1 + a} \).

Exam Tip: Always verify that the result stays within the allowed set Q - {-1}. When finding inverses, isolate the variable algebraically before stating the final inverse formula.

 

Question 9. Let A = N × N. Define * on A by (a, b) * (c, d) = (a + c, b + d). Show that (i) A is closed for *, (ii) * is commutative, (iii) * is associative, (iv) identity element does not exist in A.
Answer: (i) A set is closed under an operation when all results from that operation stay in the set. For (a, b) * (c, d) = (a + c, b + d) where A = N × N, let a = 1, b = 3, c = 8, d = 2. Then: (1, 3) * (8, 2) = (1 + 8, 3 + 2) = (9, 5) ∈ N × N. Since sums of natural numbers remain natural numbers, A is closed for *.
(ii) For commutativity, evaluate (c, d) * (a, b) = (c + a, d + b). Since addition is commutative with \( a + c = c + a \) and \( b + d = d + b \), the operation * is commutative.
(iii) For associativity, check both sides. Left: (a, b) * ((c, d) * (e, f)) = (a, b) * (c + e, d + f) = (a + c + e, b + d + f). Right: ((a, b) * (c, d)) * (e, f) = (a + c, b + d) * (e, f) = (a + c + e, b + d + f). Both equal, so * is associative.
(iv) For an identity element (e₁, e₂), we need (a, b) * (e₁, e₂) = (a, b). This gives (a + e₁, b + e₂) = (a, b), so (e₁, e₂) = (0, 0). However, (0, 0) ∉ N × N because 0 is not a natural number. Therefore, no identity element exists for *.

Exam Tip: Remember that natural numbers typically start from 1, not 0. This is why the identity fails here - check set definitions carefully when discussing identity elements.

 

Question 10. Let A = {1, -1, i, -i} be the set of four 4th roots of unity. Prepare the composition table for multiplication on A and show that (i) A is closed for multiplication, (ii) multiplication is associative on A, (iii) multiplication is commutative on A, (iv) 1 is the multiplicative identity, (v) every element in A has its multiplicative inverse.
Answer: (i) A set is closed under multiplication when all products remain in the set. The composition table for A is:

(Note: \( i^2 = -1 \)). Every entry in the table belongs to set A, confirming that A is closed for multiplication.
(ii) For associativity, test both sides with sample elements. Left: \( 1 \times (-i \times i) = 1 \times 1 = 1 \). Right: \( (1 \times -i) \times i = -i \times i = 1 \). Since both yield the same result and multiplication of complex numbers is inherently associative, A is associative for multiplication.
(iii) For commutativity, check both directions. Example: \( 1 \times -1 = -1 \) and \( -1 \times 1 = -1 \). Looking at the composition table, it is symmetric about the main diagonal, confirming \( a \times b = b \times a \). Thus A is commutative for multiplication.
(iv) For a multiplicative identity e, we need \( a \times e = e \times a = a \) for all a ∈ A. From \( a \times e = a \): \( a(e - 1) = 0 \). Since \( a \ne 0 \), we get \( e = 1 \). Testing: multiplying any element by 1 returns that element. So the multiplicative identity is e = 1.
(v) For each element, find its inverse such that \( a \times b = 1 \). For \( 1 \times b_1 = 1 \): \( b_1 = 1 \). For \( -1 \times b_2 = 1 \): \( b_2 = -1 \). For \( i \times b_3 = 1 \): \( b_3 = \frac{1}{i} = \frac{1}{i} \cdot \frac{i}{i} = \frac{i}{i^2} = \frac{i}{-1} = -i \). For \( -i \times b_4 = 1 \): \( b_4 = \frac{1}{-i} = \frac{1}{-i} \cdot \frac{i}{i} = \frac{i}{-i^2} = \frac{i}{-(-1)} = i \). The multiplicative inverses of A are {1, -1, -i, i}.

Exam Tip: For complex numbers, always rationalize denominators by multiplying by the conjugate. The composition table is a fast way to check closure and commutativity visually - a symmetric table means the operation is commutative.