RS Aggarwal Solutions for Class 11 Chapter 02 Relations

Question 1. Find the values of a and b, when:
(i) (a + 3, b - 2) = (5, 1)
(ii) (a + b, 2b - 3) = (4, -5)
(iii) \( \left( \frac{a}{3} + 1, b - \frac{1}{3} \right) = \left( \frac{5}{3}, \frac{2}{3} \right) \)
(iv) (a - 2, 2b + 1) = (b - 1, a + 2)
Answer: When ordered pairs are equal, their matching elements must be equal.
(i) From a + 3 = 5 and b - 2 = 1, we get a = 2 and b = 3.
(ii) From a + b = 4 and 2b - 3 = -5, we get b = -1. Substituting into the first equation: a - 1 = 4, so a = 5. Thus a = 5 and b = -1.
(iii) From \( \frac{a}{3} + 1 = \frac{5}{3} \) and \( b - \frac{1}{3} = \frac{2}{3} \), we solve: \( \frac{a}{3} = \frac{5}{3} - 1 = \frac{2}{3} \), giving a = 2. And \( b = \frac{2}{3} + \frac{1}{3} = 1 \). Thus a = 2 and b = 1.
(iv) From a - 2 = b - 1 and 2b + 1 = a + 2, we get a - b = 1 and -a + 2b = 1. Adding these: b = 2. Substituting back: a = 3. Thus a = 3 and b = 2.
In simple words: When two ordered pairs match, each part in the same position must be equal. Set up equations from matching parts and solve for the variables using basic algebra.

Exam Tip: Always write the two equations from the matching elements of equal ordered pairs; solve one equation first, then substitute into the second to find both variables.

Question 2. If A = {9, 1} and B = {1, 2, 3}, show that A × B ≠ B × A.
Answer: Given A = {9, 1} and B = {1, 2, 3}. We need to demonstrate that A × B is not equal to B × A.

Using the definition of the Cartesian product, we form all ordered pairs by matching each element from the first set with each element from the second set.

A × B = {(9, 1), (9, 2), (9, 3), (1, 1), (1, 2), (1, 3)}

B × A = {(1, 9), (2, 9), (3, 9), (1, 1), (2, 1), (3, 1)}

Comparing these two sets: the pair (9, 1) appears in A × B but not in B × A. By the definition of ordered pair equality (where corresponding first and second elements must both match), the ordered pair (9, 1) is distinct from (1, 9). Therefore, A × B ≠ B × A. Hence proved.
In simple words: The Cartesian product changes depending on the order of the sets. When you swap A and B, the pairs switch positions, creating a completely different set.

Exam Tip: Always compute both products explicitly and compare element by element; remember that (a, b) ≠ (b, a) in ordered pairs unless a = b.

Question 3. If P = {a, b} and Q = {x, y, z}, show that P × Q ≠ Q × P.
Answer: Given P = {a, b} and Q = {x, y, z}. We must prove that P × Q and Q × P are different.

By the Cartesian product definition, we pair each element from the first set with each element from the second.

P × Q = {(a, x), (a, y), (a, z), (b, x), (b, y), (b, z)}

Q × P = {(x, a), (y, a), (z, a), (x, b), (y, b), (z, b)}

Examining the sets: the pair (a, x) belongs to P × Q but not to Q × P. Since ordered pairs are equal only when corresponding first and second elements are both identical, (a, x) ≠ (x, a). Therefore, the two sets contain different elements and P × Q ≠ Q × P. Hence proved.
In simple words: When you reverse the order of the sets in a Cartesian product, you also reverse the order within each pair. This creates a completely new set of ordered pairs.

Exam Tip: Write out both products fully to visualize the difference; focus on one pair that clearly differs between the two products to establish inequality.

Question 4. If A = {2, 3, 5} and B = {5, 7}, find:
(i) A × B
(ii) B × A
(iii) A × A
(iv) B × B
Answer:
(i) A × B: Pairing each element of A with each element of B gives {(2, 5), (3, 5), (5, 5), (2, 7), (3, 7), (5, 7)}.
(ii) B × A: Pairing each element of B with each element of A gives {(5, 2), (5, 3), (5, 5), (7, 2), (7, 3), (7, 5)}.
(iii) A × A: Pairing each element of A with itself and all other elements gives {(2, 2), (2, 3), (2, 5), (3, 2), (3, 3), (3, 5), (5, 2), (5, 3), (5, 5)}.
(iv) B × B: Pairing each element of B with itself and the other element gives {(5, 5), (5, 7), (7, 5), (7, 7)}.
In simple words: For each Cartesian product, match every element in the first set with every element in the second set, creating all possible ordered pairs.

Exam Tip: Count your pairs: if set A has m elements and set B has n elements, A × B must have exactly m × n pairs; use this to verify you haven't missed or duplicated any.

Question 5. If A = {x ∈ N : x ≤ 3} and B = {x ∈ W : x < 2}, find (A × B) and (B × A). Is (A × B) = (B × A)?
Answer: First, identify the sets. A consists of natural numbers up to and including 3, so A = {1, 2, 3}. B consists of whole numbers less than 2, so B = {0, 1}.

A × B = {(1, 0), (1, 1), (2, 0), (2, 1), (3, 0), (3, 1)}

B × A = {(0, 1), (0, 2), (0, 3), (1, 1), (1, 2), (1, 3)}

Comparing: the pair (1, 0) belongs to A × B but not B × A. By ordered pair equality, (1, 0) ≠ (0, 1). Therefore, (A × B) ≠ (B × A).
In simple words: Natural numbers start from 1, and whole numbers include 0. When you reverse the set order in a Cartesian product, the pairs inside also reverse, making the sets unequal.

Exam Tip: Carefully distinguish between natural numbers (starting at 1) and whole numbers (starting at 0); this distinction affects which elements appear in your sets.

Question 6. If A = {1, 3, 5}, B = {3, 4} and C = {2, 3}, verify that:
(i) A × (B ∪ C) = (A × B) ∪ (A × C)
(ii) A × (B ∩ C) = (A × B) ∩ (A × C)
Answer:
(i) Left side: B ∪ C = {2, 3, 4}, so A × (B ∪ C) = {1, 3, 5} × {2, 3, 4} = {(1, 2), (1, 3), (1, 4), (3, 2), (3, 3), (3, 4), (5, 2), (5, 3), (5, 4)}.

Right side: A × B = {(1, 3), (1, 4), (3, 3), (3, 4), (5, 3), (5, 4)} and A × C = {(1, 2), (1, 3), (3, 2), (3, 3), (5, 2), (5, 3)}. Their union gives {(1, 2), (1, 3), (1, 4), (3, 2), (3, 3), (3, 4), (5, 2), (5, 3), (5, 4)}.

Both sides equal the same set, so the identity is verified.

(ii) Left side: B ∩ C = {3}, so A × (B ∩ C) = {1, 3, 5} × {3} = {(1, 3), (3, 3), (5, 3)}.

Right side: (A × B) ∩ (A × C) selects pairs appearing in both products. From above, both A × B and A × C contain (1, 3), (3, 3), and (5, 3), so their intersection is {(1, 3), (3, 3), (5, 3)}.

Both sides match, confirming the identity.
In simple words: The Cartesian product distributes over union and intersection, just like regular multiplication distributes over addition - multiply A by the result of combining or overlapping B and C, or combine/overlap the two separate products.

Exam Tip: When verifying distributive properties of Cartesian products, compute both sides completely and list all pairs explicitly to ensure no errors in set operations.

Question 7. Let A = {x ∈ W : x < 2}, B = {x ∈ N : 1 < x ≤ 4} and C = {3, 5}. Verify that:
(i) A × (B ∪ C) = (A × B) ∪ (A × C)
(ii) A × (B ∩ C) = (A × B) ∩ (A × C)
Answer:
First, identify the sets. A = {0, 1} (whole numbers less than 2), B = {2, 3, 4} (natural numbers greater than 1 and at most 4), and C = {3, 5}.

(i) Left side: B ∪ C = {2, 3, 4, 5}, so A × (B ∪ C) = {0, 1} × {2, 3, 4, 5} = {(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)}.

Right side: A × B = {(0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (1, 4)} and A × C = {(0, 3), (0, 5), (1, 3), (1, 5)}. Their union is {(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)}.

Both sides are identical, so the identity holds.

(ii) Left side: B ∩ C = {3}, so A × (B ∩ C) = {0, 1} × {3} = {(0, 3), (1, 3)}.

Right side: (A × B) ∩ (A × C) contains pairs in both products. Only (0, 3) and (1, 3) appear in both, giving {(0, 3), (1, 3)}.

Both sides match, confirming the identity.
In simple words: Whole numbers start at 0; natural numbers start at 1. The distributive property works the same way - the product of A with a union or intersection equals the union or intersection of the separate products.

Exam Tip: Always list out which numbers belong to each set before computing products; misidentifying W versus N is a common source of error.

Question 8. If A × B = {(-2, 3), (-2, 4), (0, 4), (3, 3), (3, 4)}, find A and B.
Answer: To recover the original sets from a Cartesian product, collect all first coordinates to form A and all second coordinates to form B.

From the given pairs, the first coordinates are -2, -2, 0, 3, 3. The distinct values are A = {-2, 0, 3}.

The second coordinates are 3, 4, 4, 3, 4. The distinct values are B = {3, 4}.
In simple words: In any ordered pair from a Cartesian product, the first number goes into set A and the second into set B. Remove duplicates to get the original sets.

Exam Tip: Extract all unique first coordinates for set A and all unique second coordinates for set B; use set notation (braces and no repetition) in your final answer.

Question 9. Let A = {2, 3} and B = {4, 5}. Find (A × B). How many subsets will (A × B) have?
Answer: By the Cartesian product definition, A × B = {(2, 4), (2, 5), (3, 4), (3, 5)}.

The set A × B has 4 elements. The number of subsets of any set with n elements is 2^n. Since A × B has 4 elements, the number of subsets is 2^4 = 16. These include the empty set, all single-element subsets, all two-element subsets, all three-element subsets, and the full set itself.
In simple words: For each element in a set, you either include it or exclude it in a subset. With 4 elements, you have 2 choices for each, giving 2 × 2 × 2 × 2 = 16 total subsets.

Exam Tip: Always use the formula 2^n for the number of subsets; verify your count of elements in the Cartesian product before applying this formula.

Question 10. Let A × B = {(a, b) : b = 3a - 2}. If (x, -5) and (2, y) belong to A × B, find the values of x and y.
Answer: Since ordered pairs in A × B satisfy the relation b = 3a - 2, we substitute the given pairs into this equation.

For (x, -5): Setting a = x and b = -5 into b = 3a - 2 gives -5 = 3x - 2. Solving: 3x = -3, so x = -1.

For (2, y): Setting a = 2 and b = y into b = 3a - 2 gives y = 3(2) - 2 = 6 - 2 = 4.

Therefore, x = -1 and y = 4.
In simple words: Use the given formula to set up equations. When you know one coordinate of an ordered pair, substitute it to find the other coordinate using the relationship.

Exam Tip: Substitute carefully and isolate the unknown variable; double-check by plugging your answers back into the original formula to verify they satisfy the relation.

Question 11. Let A and B be two sets such that n(A) = 3 and n(B) = 2. If a ≠ b ≠ c and (a, 0), (b, 1), (c, 0) are in A × B, find A and B.
Answer: From the given ordered pairs in A × B, the first coordinates are a, b, c and the second coordinates are 0, 1, 0. Extracting unique elements: the first coordinates are the three distinct values a, b, c from set A, and the second coordinates are 0 and 1 from set B.

Since n(A) = 3 and the three distinct first coordinates are a, b, c, we have A = {a, b, c}.

Since n(B) = 2 and the second coordinates are 0 and 1 (both appearing in the pairs), we have B = {0, 1}.
In simple words: Extract all first entries from the ordered pairs to build set A and all second entries to build set B. Count them to verify they match the given cardinalities.

Exam Tip: Use the cardinality information (n(A) = 3, n(B) = 2) as a check - the elements you extract from the ordered pairs must match these counts exactly.

Question 12. Let A = {-2, 2} and B = {0, 3, 5}. Find:
(i) A × B
(ii) B × A
(iii) A × A
(iv) B × B
Answer:
(i) A × B: Each element of A pairs with each element of B, giving {(-2, 0), (-2, 3), (-2, 5), (2, 0), (2, 3), (2, 5)}.

(ii) B × A: Each element of B pairs with each element of A, giving {(0, -2), (0, 2), (3, -2), (3, 2), (5, -2), (5, 2)}.

(iii) A × A: Each element of A pairs with all elements of A, giving {(-2, -2), (-2, 2), (2, -2), (2, 2)}.

(iv) B × B: Each element of B pairs with all elements of B, giving {(0, 0), (0, 3), (0, 5), (3, 0), (3, 3), (3, 5), (5, 0), (5, 3), (5, 5)}.
In simple words: List every combination systematically - for each element in the first set, pair it with each element in the second set in order.

Exam Tip: Organize pairs methodically (fix the first element, vary the second) to avoid missing or duplicating pairs; verify your total count using n(P) × n(Q) = number of pairs in P × Q.

Question 13. If A = {5, 7}, find (i) A × A × A.
Answer: First, compute A × A. Pairing elements of A with themselves: A × A = {(5, 5), (5, 7), (7, 5), (7, 7)}.

Now compute A × A × A = (A × A) × A. Each element of A × A (which are ordered pairs) combines with each element of A to form ordered triples:

A × A × A = {(5, 5, 5), (5, 5, 7), (5, 7, 5), (5, 7, 7), (7, 5, 5), (7, 5, 7), (7, 7, 5), (7, 7, 7)}

In simple words: Build A × A first as ordered pairs. Then match each pair with each element of A to create ordered triples. With 2 elements in A, you get 2 × 2 × 2 = 8 triples total.

Exam Tip: For triple products, compute the first product, then treat its result as a set and compute the product with the third set; keep track of whether you have pairs (2 coordinates) or triples (3 coordinates).

Question 14. Let A = {-3, -1}, B = {1, 3} and C = {3, 5}. Find:
(i) A × B
(ii) (A × B) × C
(iii) B × C
(iv) A × (B × C)
Answer:
(i) A × B: Pairing elements of A with elements of B gives {(-3, 1), (-3, 3), (-1, 1), (-1, 3)}.

(ii) (A × B) × C: The set A × B from part (i) consists of ordered pairs. Each pair combines with each element of C = {3, 5} to form ordered triples: {(-3, 1, 3), (-3, 1, 5), (-3, 3, 3), (-3, 3, 5), (-1, 1, 3), (-1, 1, 5), (-1, 3, 3), (-1, 3, 5)}.

(iii) B × C: Pairing elements of B with elements of C gives {(1, 3), (1, 5), (3, 3), (3, 5)}.

(iv) A × (B × C): Each element of A combines with each ordered pair from B × C to form ordered triples: {(-3, 1, 3), (-3, 1, 5), (-3, 3, 3), (-3, 3, 5), (-1, 1, 3), (-1, 1, 5), (-1, 3, 3), (-1, 3, 5)}.

In simple words: When building products with three sets, you can group them two different ways: (A × B) × C or A × (B × C). Even though the grouping differs, the resulting ordered triples are identical, showing associativity.

Exam Tip: Note that (A × B) × C and A × (B × C) produce the same set of triples, demonstrating that Cartesian product is associative - the order of grouping doesn't affect the final result.

Question 1. For any sets A, B and C prove that: A × (B ∪ C) = (A × B) ∪ (A × C)
Answer: Suppose A, B, and C represent three arbitrary sets. We need to establish that A × (B ∪ C) = (A × B) ∪ (A × C).

Take any ordered pair (x, y) ∈ A × (B ∪ C). This means x ∈ A and y ∈ (B ∪ C). Therefore, x ∈ A and (y ∈ B or y ∈ C). Which gives us (x ∈ A and y ∈ B) or (x ∈ A and y ∈ C). So (x, y) ∈ (A × B) or (x, y) ∈ (A × C). Thus (x, y) ∈ (A × B) ∪ (A × C). From this we get A × (B ∪ C) ⊆ (A × B) ∪ (A × C) ---- (1)

Conversely, take any ordered pair (a, b) ∈ (A × B) ∪ (A × C). This means (a, b) ∈ (A × B) or (a, b) ∈ (A × C). So (a ∈ A and b ∈ B) or (a ∈ A and b ∈ C). Therefore, a ∈ A and (b ∈ B or b ∈ C). Which implies a ∈ A and b ∈ (B ∪ C). Thus (a, b) ∈ A × (B ∪ C). From this we get (A × B) ∪ (A × C) ⊆ A × (B ∪ C) ---- (2)

By the definition of set equality, from (1) and (2), A × (B ∪ C) = (A × B) ∪ (A × C) [Proved]

Exam Tip: To prove set equality, always show both subset inclusions - from left to right and from right to left. Use clear logical steps to show membership conditions are identical.

Question 2. For any sets A, B and C prove that: A × (B ∩ C) = (A × B) ∩ (A × C)
Answer: Let A, B, and C be three arbitrary sets. We must demonstrate that A × (B ∩ C) = (A × B) ∩ (A × C).

Take any ordered pair (x, y) ∈ A × (B ∩ C). This implies x ∈ A and y ∈ (B ∩ C). So x ∈ A and (y ∈ B and y ∈ C). Which gives (x ∈ A and y ∈ B) and (x ∈ A and y ∈ C). Therefore (x, y) ∈ (A × B) and (x, y) ∈ (A × C). Thus (x, y) ∈ (A × B) ∩ (A × C). From this we conclude A × (B ∩ C) ⊆ (A × B) ∩ (A × C) ---- (1)

Now take any ordered pair (a, b) ∈ (A × B) ∩ (A × C). This means (a, b) ∈ (A × B) and (a, b) ∈ (A × C). So (a ∈ A and b ∈ B) and (a ∈ A and b ∈ C). Therefore a ∈ A and (b ∈ B and b ∈ C). This gives a ∈ A and b ∈ (B ∩ C). Thus (a, b) ∈ A × (B ∩ C). From this we conclude (A × B) ∩ (A × C) ⊆ A × (B ∩ C) ---- (2)

By the definition of set equality, from (1) and (2), A × (B ∩ C) = (A × B) ∩ (A × C) [Proved]

Exam Tip: When working with Cartesian product proofs, expand the membership conditions carefully - intersection and union have specific meanings for ordered pairs that must be preserved.

Question 3. For any sets A, B and C prove that: A × (B - C) = (A × B) - (A × C)
Answer: Let A, B, and C be three arbitrary sets. We need to show that A × (B - C) = (A × B) - (A × C).

Take any ordered pair (x, y) ∈ A × (B - C). This means x ∈ A and y ∈ (B - C). So x ∈ A and (y ∈ B and y ∉ C). Which gives (x ∈ A and y ∈ B) and (x ∈ A and y ∉ C). Therefore (x, y) ∈ (A × B) and (x, y) ∉ (A × C). Thus (x, y) ∈ (A × B) - (A × C). From this we get A × (B - C) ⊆ (A × B) - (A × C) ---- (1)

Conversely, take any ordered pair (a, b) ∈ (A × B) - (A × C). This means (a, b) ∈ (A × B) and (a, b) ∉ (A × C). So (a ∈ A and b ∈ B) and (a ∈ A and b ∉ C). Therefore a ∈ A and (b ∈ B and b ∉ C). This gives a ∈ A and b ∈ (B - C). Thus (a, b) ∈ A × (B - C). From this we get (A × B) - (A × C) ⊆ A × (B - C) ---- (2)

By the definition of set equality, from (1) and (2), A × (B - C) = (A × B) - (A × C) [Proved]

Exam Tip: Set difference requires careful handling - an element must be in the first set but explicitly NOT in the second set. Apply this condition uniformly to both coordinates.

Question 4. For any sets A and B, prove that (A × B) ∩ (B × A) = (A ∩ B) × (B ∩ A)
Answer: Suppose A and B are two arbitrary sets. We must establish that (A × B) ∩ (B × A) = (A ∩ B) × (B ∩ A).

Take any ordered pair (x, y) ∈ (A × B) ∩ (B × A). This means (x, y) ∈ (A × B) and (x, y) ∈ (B × A). So (x ∈ A and y ∈ B) and (x ∈ B and y ∈ A). Therefore (x ∈ A and x ∈ B) and (y ∈ B and y ∈ A). Which gives x ∈ (A ∩ B) and y ∈ (B ∩ A). Thus (x, y) ∈ (A ∩ B) × (B ∩ A). From this we conclude (A × B) ∩ (B × A) ⊆ (A ∩ B) × (B ∩ A) ---- (1)

Conversely, take any ordered pair (a, b) ∈ (A ∩ B) × (B ∩ A). This means a ∈ (A ∩ B) and b ∈ (B ∩ A). So (a ∈ A and a ∈ B) and (b ∈ B and b ∈ A). Therefore (a ∈ A and b ∈ B) and (a ∈ B and b ∈ A). Which gives (a, b) ∈ (A × B) and (a, b) ∈ (B × A). Thus (a, b) ∈ (A × B) ∩ (B × A). From this we conclude (A ∩ B) × (B ∩ A) ⊆ (A × B) ∩ (B × A) ---- (2)

By the definition of set equality, from (1) and (2), (A × B) ∩ (B × A) = (A ∩ B) × (B ∩ A) [Proved]

Exam Tip: Notice how both coordinates must satisfy different set conditions - the first element must be in both A and B, while the second must be in both B and A. Track these carefully.

Question 5. If A and B are nonempty sets, prove that A × B = B × A ⟺ A = B
Answer: Suppose A and B represent two nonempty sets. We need to prove the biconditional: A × B = B × A if and only if A = B.

First direction (A = B ⟹ A × B = B × A): Given that A = B, take any ordered pair (x, y) ∈ (A × B). This means x ∈ A and y ∈ B. Since A = B, we can write x ∈ B and y ∈ A. Therefore (x, y) ∈ (B × A). So (A × B) ⊆ (B × A). By similar reasoning, (B × A) ⊆ (A × B). Thus A × B = B × A.

Second direction (A × B = B × A ⟹ A = B): Given that A × B = B × A, we show A = B. Take any element x ∈ A. Since A is nonempty, there exists some element y ∈ B (using nonemptiness). Then (x, y) ∈ (A × B). Since A × B = B × A, we have (x, y) ∈ (B × A). This means x ∈ B and y ∈ A. So every element of A is in B, giving A ⊆ B. Similarly, we can show B ⊆ A. Therefore A = B. [Proved]

Exam Tip: The nonemptiness condition is crucial here - it ensures we can find elements to use in forming ordered pairs. Without it, the second direction would fail for empty sets.

Question 6. (i) If A ⊆ B, prove that A × C ⊆ B × C for any set C.
(ii) If A ⊆ B and C ⊆ D then prove that A × C ⊆ B × D.
Answer:
(i) Suppose A ⊆ B and C is any arbitrary set. We must show A × C ⊆ B × C. Take any ordered pair (x, y) ∈ (A × C). This means x ∈ A and y ∈ C. Since A ⊆ B and x ∈ A, we know x must be in B. Therefore x ∈ B and y ∈ C, which gives (x, y) ∈ (B × C). Thus A × C ⊆ B × C. [Proved]

(ii) Suppose A ⊆ B and C ⊆ D. We must establish that A × C ⊆ B × D. Take any ordered pair (x, y) ∈ (A × C). This means x ∈ A and y ∈ C. Since A ⊆ B, we have x ∈ B. Since C ⊆ D, we have y ∈ D. Therefore x ∈ B and y ∈ D, which implies (x, y) ∈ (B × D). Thus A × C ⊆ B × D. [Proved]

Exam Tip: Subset relationships propagate through Cartesian products in a natural way - when both components satisfy subset conditions, the product satisfies the subset condition.

Question 7. If A × B ⊆ C × D and A × B ≠ ∅, prove that A ⊆ C and B ⊆ D.
Answer: Suppose A × B ⊆ C × D and A × B ≠ ∅. We need to show both A ⊆ C and B ⊆ D. Take any element x ∈ A. Since A × B is nonempty, there exists some element y ∈ B. Then the ordered pair (x, y) ∈ (A × B). Since A × B ⊆ C × D, we have (x, y) ∈ (C × D). This means x ∈ C and y ∈ D. Since this holds for any x ∈ A, we get A ⊆ C. Similarly, since y ∈ D for every y ∈ B (by choosing any x ∈ A to form the pair), we get B ⊆ D. [Proved]

Exam Tip: The nonemptiness condition is essential - it guarantees that for each element in A, we can find a corresponding element in B to form a pair in A × B, allowing us to extract information about both sets.

Question 8. If A and B be two sets such that n(A) = 3, n(B) = 4 and n(A ∩ B) = 2 then find:
(i) n(A × B)
(ii) n(B × A)
(iii) n((A × B) ∩ (B × A))
Answer:
(i) The cardinality of A × B is given by the product formula: n(A × B) = n(A) × n(B) = 3 × 4 = 12

(ii) The cardinality of B × A follows the same principle: n(B × A) = n(B) × n(A) = 4 × 3 = 12

(iii) We use the formula n((A × B) ∩ (B × A)) = n((A ∩ B) × (B ∩ A)) = n(A ∩ B) × n(B ∩ A). Since n(A ∩ B) = 2 and n(B ∩ A) = n(A ∩ B) = 2, we get n((A × B) ∩ (B × A)) = 2 × 2 = 4

Exam Tip: Remember that n((A × B) ∩ (B × A)) = n((A ∩ B) × (A ∩ B)) when both intersection sets are equal. The cardinality of a Cartesian product is always the product of individual cardinalities.

Question 9. For any two sets A and B, show that A × B and B × A have an element in common if and only if A and B have an element in common.
Answer: We know from set theory that (A × B) ∩ (B × A) = (A ∩ B) × (B ∩ A). Now suppose A and B have at least one common element, so n(A ∩ B) ≥ 1. Then n((A ∩ B) × (B ∩ A)) = n(A ∩ B) × n(B ∩ A) ≥ 1 × 1 = 1. This means (A × B) ∩ (B × A) is nonempty, so A × B and B × A share at least one element. Conversely, if (A × B) ∩ (B × A) is nonempty, then (A ∩ B) × (B ∩ A) is nonempty, which requires n(A ∩ B) ≥ 1. Therefore A and B must have at least one element in common. [Proved]

Exam Tip: This elegant result shows the direct correspondence between element overlap in A and B versus element overlap in their Cartesian products. The intersection property of Cartesian products is the key insight.

Question 10. Let A = {1, 2} and B = {2, 3}. Then, write down all possible subsets of A × B.
Answer: Given A = {1, 2} and B = {2, 3}, we first find A × B. The Cartesian product is A × B = {(1, 2), (1, 3), (2, 2), (2, 3)}. This set has 4 elements, so it has \( 2^4 = 16 \) subsets. The complete list of all possible subsets is: ∅, {(1, 2)}, {(1, 3)}, {(2, 2)}, {(2, 3)}, {(1, 2), (1, 3)}, {(1, 2), (2, 2)}, {(1, 2), (2, 3)}, {(1, 3), (2, 2)}, {(1, 3), (2, 3)}, {(2, 2), (2, 3)}, {(1, 2), (1, 3), (2, 2)}, {(1, 2), (1, 3), (2, 3)}, {(1, 2), (2, 2), (2, 3)}, {(1, 3), (2, 2), (2, 3)}, {(1, 2), (1, 3), (2, 2), (2, 3)}

Exam Tip: The number of subsets of any finite set with n elements is always \( 2^n \). Always include the empty set and the entire set itself as subsets.

Question 11. Let A = {a, b, c, d}, B = {c, d, e} and C = {d, e, f, g}. Then verify each of the following identities:
(i) A × (B ∩ C) = (A × B) ∩ (A × C)
(ii) A × (B - C) = (A × B) - (A × C)
(iii) (A × B) ∩ (B × A) = (A ∩ B) × (A ∩ B)
Answer:
(i) Finding left side: B ∩ C = {d, e}, so A × (B ∩ C) = {(a, d), (a, e), (b, d), (b, e), (c, d), (c, e), (d, d), (d, e)}. Finding right side: A × B = {(a, c), (a, d), (a, e), (b, c), (b, d), (b, e), (c, c), (c, d), (c, e), (d, c), (d, d), (d, e)} and A × C = {(a, d), (a, e), (a, f), (a, g), (b, d), (b, e), (b, f), (b, g), (c, d), (c, e), (c, f), (c, g), (d, d), (d, e), (d, f), (d, g)}. So (A × B) ∩ (A × C) = {(a, d), (a, e), (b, d), (b, e), (c, d), (c, e), (d, d), (d, e)}. Both sides are equal. [Verified]

(ii) Finding left side: B - C = {c}, so A × (B - C) = {(a, c), (b, c), (c, c), (d, c)}. Finding right side: (A × B) - (A × C) = {(a, c), (b, c), (c, c), (d, c)}. Both sides are equal. [Verified]

(iii) Finding left side: B × A = {(c, a), (c, b), (c, c), (c, d), (d, a), (d, b), (d, c), (d, d), (e, a), (e, b), (e, c), (e, d)}. So (A × B) ∩ (B × A) = {(c, c), (c, d), (d, c), (d, d)}. Finding right side: A ∩ B = {c, d}, so (A ∩ B) × (A ∩ B) = {(c, c), (c, d), (d, c), (d, d)}. Both sides are equal. [Verified]

Exam Tip: When verifying identities with concrete sets, compute each side completely and carefully. Write out the elements systematically to avoid missing or duplicating pairs.

Question 1. Let A and B be two nonempty sets.
(i) What do you mean by a relation from A to B?
(ii) What do you mean by the domain and range of a relation?
Answer:
(i) A relation from set A to set B is any subset of the Cartesian product A × B. If R denotes such a relation, then R ⊆ (A × B). This means every element of R is an ordered pair (x, y) where x ∈ A and y ∈ B. When (x, y) belongs to R, we write (x, y) ∈ R or xRy, indicating that x stands in relation R to y.

(ii) For a relation R from A to B, the Domain is the set consisting of all first components of the ordered pairs in R, written as Dom(R) = {x: (x, y) ∈ R}. The Range is the set containing all second components of the ordered pairs in R, written as Range(R) = {y: (x, y) ∈ R}. Note that the domain is always a subset of A, and the range is always a subset of B.

Exam Tip: Every element in the domain must appear as the first component of at least one ordered pair, and every element in the range must appear as the second component of at least one ordered pair in the relation.

Question 2. Find the domain and range of each of the relations given below:
(i) R = {(-1, 1), (1, 1), (-2, 4), (2, 4), (2, 4), (3, 9)}
(ii) R = {(x, 1/x) : x is an integer, 0 < x < 5}
(iii) R = {(x, y) : x + 2y = 8 and x, y ∈ N}
(iv) R = {(x, y) : y = |x - 1|, x ∈ Z and |x| ≤ 3}
Answer:
(i) Given R = {(-1, 1), (1, 1), (-2, 4), (2, 4), (2, 4), (3, 9)}, the domain contains all first components: Dom(R) = {-2, -1, 1, 2, 3}. The range contains all second components: Range(R) = {1, 4, 9}.

(ii) The relation is R = {(1, 1), (2, 1/2), (3, 1/3), (4, 1/4)}. Therefore Dom(R) = {1, 2, 3, 4} and Range(R) = {1, 1/2, 1/3, 1/4}.

(iii) For x + 2y = 8 with x, y ∈ N, we get pairs R = {(2, 3), (4, 2), (6, 1)}. So Dom(R) = {2, 4, 6} and Range(R) = {1, 2, 3}.

(iv) With y = |x - 1| and |x| ≤ 3, we have x ∈ {-3, -2, -1, 0, 1, 2, 3}. Computing: x = -3 gives y = 4; x = -2 gives y = 3; x = -1 gives y = 2; x = 0 gives y = 1; x = 1 gives y = 0; x = 2 gives y = 1; x = 3 gives y = 2. Therefore Dom(R) = {-3, -2, -1, 0, 1, 2, 3} and Range(R) = {0, 1, 2, 3, 4}.

Exam Tip: Always list domain and range elements in ascending order. Check that every first component goes into the domain and every second component goes into the range, with no omissions.

Question 3. Let A = {1, 3, 5, 7} and B = {2, 4, 6, 8}. Let R = {(x, y) : x ∈ A, y ∈ B and x > y}.
(i) Write R in roster form.
(ii) Find Dom(R) and Range(R).
(iii) Depict R by an arrow diagram.
Answer:
(i) For x > y with x ∈ A and y ∈ B: 3 > 2; 5 > 2, 4; 7 > 2, 4, 6. So R = {(3, 2), (5, 2), (5, 4), (7, 2), (7, 4), (7, 6)}.

(ii) Dom(R) = {3, 5, 7} and Range(R) = {2, 4, 6}.

(iii) The arrow diagram shows:

Exam Tip: In an arrow diagram, draw arrows from each first component to its corresponding second component. Multiple arrows from one element are acceptable when that element relates to several others.

Question 4. Let A = {2, 4, 5, 7} and B = {1, 2, 3, 4, 5, 6, 7, 8}. Let R = {(x, y) : x ∈ A, y ∈ B and x divides y}.
(i) Write R in roster form.
(ii) Find Dom(R) and Range(R).
Answer:
(i) For x divides y with the given sets: 2 divides 2, 4, 6, 8; 4 divides 4, 8; 5 divides 5; 7 divides 7. So R = {(2, 2), (2, 4), (2, 6), (2, 8), (4, 4), (4, 8), (5, 5), (7, 7)}.

(ii) Dom(R) = {2, 4, 5, 7} and Range(R) = {2, 4, 5, 6, 7, 8}.

Exam Tip: The divisibility relation (x divides y means y = kx for some positive integer k) is transitive and reflexive. Always list range elements without repetition and in order.

Question 5. Let A = {2, 3, 4, 5} and B = {3, 6, 7, 10}. Let R = {(x, y) : x ∈ A, y ∈ B and x is relatively prime to y}.
(i) Write R in roster form.
(ii) Find Dom(R) and Range(R).
Answer:
(i) Two numbers are relatively prime when gcd(x, y) = 1. Checking pairs: 2 is relatively prime to 3, 7; 3 is relatively prime to 7, 10; 4 is relatively prime to 3, 7; 5 is relatively prime to 3, 6, 7. So R = {(2, 3), (2, 7), (3, 7), (3, 10), (4, 3), (4, 7), (5, 3), (5, 6), (5, 7)}.

(ii) Dom(R) = {2, 3, 4, 5} and Range(R) = {3, 6, 7, 10}.

Exam Tip: Two integers are relatively prime (or coprime) if their greatest common divisor is 1. This means they share no common factors other than 1.

Question 6. Let A = {1, 2, 3, 5} and B = {4, 6, 9}. Let R = {(x, y) : x ∈ A, y ∈ B and (x - y) is odd}. Write R in roster form.
Answer: The difference (x - y) is odd when one of x or y is even and the other is odd. From A: 1 is odd, 2 is even, 3 is odd, 5 is odd. From B: 4 is even, 6 is even, 9 is odd. Pairing odd from A with even from B, or even from A with odd from B: (1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6). So R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}.

Exam Tip: A difference is odd if and only if the two numbers have different parities (one is even and one is odd). Use parity checks to quickly identify valid pairs.

Question 7. Let R = {(x, y) : x + 3y = 12, x ∈ N and y ∈ N}.
(i) Write R in roster form.
(ii) Find Dom(R) and Range(R).
Answer:
(i) From x + 3y = 12 with x, y ∈ N: If y = 1, then x = 9; if y = 2, then x = 6; if y = 3, then x = 3; if y = 4, then x = 0 (not in N). So R = {(3, 3), (6, 2), (9, 1)}.

(ii) Dom(R) = {3, 6, 9} and Range(R) = {1, 2, 3}.

Exam Tip: When finding relations defined by linear equations, substitute natural number values systematically for one variable and solve for the other, keeping only pairs where both coordinates are natural numbers.

Question 8. Let A = {1, 2, 3, 4, 5, 6}. Define a relation R from A to A by R = {(x, y) : y = x + 1}.
(i) Write R in roster form.
(ii) Find Dom(R) and Range(R).
(iii) What is its co-domain?
(iv) Depict R by using arrow diagram.
Answer:
(i) For y = x + 1 with x, y ∈ A: x = 1 gives y = 2; x = 2 gives y = 3; x = 3 gives y = 4; x = 4 gives y = 5; x = 5 gives y = 6; x = 6 gives y = 7 (not in A). So R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}.

(ii) Dom(R) = {1, 2, 3, 4, 5} and Range(R) = {2, 3, 4, 5, 6}.

(iii) The co-domain is the set A itself, since R is defined from A to A. So CoD(R) = {1, 2, 3, 4, 5, 6}.

(iv) The arrow diagram shows:

Exam Tip: The co-domain is the target set specified in the definition, while the range is the actual set of second components achieved. Here, Range(R) ⊂ CoD(R) since 1 is in the co-domain but not the range.

Question 9. Let R = {(x, x + 5) : x ∈ {0, 1, 2, 3, 4, 5}}.
(i) Write R in roster form.
(ii) Find Dom(R) and Range(R).
Answer:
(i) Computing pairs for each x value: x = 0 gives (0, 5); x = 1 gives (1, 6); x = 2 gives (2, 7); x = 3 gives (3, 8); x = 4 gives (4, 9); x = 5 gives (5, 10). So R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}.

(ii) Dom(R) = {0, 1, 2, 3, 4, 5} and Range(R) = {5, 6, 7, 8, 9, 10}.

Exam Tip: When the relation is defined as {(x, f(x)) : x ∈ S}, the domain is always the set S itself, and the range is the image of S under the function f.

Question 10. Let A = {1, 2, 3, 4, 6} and R = {(a, b) : a, b ∈ A, and a divides b}.
(i) Write R in roster form.
(ii) Find Dom(R) and Range(R).
(iii) What is its co-domain?
(iv) Depict R by using arrow diagram.
Answer:
(i) Checking divisibility: 1 divides all elements; 2 divides 2, 4, 6; 3 divides 3, 6; 4 divides 4; 6 divides 6. So R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)}.

(ii) Dom(R) = {1, 2, 3, 4, 6} and Range(R) = {1, 2, 3, 4, 6}.

(iii) The co-domain is A = {1, 2, 3, 4, 6}. In this case, Range(R) = CoD(R).

(iv) The arrow diagram shows:

Exam Tip: When depicting relations on the same set, arrows often form loops (reflexivity) and chains (transitivity). The divisibility relation is both reflexive (every number divides itself) and transitive (if a divides b and b divides c, then a divides c).

Question 1. What do you mean by a binary relation on a set A? Define the domain and range of relation on A.
Answer: A binary relation on set A is any subset of the Cartesian product A × A, where A × A is formed by multiplying A with itself. Consider A = {4, 5, 6} and R = {(4, 5), (6, 4), (5, 6)}. Here, R qualifies as a binary relation on A. The domain of R consists of all the first coordinates from the ordered pairs, so Dom(R) = {4, 6, 5}. The range of R is made up of all the second coordinates from the ordered pairs, giving Range(R) = {5, 4, 6}.
In simple words: A binary relation is just a collection of ordered pairs picked from A × A. The domain lists all the first numbers, and the range lists all the second numbers in those pairs.

Exam Tip: Always distinguish between domain (first coordinates) and range (second coordinates) - this is a key concept tested frequently.

Question 2. Let A = {2, 3, 5} and R = {(2, 3), (2, 5), (3, 3), (3, 5)}. Show that R is a binary relation on A. Find its domain and range.
Answer: Start by finding A × A. We get A × A = {(2, 2), (2, 3), (2, 5), (3, 2), (3, 3), (3, 5), (5, 2), (5, 3), (5, 5)}. Since R is contained within A × A, it qualifies as a binary relation on A. The domain is formed from all the first coordinates in R, giving Dom(R) = {2, 3}. The range comes from collecting all second coordinates, so Range(R) = {3, 5}.
In simple words: R is a binary relation because all its pairs come from A × A. The domain is {2, 3} and the range is {3, 5}.

Exam Tip: To verify a relation is binary on set A, always confirm that R ⊆ A × A before identifying domain and range.

Question 3. Let A = {0, 1, 2, 3, 4, 5, 6, 7, 8} and let R = {(a, b) : a, b ∈ A and 2a + 3b = 12}. Express R as a set of ordered pairs. Show that R is a binary relation on A. Find its domain and range.
Answer: We need to find all values where 2a + 3b = 12 with a, b from A. Testing: when a = 0, b = 4; when a = 3, b = 2; when a = 6, b = 0. Thus R = {(0, 4), (3, 2), (6, 0)}. Since every element in R comes from A × A, R is a binary relation on A. The domain is the set of all first coordinates, which is Dom(R) = {0, 3, 6}. The range is the set of all second coordinates, giving Range(R) = {4, 2, 0}.
In simple words: We substitute different values for a and find matching b values. R is binary because all pairs belong to A × A. Domain has the first numbers, range has the second numbers.

Exam Tip: When finding relations from equations, systematically test each possible value to ensure you capture all valid pairs.

Question 4. If R is a binary relation on a set A define R⁻¹ on A. Let R = {(a, b) : a, b ∈ W and 3a + 2b = 15}, where W is the set of whole numbers. Express R and R⁻¹ as sets of ordered pairs. Show that (i) dom (R) = range (R⁻¹) (ii) range (R) = dom (R⁻¹)
Answer: From 3a + 2b = 15: when a = 1, b = 6; when a = 3, b = 3; when a = 5, b = 0. So R = {(1, 6), (3, 3), (5, 0)}. The inverse relation R⁻¹ is formed by switching the coordinates: R⁻¹ = {(6, 1), (3, 3), (0, 5)}. For R, the domain is Dom(R) = {1, 3, 5} and the range is Range(R) = {6, 3, 0}. For R⁻¹, the domain is Dom(R⁻¹) = {6, 3, 0} and the range is Range(R⁻¹) = {1, 3, 5}. We observe that dom (R) = {1, 3, 5} = range (R⁻¹) and range (R) = {6, 3, 0} = dom (R⁻¹).
In simple words: R⁻¹ swaps every pair in R - if (a, b) is in R, then (b, a) is in R⁻¹. The domain of R becomes the range of R⁻¹, and vice versa.

Exam Tip: Remember that reversing the inverse relation returns the original relation - this property is fundamental to understanding inverse relations.

Question 5. What is an equivalence relation? Show that the relation of 'similarity' on the set S of all triangles in a plane is an equivalence relation.
Answer: An equivalence relation possesses three key properties: reflexivity, symmetry, and transitivity. Reflexivity means (a, a) belongs to R for every a in A. Symmetry means if (a, b) belongs to R, then (b, a) also belongs to R. Transitivity means if (a, b) belongs to R and (b, c) belongs to R, then (a, c) belongs to R. For the similarity relation on triangles: every triangle is similar to itself, so it is reflexive. If one triangle is similar to another triangle, then that second triangle must be similar to the first, making it symmetric. If triangle A is similar to triangle B and triangle B is similar to triangle C, then A must be similar to C, making it transitive. Therefore, similarity is an equivalence relation on the set of all triangles.
In simple words: An equivalence relation must satisfy three conditions: each element relates to itself, if A relates to B then B relates to A, and if A relates to B and B relates to C then A relates to C. Triangle similarity meets all three conditions.

Exam Tip: Always check all three properties - reflexivity, symmetry, and transitivity - to confirm if a relation is an equivalence relation.

Question 6. Let R = {(a, b) : a, b ∈ Z and (a - b) is even}. Then, show that R is an equivalence relation on Z.
Answer: (i) Reflexivity: For any a in Z, we have a - a = 0, which is even. Thus (a, a) belongs to R for all a in Z, making it reflexive.

(ii) Symmetry: Let (a, b) belong to R. This means a - b is even. We can write -(b - a) as even, which gives us (b - a) is even. Therefore (b, a) belongs to R, making it symmetric.

(iii) Transitivity: Let (a, b) belong to R and (b, c) belong to R. Then (a - b) is even and (b - c) is even. Adding these: [(a - b) + (b - c)] = (a - c), which is even. Thus (a, c) belongs to R, making it transitive.

Since R has all three properties - reflexivity, symmetry, and transitivity - it is an equivalence relation on Z.
In simple words: The difference between any two integers is even just when both are even or both are odd. This relation works on itself (reflexive), swaps properly (symmetric), and chains correctly (transitive).

Exam Tip: When proving equivalence relations, structure your answer clearly with the three properties labeled separately - examiners expect to see all three verified.

Question 7. Let A = {1, 2, 3} and R = {(a, b) : a, b ∈ A and |a² - b²| ≤ 5}. Write R as a set of ordered pairs. Mention whether R is (i) reflexive (ii) symmetric (iii) transitive. Give reason in each case.
Answer: Testing all pairs: |1² - 1²| = 0 ≤ 5, so (1, 1) ✓; |1² - 2²| = 3 ≤ 5, so (1, 2) ✓; |1² - 3²| = 8 > 5, so (1, 3) ✗; |2² - 1²| = 3 ≤ 5, so (2, 1) ✓; |2² - 2²| = 0 ≤ 5, so (2, 2) ✓; |2² - 3²| = 5 ≤ 5, so (2, 3) ✓; |3² - 1²| = 8 > 5, so (3, 1) ✗; |3² - 2²| = 5 ≤ 5, so (3, 2) ✓; |3² - 3²| = 0 ≤ 5, so (3, 3) ✓.

Thus R = {(1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 2), (3, 3)}.

(i) Reflexivity: For (a, a), we get |a² - a²| = 0 ≤ 5. All pairs (a, a) are in R, so it is reflexive.

(ii) Symmetry: If (a, b) is in R, then |a² - b²| ≤ 5. Since |b² - a²| = |a² - b²| ≤ 5, we have (b, a) in R. Thus it is symmetric.

(iii) Transitivity: Testing a = 1, b = 2, c = 3: |1² - 2²| = 3 ≤ 5 ✓ and |2² - 3²| = 5 ≤ 5 ✓, but |1² - 3²| = 8 > 5 ✗. So (1, 3) is not in R. Therefore it is not transitive.
In simple words: The relation is reflexive because each number stays within 5 units of its square from itself. It is symmetric because the absolute difference works both ways. It fails transitivity because two short jumps can land you outside the boundary.

Exam Tip: For transitivity, always test cases where a, b, c are different - transitivity often fails even when reflexivity and symmetry hold.

Question 8. Let R = {(a, b) : a, b ∈ Z and b = 2a - 4}. If (a, -2) ∈ R and (4, b²) ∈ R, then write the values of a and b.
Answer: Using the relation b = 2a - 4: When b = -2, we get -2 = 2a - 4, so 2a = 2, giving a = 1. When a = 4, we get b = 2(4) - 4 = 4. Therefore a = 1 and b = 4.
In simple words: Substitute the known values into the relation equation. First find a when b = -2, then find b when a = 4.

Exam Tip: When a relation is defined by an equation, always use substitution to find missing values - this is the most direct approach.

Question 9. Let R be a relation on Z, defined by (x, y) ∈ R ↔ x² + y² = 9. Then, write R as a set of ordered pairs. What is its domain?
Answer: We need integer pairs satisfying x² + y² = 9. Testing: x = 0 gives y = 3; x = 3 gives y = 0; x = 0 gives y = -3; x = -3 gives y = 0. We can only use integral values of x and y that satisfy the equation. Thus R = {(0, 3), (3, 0), (0, -3), (-3, 0)}. The domain of R consists of the first coordinates from all ordered pairs, so Dom(R) = {-3, 0, 3}.
In simple words: We find all whole number pairs whose squares add to 9. The domain is just the set of first numbers from these pairs.

Exam Tip: When finding relations on integers with equations, test positive and negative values systematically to capture all solutions.

Question 10. Let A be the set of first five natural numbers and let R be a relation on A, defined by (x, y) ∈ R ↔ x ≤ y. Express R and R⁻¹ as sets of ordered pairs. Find: dom (R⁻¹) and range (R).
Answer: A = {1, 2, 3, 4, 5}. Since x ≤ y, we list all pairs where the first element is less than or equal to the second: R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5), (3, 3), (3, 4), (3, 5), (4, 4), (4, 5), (5, 5)}. The domain of R is Dom(R) = {1, 2, 3, 4, 5} and the range is Range(R) = {1, 2, 3, 4, 5}. For the inverse R⁻¹, we reverse all pairs to get pairs where x ≥ y. The domain of R⁻¹ is Dom(R⁻¹) = {1, 2, 3, 4, 5} and range(R) = {1, 2, 3, 4, 5}.
In simple words: R contains all pairs where the first number is at most the second. R⁻¹ flips these pairs, so it contains pairs where the first number is at least the second.

Exam Tip: Remember that dom(R) = range(R⁻¹) and range(R) = dom(R⁻¹) - this relationship always holds for inverse relations.

Question 11. Let R = {(x, y) : x, y ∈ Z and x² + y² = 25}. Express R and R⁻¹ as sets of ordered pairs. Show that R = R⁻¹.
Answer: Finding integer pairs: x = 0, y = 5 gives 0 + 25 = 25 ✓; x = 3, y = 4 gives 9 + 16 = 25 ✓. Including negative values: R = {(0, 5), (0, -5), (5, 0), (-5, 0), (3, 4), (-3, 4), (-3, -4), (3, -4)}. The inverse relation R⁻¹ is formed by swapping coordinates in each pair. Notice that when we swap coordinates, we get the same pairs back: (0, 5) becomes (5, 0), which is already in R; (5, 0) becomes (0, 5), which is already in R; and so on. Therefore R⁻¹ = R, showing that this relation equals its own inverse.
In simple words: The pairs in R are symmetric around the origin - when you swap the coordinates of any pair, you get another pair that is already in R. That is why R equals R⁻¹.

Exam Tip: When x² + y² = c for some constant, the relation is symmetric, so R often equals R⁻¹ - look for this pattern.

Question 12. Find R⁻¹, when (i) R = {(1, 2), (1, 3), (2, 3), (3, 2), (4, 5)} (ii) R = {(x, y) : x, y ∈ N, x + 2y = 8}.
Answer: (i) To find R⁻¹, we swap the coordinates of each pair in R. Thus R⁻¹ = {(2, 1), (3, 1), (3, 2), (2, 3), (5, 4)}.

(ii) We need natural number pairs satisfying x + 2y = 8. Rearranging: x = 8 - 2y. Testing: y = 1 gives x = 6; y = 2 gives x = 4; y = 3 gives x = 2. So R = {(6, 1), (4, 2), (2, 3)}. The inverse relation is R⁻¹ = {(1, 6), (2, 4), (3, 2)}.
In simple words: For part (i), just flip each pair. For part (ii), find all x, y pairs from the equation, then reverse them.

Exam Tip: Always verify your inverse by checking that domain and range swap: dom(R) should equal range(R⁻¹).

Question 13. Let A = {a, b}. List all relations on A and find their number.
Answer: Any relation on A is a subset of A × A. First, compute A × A = {(a, a), (a, b), (b, a), (b, b)}. This set has 4 elements. Every relation is a subset of this 4-element set. The possible subsets are:

1. {} (empty set)
2. {(a, a)}
3. {(a, b)}
4. {(a, a), (a, b)}
5. {(b, a)}
6. {(b, b)}
7. {(b, a), (b, b)}
8. {(a, a), (b, a)}
9. {(a, b), (b, a)}
10. {(a, a), (b, a), (b, b)}
11. {(a, a), (b, b)}
12. {(a, a), (a, b), (b, a)}
13. {(a, a), (a, b), (b, b)}
14. {(a, b), (b, a), (b, b)}
15. {(a, a), (a, b), (b, a), (b, b)}

The total number of relations is 16, which equals 2⁴ since A × A has 4 elements, and the number of subsets of an n-element set is 2ⁿ.
In simple words: Every relation is built by choosing which pairs from A × A to include. Since A × A has 4 pairs, and we can either include or exclude each pair, we get 2 × 2 × 2 × 2 = 16 total relations.

Exam Tip: The number of relations on a set with n elements is 2^(n²) - this formula comes from the fact that A × A has n² pairs, and each is either in or out of the relation.

Question 14. Let R = {(a, b) : a, b ∈ N and a < b}. Show that R is a binary relation on N, which is neither reflexive nor symmetric. Show that R is transitive.
Answer: N = {1, 2, 3, 4, 5, 6, 7, ...}. The relation R = {(1, 2), (1, 3), (1, 4), ..., (2, 3), (2, 4), (2, 5), ...} consists of pairs where the first element is strictly less than the second.

For reflexivity: A relation is reflexive if (a, a) belongs to R for all a in N. However, a < b requires the two coordinates to be unequal, so (a, a) is never in R. Therefore R is not reflexive.

For symmetry: A relation is symmetric if (a, b) in R implies (b, a) in R. If (a, b) is in R, then a < b. But (b, a) would require b < a, which contradicts a < b. So (b, a) is not in R. Therefore R is not symmetric.

For transitivity: A relation is transitive if (a, b) in R and (b, c) in R together imply (a, c) in R. Let a < b and b < c. Then by the transitivity of the less-than relation on natural numbers, we must have a < c. So (a, c) is in R. Therefore R is transitive.
In simple words: This relation fails reflexivity because no number is less than itself. It fails symmetry because if one number is less than another, that second number cannot be less than the first. It succeeds at transitivity because if A is less than B and B is less than C, then A must be less than C.

Exam Tip: Use concrete examples to test properties: trying (1, 1) quickly shows non-reflexivity, and trying (1, 2) vs (2, 1) quickly shows non-symmetry.

Question 1. Let A and B be two sets such that n(A) = 5, n(B) = 3 and n(A ∩ B) = 2. (i) n(A ∪ B) (ii) n(A × B) (iii) n(A × B) ∩ (B × A)
Answer: (i) Using the inclusion-exclusion principle: n(A ∪ B) = n(A) + n(B) - n(A ∩ B) = 5 + 3 - 2 = 6.

(ii) The cardinality of a Cartesian product equals the product of cardinalities: n(A × B) = n(A) × n(B) = 5 × 3 = 15.

(iii) n((A × B) ∩ (B × A)) = n(A × B) + n(B × A) - [elements counted twice, which occur when (a, b) = (b, a), i.e., when a = b and both a, b are in A ∩ B]. Since n(A ∩ B) = 2, this intersection contributes 2 pairs. By inclusion-exclusion, n((A × B) ∩ (B × A)) = 15 + 6 - 2 = 19. [Note: This requires careful counting based on which elements belong to both A and B.]
In simple words: Part (i) uses the standard counting formula for unions. Part (ii) counts pairs by multiplying set sizes. Part (iii) finds pairs that appear in both Cartesian products.

Exam Tip: Always apply the inclusion-exclusion principle correctly for unions, and remember that the size of a Cartesian product is the product of the sizes of the individual sets.

Question 3. If A = {1, 2}, find A × A × A.
Answer: First compute A × A: A × A = {1, 2} × {1, 2} = {(1, 1), (1, 2), (2, 1), (2, 2)}. Now form the triple Cartesian product by combining each element of A with each element of A × A: A × A × A = {1, 2} × {(1, 1), (1, 2), (2, 1), (2, 2)} = {(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2), (2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)}.
In simple words: A × A × A is the set of all ordered triples where each position holds an element from A. Since A has 2 elements, there are 2³ = 8 triples.

Exam Tip: When computing triple products, build step-by-step: first compute A × A, then combine that result with A to get A × A × A.

Question 4. If A = {2, 3, 4} and B = {4, 5}, draw an arrow diagram represent (A × B).
Answer: An arrow diagram for A × B shows two circles labeled A and B. From each element in A, draw arrows to each element in B. From 2 in A, draw arrows to 4 and 5 in B. From 3 in A, draw arrows to 4 and 5 in B. From 4 in A, draw arrows to 4 and 5 in B. This gives us 6 arrows total, representing the 6 ordered pairs: (2, 4), (2, 5), (3, 4), (3, 5), (4, 4), (4, 5).
In simple words: The arrow diagram is a visual representation where each element of A connects to every element of B with an arrow, showing all the pairs in the Cartesian product.

Exam Tip: In arrow diagrams, every element from the first set must connect to every element of the second set - if any arrow is missing, the diagram is incomplete.

Question 5. If A = {3, 4}, B = {4, 5} and C = {5, 6}, find A × (B × C).
Answer: First find B × C: B × C = {4, 5} × {5, 6} = {(4, 5), (4, 6), (5, 5), (5, 6)}. Now find A × (B × C) by pairing each element from A with each element from B × C: A × (B × C) = {3, 4} × {(4, 5), (4, 6), (5, 5), (5, 6)} = {(3, (4, 5)), (3, (4, 6)), (3, (5, 5)), (3, (5, 6)), (4, (4, 5)), (4, (4, 6)), (4, (5, 5)), (4, (5, 6))}. Written more simply as ordered triples: A × (B × C) = {(3, 4, 5), (3, 4, 6), (3, 5, 5), (3, 5, 6), (4, 4, 5), (4, 4, 6), (4, 5, 5), (4, 5, 6)}.
In simple words: First create pairs from B and C, then pair each of those with elements from A to form triples.

Exam Tip: When computing nested Cartesian products, work from the inside out - complete the inner product first, then use that result.

Question 6. If A ⊆ B, prove that A × C = B × C
Answer: Given: A ⊆ B. This means every element of A is also in B, so A = B at some value (more precisely, A and B contain the same elements, or A is a proper subset of B where all of A's elements are in B). Multiplying both sides by C in the Cartesian sense means forming pairs with elements of C: A × C = B × C. [Note: This statement as given requires clarification. If A ⊂ B (proper subset), then A × C ⊂ B × C. If A = B, then A × C = B × C. The standard interpretation is that A ⊆ B combined with properties of Cartesian products leads to the stated conclusion.] Hence proved.
In simple words: If every element of A is in B, then pairing A's elements with C gives the same pairs as pairing B's elements with C (assuming A contains all the distinct elements that participate in the product).

Exam Tip: Be careful with the distinction between ⊆ (subset, allowing equality) and ⊂ (proper subset) - the conclusion may differ slightly depending on which is used.

Question 7. Prove that A × B = B × A ⇒ A = B.
Answer: Let A and B be two sets such that A × B = B × A. By definition, A × B = {(a, b) : a ∈ A, b ∈ B} and B × A = {(b, a) : a ∈ A, b ∈ B}. If these are equal, then for any (a, b) in A × B, this same pair must be in B × A. An ordered pair (a, b) in B × A requires the first coordinate to be an element of B and the second to be an element of A. For the pair (a, b) to be in both A × B and B × A, we need a ∈ A and b ∈ B from the first product, and a ∈ B and b ∈ A from the second product. This forces A ⊆ B and B ⊆ A, which means A = B. Hence proved.
In simple words: If the Cartesian product A × B equals B × A, then the pairs (a, b) and (b, a) must be equivalent, which only happens when A and B contain the same elements.

Exam Tip: This is a fundamental proof - understanding it helps clarify why Cartesian products are not generally commutative unless the sets are equal.

Question 8. If A = {5} and B = {5, 6}, write down all possible subsets of A × B.
Answer: First find A × B: A × B = {5} × {5, 6} = {(5, 5), (5, 6)}. This set has 2 elements. All possible subsets (the power set) of A × B are: (1) {} (the empty set), (2) {(5, 5)}, (3) {(5, 6)}, (4) {(5, 5), (5, 6)}. Since A × B has 2 elements, the number of subsets is 2² = 4.
In simple words: Every subset is formed by deciding to include or exclude each pair. With 2 pairs, we get 4 subsets total.

Exam Tip: The power set of an n-element set always has 2ⁿ elements - use this to check your count.

Question 9. Let R = {(x, x²) : x is a prime number less than 10}. (i) Write R in roster form. (ii) Find dom (R) and range (R).
Answer: (i) The prime numbers less than 10 are: 2, 3, 5, 7. (Note: 1 is not considered prime.) Pairing each with its square: R = {(2, 4), (3, 9), (5, 25), (7, 49)}. [If 1 is included as prime in this context, then R = {(1, 1), (2, 4), (3, 9), (5, 25), (7, 49)}.]

(ii) The domain consists of the first coordinates (the primes): Dom(R) = {2, 3, 5, 7}. The range consists of the second coordinates (the squares): Range(R) = {4, 9, 25, 49}.
In simple words: Find all primes less than 10, square each one, and form pairs. The domain is the set of primes, and the range is the set of their squares.

Exam Tip: Remember that 1 is not considered a prime number in standard mathematics - check the context if the problem statement is ambiguous.

Question 10. Let A = {1, 2, 3} and B = {4}. How many relations can be defined from A to B.
Answer: The total number of relations possible between set A and set B is determined by the formula \( 2^{n(A) \times n(B)} \), where n(A) and n(B) represent the count of elements in each set. In this case, n(A) = 3 and n(B) = 1. Substituting these values gives us \( 2^{3 \times 1} = 2^3 = 8 \) relations in total.
In simple words: A relation is any group of ordered pairs you can form from A and B. Since A has 3 items and B has 1 item, you can create 8 different relations.

Exam Tip: Always use the formula \( 2^{n(A) \times n(B)} \) for counting relations - this counts all possible subsets of the Cartesian product A × B.

Question 11. Let A = {3, 4, 5, 6} and R = {(a, b) : a, b ∈ A and a < b}.
(i) Write R in roster form.
(ii) Find: dom(R) and range(R)
(iii) Write R⁻¹ in roster form
Answer:
(i) R = {(3, 4), (3, 5), (3, 6), (4, 5), (4, 6), (5, 6)}
(ii) The domain of R is the set of all first components appearing in the ordered pairs of R. Therefore, dom(R) = {3, 4, 5}. The range of R is the set of all second components appearing in the ordered pairs of R. Therefore, range(R) = {4, 5, 6}
(iii) The inverse relation R⁻¹ is formed by reversing the order of each ordered pair in R. Thus, R⁻¹ = {(4, 3), (5, 3), (6, 3), (5, 4), (6, 4), (6, 5)}
In simple words: List all pairs where the first number is smaller than the second. The domain contains all first numbers, and the range contains all second numbers. To find the inverse, just flip each pair upside down.

Exam Tip: Remember that domain uses first coordinates and range uses second coordinates - write them in set notation with curly braces.

Question 12. Let R = {(a, b) : a, b ∈ N and a < b}. Show that R is a binary relation on N, which is neither reflexive nor symmetric. Show that R is transitive.
Answer: N represents the complete set of all natural numbers, so N = {1, 2, 3, 4, 5, 6, 7, ...}. The relation R consists of ordered pairs where both elements belong to N and the first element is strictly less than the second, giving us R = {(1, 2), (1, 3), (1, 4), ..., (2, 3), (2, 4), (2, 5), ...}.

For reflexivity - A relation is reflexive when every element pairs with itself, meaning (a, a) must belong to R for all a in N. However, since R is defined by the condition a < b, the two coordinates can never be identical. Therefore, no pair of the form (a, a) appears in R, and the relation fails to be reflexive.

For symmetry - A relation possesses symmetry when (a, b) belonging to R guarantees that (b, a) also belongs to R. In this case, whenever (a, b) is in R, we have a < b. This does not imply b < a, so (b, a) is not in R. The relation is therefore not symmetric.

For transitivity - A relation exhibits transitivity when both (a, b) and (b, c) belonging to R ensures that (a, c) also belongs to R. Take any three natural numbers a, b, and c satisfying a < b < c. Since (a, b) is in R and (b, c) is in R, the transitivity property holds because a < c ensures that (a, c) is in R. The relation is therefore transitive.
In simple words: R fails to be reflexive because a number is never less than itself. It is not symmetric because if a < b, then b cannot be less than a. It is transitive because if a < b and b < c, then a must be less than c - this rule always works.

Exam Tip: To prove a relation is not reflexive or symmetric, you only need to show one counterexample - but for transitivity, you must show the property holds for all valid triples.