Samacheer Kalvi Class 11 Maths Solutions Chapter 1 Sets, Relations and Functions Exercise 1.2

Question 1. Discuss the following relations for reflexivity, symmetricity and transitivity:
(i) The relation R defined on the set of all positive integers by "m R n if m divides n".
(ii) Let P denote the set of all straight lines in a plane. The relation R defined by " I R m if I is perpendicular to m”.
(iii) Let A be the set consisting of all the members of a family. The relation R defined by "a R b if a is not a sister of b".
(iv) Let A be the set consisting of all the female members of a family. The relation R defined by "a R b if a is not a sister of b".
(v) On the set of natural numbers the relation R defined by "x R y if x + 2y = 1"
Answer:
(i) Relation R is defined on the set of all positive integers such that \( m R n \) if \( m \) divides \( n \).
(a) Reflexive: If \( m \) divides \( m \), which is true for all positive integers \( m \). Thus, \( m R m \) means \( m \) divides \( m \), so R is reflexive. For instance, 5 divides 5.
(b) Symmetric: If \( m R n \), then \( m \) divides \( n \). However, this does not mean \( n \) divides \( m \). For example, 2 divides 4, but 4 does not divide 2. So, \( m R n \) does not imply \( n R m \). Therefore, R is not symmetric.
(c) Transitive: If \( m R n \) and \( n R r \), then \( m \) divides \( n \) and \( n \) divides \( r \). This means \( m \) must also divide \( r \). For example, if 2 divides 4 and 4 divides 8, then 2 divides 8. Thus, R is transitive.
In simple words: A relation is like a rule. For "m divides n", it's reflexive because any number divides itself. It's not symmetric because if 2 divides 4, 4 doesn't divide 2. It is transitive because if 2 divides 4 and 4 divides 8, then 2 divides 8.

🎯 Exam Tip: When checking for symmetry or transitivity, always look for counter-examples if the relation is not true for all elements. A single counter-example is enough to disprove the property.

Answer:
(ii) Relation R is defined on the set P of all straight lines in a plane such that \( l R m \) if \( l \) is perpendicular to \( m \).
(a) Reflexive: A line \( l \) cannot be perpendicular to itself. So, \( (l, l) \notin R \). Therefore, R is not reflexive. This is because a line forms a 0-degree angle with itself, not 90 degrees.
(b) Symmetric: If line \( l \) is perpendicular to line \( m \), then line \( m \) is also perpendicular to line \( l \). So, if \( (l, m) \in R \), then \( (m, l) \in R \). Therefore, R is symmetric.
(c) Transitive: If line \( l \) is perpendicular to line \( m \), and line \( m \) is perpendicular to line \( n \), it means \( l \) is parallel to \( n \). For example, if line A is vertical and line B is horizontal, and line C is vertical, then A is perpendicular to B, and B is perpendicular to C. But A is parallel to C, not perpendicular. So, \( (l, n) \notin R \). Therefore, R is not transitive.
In simple words: For "perpendicular lines," a line cannot be perpendicular to itself, so it's not reflexive. If line A is perpendicular to B, then B is also perpendicular to A, so it's symmetric. But if A is perpendicular to B, and B is perpendicular to C, then A and C are parallel, not perpendicular, so it's not transitive.

🎯 Exam Tip: Visualize geometric relations. Drawing simple diagrams of lines for perpendicularity and parallelism can quickly clarify reflexivity, symmetry, and transitivity.

Answer:
(iii) Relation R is defined on a set A of all family members such that \( a R b \) if \( a \) is not a sister of \( b \).
(a) Reflexive: If \( a \) is a member of the family, can \( a \) be "not a sister of \( a \)"? Yes, because one cannot be their own sister. So, \( a R a \) is true. Thus, R is reflexive. Everyone is "not a sister of themselves."
(b) Symmetric: If \( a \) is not a sister of \( b \), does that mean \( b \) is not a sister of \( a \)? Yes, if \( a \) is male and \( b \) is female, \( a \) is not a sister of \( b \). Also, \( b \) (female) is not a sister of \( a \) (male). Consider if \( a \) and \( b \) are brothers, then \( a \) is not a sister of \( b \), and \( b \) is not a sister of \( a \). The relation is symmetric.
(c) Transitive: If \( a \) is not a sister of \( b \), and \( b \) is not a sister of \( c \), does that mean \( a \) is not a sister of \( c \)? Not always. For example, if \( a \) and \( c \) are sisters, and \( b \) is their brother. \( a \) is not a sister of \( b \) (because \( b \) is male), and \( b \) is not a sister of \( c \) (because \( b \) is male). But \( a \) IS a sister of \( c \). In this case, \( a R b \) and \( b R c \) are true, but \( a R c \) is false. Therefore, R is not transitive.
In simple words: For "not a sister of," it's reflexive because you are not your own sister. It's symmetric because if A is not a sister of B, B is also not a sister of A (e.g., if one is male). It's not transitive because if A is not sister of B, and B is not sister of C, A could still be a sister of C (e.g., A and C are sisters, B is their brother).

🎯 Exam Tip: For relations involving family members, always consider all gender combinations (brother/sister, brother/brother, sister/sister) to thoroughly test the properties.

Answer:
(iv) Relation R is defined on a set A of all female members of a family such that \( a R b \) if \( a \) is not a sister of \( b \).
Let A = {M (Mother), H (Female Child)}. The relation R on A is defined as \( a R b \) if \( a \) is not a sister of \( b \).
If M and H are mother and daughter, they are not sisters. So \( (M, H) \in R \) and \( (H, M) \in R \).
If there is only one female child, she is not a sister of herself. So \( (H, H) \in R \). A mother is also not a sister of herself, so \( (M, M) \in R \).
So, \( R = \{(M, M), (M, H), (H, H), (H, M)\} \).
(a) Reflexive: \( (M, M) \in R \) and \( (H, H) \in R \). A female member is not her own sister. So, R is reflexive.
(b) Symmetric: If \( (M, H) \in R \), we have \( (H, M) \in R \). This holds true because if Mother is not a sister of Child, then Child is not a sister of Mother. So, R is symmetric.
(c) Transitive: Let's consider \( (M, H) \in R \) and \( (H, M) \in R \). Then we look for \( (M, M) \in R \), which is true. This checks out for the elements in our example set. If there were two sisters, \( S_1 \) and \( S_2 \), then \( S_1 \) is not a sister of \( S_2 \) would be false. So, if we take two distinct sisters, say \( S_1 \) and \( S_2 \). Then \( S_1 \) is a sister of \( S_2 \), so \( (S_1, S_2) \notin R \). If \( a, b, c \) are three distinct female members and \( a \) is not a sister of \( b \), and \( b \) is not a sister of \( c \), it does not mean \( a \) is not a sister of \( c \). For example, if \( a \) and \( c \) are distinct sisters but \( b \) is their female cousin. Then \( a \) is not a sister of \( b \), and \( b \) is not a sister of \( c \), but \( a \) IS a sister of \( c \). This violates transitivity. Hence, R is not transitive.
In simple words: For "not a sister of" among only female family members: It's reflexive because no one is their own sister. It's symmetric because if Mother is not sister of daughter, daughter is not sister of mother. It's not transitive if, for example, two sisters have a female cousin; then sister A is not a sister of the cousin, the cousin is not a sister of sister B, but sister A IS a sister of B.

🎯 Exam Tip: Pay close attention to the definition of the set. When the set is restricted to "female members," the dynamics of "sister" relationships change compared to "all members."

Answer:
(v) Relation R is defined on the set of natural numbers N such that \( x R y \) if \( x + 2y = 1 \).
(a) Reflexive: For \( x R x \), we need \( x + 2x = 1 \), which simplifies to \( 3x = 1 \), so \( x = \frac{1}{3} \). Since \( \frac{1}{3} \) is not a natural number, \( x R x \) is not true for any \( x \in N \). Therefore, R is not reflexive. No natural number can satisfy this condition.
(b) Symmetric: If \( x R y \), then \( x + 2y = 1 \). If R were symmetric, then \( y R x \) would mean \( y + 2x = 1 \). However, \( x + 2y = 1 \) does not imply \( y + 2x = 1 \). For example, if \( x=1, y=0 \), then \( 1+2(0)=1 \). But \( 0 \notin N \). If we try to find any pair, say \( x=1 \), then \( 1+2y=1 \implies 2y=0 \implies y=0 \), which is not in N. In fact, if \( x \in N \), then \( x \ge 1 \), so \( 1-x \le 0 \). For \( 2y = 1-x \), \( y \) would have to be 0 or negative. So, there are no natural number pairs satisfying \( x+2y=1 \). The relation R is an empty set. An empty relation is trivially symmetric, but since no pairs exist, we can state it's not symmetric in the practical sense as there's nothing to test. An empty relation is often considered symmetric and transitive by convention because there are no counterexamples. However, as it's typically understood for non-empty relations, we conclude it's not symmetric given the problem context of natural numbers.
(c) Transitive: Since there are no elements in the relation R (it's an empty set), the condition for transitivity (if \( (x, y) \in R \) and \( (y, z) \in R \), then \( (x, z) \in R \)) cannot be violated. Thus, an empty relation is considered transitive. However, similar to symmetry, from a practical standpoint, it offers no connected pairs within natural numbers. If we are strict, an empty relation on any non-empty set is both symmetric and transitive. But as a practical answer, it means there are no numbers for which this chain holds.
In simple words: The rule \( x + 2y = 1 \) doesn't work for natural numbers because \( x \) and \( y \) must be at least 1, which would make the sum at least \( 1+2(1)=3 \). So, there are no pairs that fit this rule at all. Because no pairs exist, it cannot be reflexive. It is technically symmetric and transitive because there are no examples to prove otherwise, but for practical understanding, it means it doesn't show these properties within the natural numbers.

🎯 Exam Tip: When a relation definition leads to an empty set for the given domain, it's not reflexive. For symmetry and transitivity, an empty relation is technically considered both symmetric and transitive because there are no counterexamples, but clarity is needed on whether "not symmetric/transitive" refers to having no elements to demonstrate the property, or strictly violating it.

Question 2. Let X = { a, b, c, d } and R = { (a, a), (b, b ), (a, c)}. Write down the minimum number of ordered pairs to be included to R to make it
(i) reflexive
(ii) symmetric
(iii) transitive
(iv) equivalence
Answer:
Given set \( X = \{a, b, c, d\} \).
Given relation \( R = \{(a, a), (b, b), (a, c)\} \).
(i) To make R reflexive, all elements in X must have a pair with themselves. The missing pairs are \( (c, c) \) and \( (d, d) \). So, the minimum ordered pairs to be included are \( (c, c) \) and \( (d, d) \). These ensure every element relates to itself.
(ii) To make R symmetric, if \( (x, y) \in R \), then \( (y, x) \) must also be in R. We have \( (a, c) \in R \), but \( (c, a) \notin R \). So, the minimum ordered pair to be included is \( (c, a) \). This makes the relation two-way.
(iii) R is currently transitive. Let's check: \( (a, a) \in R \), \( (b, b) \in R \). For \( (a, c) \in R \), there is no \( (c, x) \in R \), so no need to check \( (a, x) \). Also, \( (a, a) \in R \) and \( (a, c) \in R \). Then we need \( (a, c) \in R \), which is already present. So, no additional pairs are needed to make R transitive. It already satisfies the condition for all existing paths.
(iv) To make R an equivalence relation, it must be reflexive, symmetric, and transitive.
From (i), we need to add \( (c, c) \) and \( (d, d) \).
From (ii), we need to add \( (c, a) \).
The relation is already transitive.
After adding these pairs, the new relation \( R_1 \) becomes: \( R_1 = \{(a, a), (b, b), (c, c), (d, d), (a, c), (c, a)\} \). This \( R_1 \) is reflexive, symmetric, and transitive. So, the minimum ordered pairs to be included are \( (c, c), (d, d), (c, a) \).
In simple words: To make the relation complete for all types: For reflexive, add pairs so each element relates to itself (like (c,c) and (d,d)). For symmetric, if (a,c) is there, add (c,a). For transitive, no new pairs are needed here. To be an equivalence relation, it needs all three, so we add all the needed pairs.

🎯 Exam Tip: When building an equivalence relation, systematically check each property (reflexive, symmetric, transitive) in order. Adding pairs for one property might inadvertently satisfy or require more pairs for another, so verify the final set of additions carefully.

Question 3. Let A = { a, b, c } and R = { (a, a ), (b, b ), (a, c) }. Write down the minimum number of ordered pairs to be included to R to make it
(i) reflexive
(ii) symmetric
(iii) transitive
(iv) equivalence
Answer:
Given set \( A = \{a, b, c\} \).
Given relation \( R = \{(a, a), (b, b), (a, c)\} \).
(i) To make R reflexive, every element in A must be related to itself. The missing pair is \( (c, c) \). So, the minimum ordered pair to be included is \( (c, c) \). This ensures all elements \( a, b, c \) are related to themselves.
(ii) To make R symmetric, if \( (x, y) \in R \), then \( (y, x) \) must also be in R. We have \( (a, c) \in R \), but \( (c, a) \notin R \). So, the minimum ordered pair to be included is \( (c, a) \). This makes the relation reciprocal for all existing links.
(iii) R is currently transitive. Let's check: We have \( (a, a) \in R \) and \( (a, c) \in R \), which implies \( (a, c) \in R \) (which is already there). All other elements are self-related or don't form chains. Thus, no additional pairs are needed to make R transitive. The existing structure already supports transitivity.
(iv) To make R an equivalence relation, it needs to be reflexive, symmetric, and transitive.
From (i), we need to add \( (c, c) \).
From (ii), we need to add \( (c, a) \).
The relation is already transitive.
After including these pairs, the new relation \( R_1 \) becomes: \( R_1 = \{(a, a), (b, b), (c, c), (a, c), (c, a)\} \). This relation \( R_1 \) is reflexive, symmetric, and transitive. So, the minimum ordered pairs to be included are \( (c, c) \) and \( (c, a) \).
In simple words: We start with a set A and a relation R. To make R reflexive, add (c,c) so all elements relate to themselves. To make it symmetric, add (c,a) because (a,c) is already there. It is already transitive. To make it an equivalence relation, we need to add all the pairs needed for reflexivity and symmetry, which are (c,c) and (c,a).

🎯 Exam Tip: Remember that an equivalence relation requires all three properties: reflexivity, symmetry, and transitivity. If any one is missing, it's not an equivalence relation.

Question 4. Let P be the set of all triangles in a plane and R be the relation defined on P as a R b if a is similar to b. Prove that R is an equivalence relation.
Answer:
Given P is the set of all triangles in a plane. The relation R on P is defined as \( a R b \) if triangle \( a \) is similar to triangle \( b \). We need to prove that R is an equivalence relation, which means it must be reflexive, symmetric, and transitive.
(a) Reflexive: For any triangle \( a \in P \), triangle \( a \) is always similar to itself. This is a fundamental property of similarity. So, \( (a, a) \in R \). Therefore, R is reflexive.
(b) Symmetric: Let \( (a, b) \in R \). This means triangle \( a \) is similar to triangle \( b \). If triangle \( a \) is similar to triangle \( b \), it naturally follows that triangle \( b \) is also similar to triangle \( a \). So, \( (b, a) \in R \). Therefore, R is symmetric.
(c) Transitive: Let \( (a, b) \in R \) and \( (b, c) \in R \). This means triangle \( a \) is similar to triangle \( b \), and triangle \( b \) is similar to triangle \( c \). By the property of similarity, if \( a \) is similar to \( b \) and \( b \) is similar to \( c \), then triangle \( a \) must also be similar to triangle \( c \). So, \( (a, c) \in R \). Therefore, R is transitive.
Since R is reflexive, symmetric, and transitive, it is an equivalence relation. This shows that "similarity" groups triangles that share the same shape, regardless of size or orientation.
In simple words: We are checking if "is similar to" for triangles is an equivalence relation. It is reflexive because any triangle is similar to itself. It is symmetric because if triangle A is similar to B, then B is similar to A. It is transitive because if A is similar to B, and B is similar to C, then A is also similar to C. Since all three are true, it's an equivalence relation.

🎯 Exam Tip: When proving a relation is an equivalence relation, clearly state and demonstrate each of the three properties (reflexive, symmetric, transitive) separately using the given definition of the relation.

Question 5. On the set of natural numbers let R be the relation defined by a R b if 2a + 3b = 30. Write down the relation by listing all the pairs. Cheek whether it is
(i) reflexive
(ii) symmetric
(iii) transitive
(iv) equivalence
Answer:
Given N is the set of natural numbers. The relation R is defined by \( a R b \) if \( 2a + 3b = 30 \). We need to find all pairs \( (a, b) \in N \times N \) that satisfy this equation.
Since \( a, b \) are natural numbers, \( a \ge 1 \) and \( b \ge 1 \).
From \( 2a + 3b = 30 \), we can write \( 3b = 30 - 2a \), so \( b = \frac{30 - 2a}{3} \).
Since \( b \) must be a natural number, \( (30 - 2a) \) must be divisible by 3 and positive.
Let's test values for \( a \):
If \( a = 1 \), \( b = \frac{30 - 2(1)}{3} = \frac{28}{3} \notin N \).
If \( a = 2 \), \( b = \frac{30 - 2(2)}{3} = \frac{26}{3} \notin N \).
If \( a = 3 \), \( b = \frac{30 - 2(3)}{3} = \frac{24}{3} = 8 \in N \). So \( (3, 8) \in R \).
If \( a = 4 \), \( b = \frac{30 - 2(4)}{3} = \frac{22}{3} \notin N \).
If \( a = 5 \), \( b = \frac{30 - 2(5)}{3} = \frac{20}{3} \notin N \).
If \( a = 6 \), \( b = \frac{30 - 2(6)}{3} = \frac{18}{3} = 6 \in N \). So \( (6, 6) \in R \).
If \( a = 7 \), \( b = \frac{30 - 2(7)}{3} = \frac{16}{3} \notin N \).
If \( a = 8 \), \( b = \frac{30 - 2(8)}{3} = \frac{14}{3} \notin N \).
If \( a = 9 \), \( b = \frac{30 - 2(9)}{3} = \frac{12}{3} = 4 \in N \). So \( (9, 4) \in R \).
If \( a = 10 \), \( b = \frac{30 - 2(10)}{3} = \frac{10}{3} \notin N \).
If \( a = 11 \), \( b = \frac{30 - 2(11)}{3} = \frac{8}{3} \notin N \).
If \( a = 12 \), \( b = \frac{30 - 2(12)}{3} = \frac{6}{3} = 2 \in N \). So \( (12, 2) \in R \).
If \( a = 13 \), \( b = \frac{30 - 2(13)}{3} = \frac{4}{3} \notin N \).
If \( a = 14 \), \( b = \frac{30 - 2(14)}{3} = \frac{2}{3} \notin N \).
If \( a = 15 \), \( b = \frac{30 - 2(15)}{3} = \frac{0}{3} = 0 \notin N \).
For \( a > 15 \), \( 30 - 2a \) will be negative, making \( b \) negative, which is not in N.
So, the relation is \( R = \{(3, 8), (6, 6), (9, 4), (12, 2)\} \).

Now, let's check the properties of R:
(i) Reflexive: For R to be reflexive, \( (a, a) \) must be in R for all \( a \in N \). However, for \( (3, 3) \), \( 2(3) + 3(3) = 6 + 9 = 15 \ne 30 \). Only \( (6, 6) \) satisfies the condition. Since not all \( (a, a) \) are in R (e.g., \( (1,1) \notin R \), \( (2,2) \notin R \), \( (3,3) \notin R \)), R is not reflexive.
(ii) Symmetric: For R to be symmetric, if \( (a, b) \in R \), then \( (b, a) \) must also be in R.
We have \( (3, 8) \in R \). If R is symmetric, then \( (8, 3) \) should also be in R. Let's check: \( 2(8) + 3(3) = 16 + 9 = 25 \ne 30 \). So, \( (8, 3) \notin R \). Therefore, R is not symmetric.
(iii) Transitive: For R to be transitive, if \( (a, b) \in R \) and \( (b, c) \in R \), then \( (a, c) \) must also be in R.
Let's check pairs: We have \( (3, 8) \in R \). Is there any pair starting with 8? No. We have \( (6, 6) \in R \). Is there any pair starting with 6 other than \( (6,6) \)? No. We have \( (9, 4) \in R \). Is there any pair starting with 4? No. We have \( (12, 2) \in R \). Is there any pair starting with 2? No.
Since we cannot find any \( (a, b) \) and \( (b, c) \) pairs where \( b \) is different for the second element of the first pair and the first element of the second pair, and where \( a \ne c \), we do not have a chain to test for non-transitivity. For a relation like this, where no such \( (a,b), (b,c) \) chains exist (apart from self-loops), it is considered vacuously transitive. However, it's not a rich transitive relation. So, we say it is transitive, as there are no counter-examples that violate the definition.
(iv) Equivalence: Since R is neither reflexive nor symmetric, it cannot be an equivalence relation.
In simple words: First, we find all pairs of natural numbers \( (a,b) \) where \( 2a + 3b = 30 \). These are \( (3,8), (6,6), (9,4), (12,2) \). Then we check: It's not reflexive because not all numbers relate to themselves (only (6,6) does). It's not symmetric because (3,8) is in R but (8,3) is not. It is transitive because there are no chain-like pairs to break the rule (like (a,b) and (b,c) where a, b, c are different). Since it's not reflexive or symmetric, it cannot be an equivalence relation.

🎯 Exam Tip: Listing all pairs that satisfy the relation is crucial. From this list, systematically checking each property becomes straightforward. For transitivity, if no suitable "chains" (a,b) and (b,c) exist, the relation is vacuously transitive.

Question 6. Prove that the relation 'friendship' is not an equivalence relation on the set of all people in Chennai.
Answer:
Let R be the relation 'friendship' defined on the set of all people in Chennai. We need to check if it is reflexive, symmetric, and transitive.
(a) Reflexive: A person can be a friend to themselves. While some might argue this philosophically, mathematically, for a relation like "friendship," it's generally accepted that \( a R a \) is true if a person considers themselves a friend. However, in common understanding, friendship is between two distinct people. If we consider it as self-friendship, it could be reflexive. If we consider friendship only between distinct individuals, it would not be reflexive. For the purpose of disproving equivalence, we can consider reflexivity to be true, or look at other properties.
(b) Symmetric: If person \( a \) is a friend of person \( b \), then person \( b \) is also a friend of person \( a \). This is a generally accepted understanding of friendship. So, if \( a R b \), then \( b R a \). Therefore, the relation 'friendship' is symmetric.
(c) Transitive: If person \( a \) is a friend of person \( b \), and person \( b \) is a friend of person \( c \), it does not necessarily mean that person \( a \) is a friend of person \( c \). For example, \( a \) might be friends with \( b \), and \( b \) might be friends with \( c \), but \( a \) and \( c \) might not even know each other, or they might dislike each other. So, if \( a R b \) and \( b R c \), it does not imply \( a R c \). Therefore, the relation 'friendship' is not transitive.
Since the relation 'friendship' is not transitive (and its reflexivity can be debated depending on interpretation), it is not an equivalence relation. The lack of transitivity is the clearest reason to disprove it.
In simple words: The friendship relation is usually symmetric (if A is B's friend, B is A's friend). Its reflexivity (being a friend to yourself) is debatable. But it is definitely NOT transitive because if A is friends with B, and B is friends with C, it doesn't mean A is friends with C. Because it's not transitive, it cannot be an equivalence relation.

🎯 Exam Tip: When dealing with real-world relations like "friendship," use a common-sense understanding. For mathematical rigor, one counter-example for a property (like transitivity here) is sufficient to disprove it.

Question 7. On the set of natural numbers let R be the relation defined by a R b if a + b < 6. Write down the relation by listing all the pairs. Check whether it is
(i) reflexive
(ii) symmetric
(iii) transitive
(iv) equivalence
Answer:
Given N is the set of natural numbers. The relation R is defined by \( a R b \) if \( a + b < 6 \). We need to list all pairs \( (a, b) \in N \times N \) that satisfy this condition.
For \( a, b \in N \), the smallest value for \( a \) or \( b \) is 1.
If \( a = 1 \):
\( 1 + b < 6 \implies b < 5 \). So \( b \) can be 1, 2, 3, 4.
Pairs: \( (1, 1), (1, 2), (1, 3), (1, 4) \).
If \( a = 2 \):
\( 2 + b < 6 \implies b < 4 \). So \( b \) can be 1, 2, 3.
Pairs: \( (2, 1), (2, 2), (2, 3) \).
If \( a = 3 \):
\( 3 + b < 6 \implies b < 3 \). So \( b \) can be 1, 2.
Pairs: \( (3, 1), (3, 2) \).
If \( a = 4 \):
\( 4 + b < 6 \implies b < 2 \). So \( b \) can be 1.
Pair: \( (4, 1) \).
If \( a = 5 \):
\( 5 + b < 6 \implies b < 1 \). No natural number \( b \) satisfies this. So no pairs for \( a = 5 \) or greater.
So, the relation is \( R = \{(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)\} \).

Now, let's check the properties of R:
(i) Reflexive: For R to be reflexive, \( (a, a) \) must be in R for all \( a \in N \). We need \( a + a < 6 \implies 2a < 6 \implies a < 3 \). So, only \( (1, 1) \) and \( (2, 2) \) are in R. For \( a = 3 \), \( (3, 3) \notin R \) because \( 3+3=6 \not< 6 \). Thus, R is not reflexive.
(ii) Symmetric: For R to be symmetric, if \( (a, b) \in R \), then \( (b, a) \) must also be in R.
If \( a + b < 6 \), then \( b + a < 6 \) is also true. So, for every pair \( (a, b) \) in R, the pair \( (b, a) \) is also in R. For example, \( (1, 2) \in R \) and \( (2, 1) \in R \). \( (1, 3) \in R \) and \( (3, 1) \in R \). \( (1, 4) \in R \) and \( (4, 1) \in R \). Thus, R is symmetric.
(iii) Transitive: For R to be transitive, if \( (a, b) \in R \) and \( (b, c) \in R \), then \( (a, c) \) must also be in R.
Consider \( (3, 1) \in R \) (because \( 3+1=4 < 6 \)) and \( (1, 4) \in R \) (because \( 1+4=5 < 6 \)).
If R were transitive, then \( (3, 4) \) should be in R. But \( 3+4=7 \not< 6 \). So, \( (3, 4) \notin R \). Therefore, R is not transitive.
(iv) Equivalence: Since R is neither reflexive nor transitive, it is not an equivalence relation.
In simple words: First, we list all pairs of natural numbers that add up to less than 6. These are (1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1). Then, it's not reflexive because (3,3) (for example) is not in the list (3+3 is not less than 6). It is symmetric because if (a,b) is there, (b,a) is also there (e.g., (1,2) and (2,1)). It's not transitive because (3,1) and (1,4) are there, but (3,4) is not (3+4 is not less than 6). Since it's not fully reflexive and not transitive, it's not an equivalence relation.

🎯 Exam Tip: When listing pairs for a numerical relation, systematically test values for one variable and deduce the possible values for the other. This ensures no pairs are missed and helps in checking properties.

Question 8. Let A = { a, b, c }. What is the equivalence relation of smallest cardinality on A? What is the equivalence relation of largest cardinality on A?
Answer:
Given set \( A = \{a, b, c\} \).
An equivalence relation must be reflexive, symmetric, and transitive.

**Equivalence relation of smallest cardinality on A:**
For a relation to be reflexive, it must contain all self-loops: \( (a, a), (b, b), (c, c) \).
If we only include these, \( R_{min} = \{(a, a), (b, b), (c, c)\} \).
Let's check if this is an equivalence relation:
- Reflexive: Yes, all self-loops are present.
- Symmetric: Yes, for each \( (x, x) \in R_{min} \), \( (x, x) \) is also in \( R_{min} \).
- Transitive: Yes, if \( (x, x) \in R_{min} \) and \( (x, x) \in R_{min} \), then \( (x, x) \in R_{min} \). There are no chains of distinct elements to violate transitivity.
So, \( R_{min} = \{(a, a), (b, b), (c, c)\} \) is the equivalence relation with the smallest cardinality. Its cardinality (number of pairs) is \( n(R_{min}) = 3 \). This relation represents the identity relation, where each element is only related to itself.

**Equivalence relation of largest cardinality on A:**
The equivalence relation with the largest cardinality is the universal relation, where every element is related to every other element. This means all possible pairs in \( A \times A \) are included.
The total number of pairs in \( A \times A \) is \( |A| \times |A| = 3 \times 3 = 9 \).
So, \( R_{max} = \{(a, a), (a, b), (a, c), (b, a), (b, b), (b, c), (c, a), (c, b), (c, c)\} \).
Let's check if this is an equivalence relation:
- Reflexive: Yes, \( (a, a), (b, b), (c, c) \) are all present.
- Symmetric: Yes, for every \( (x, y) \in R_{max} \), \( (y, x) \) is also in \( R_{max} \). (e.g., \( (a, b) \) and \( (b, a) \) are both present).
- Transitive: Yes, if \( (x, y) \in R_{max} \) and \( (y, z) \in R_{max} \), then \( (x, z) \) is also in \( R_{max} \), because every possible pair is included. This is trivially true.
So, \( R_{max} = \{(a, a), (a, b), (a, c), (b, a), (b, b), (b, c), (c, a), (c, b), (c, c)\} \) is the equivalence relation with the largest cardinality. Its cardinality is \( n(R_{max}) = 9 \). This relation means all elements are considered equivalent to each other.
In simple words: For a set with elements {a, b, c}: The smallest equivalence relation just connects each element to itself, like {(a,a), (b,b), (c,c)}. This is the simplest way to be reflexive, symmetric, and transitive. The largest equivalence relation connects every element to every other element, including itself. This means all possible pairs are included, making it fully reflexive, symmetric, and transitive.

🎯 Exam Tip: The identity relation (all \( (x,x) \) pairs) is always the smallest equivalence relation, and the universal relation (all possible \( (x,y) \) pairs) is always the largest equivalence relation on any given set.

Question 9. In the set Z of integers, define m R n if m – n is divisible by 7. Prove that R is an equivalence relation.
Answer:
Given Z is the set of integers. The relation R is defined as \( m R n \) if \( m - n \) is divisible by 7. This means \( m - n = 7k \) for some integer \( k \). We need to prove that R is an equivalence relation.

(a) Reflexive: For any integer \( m \in Z \), we need to check if \( m R m \).
\( m - m = 0 \).
Since \( 0 = 7 \times 0 \), and 0 is an integer, \( m - m \) is divisible by 7.
Thus, \( m R m \) for all \( m \in Z \). Therefore, R is reflexive.

(b) Symmetric: Let \( m R n \), where \( m, n \in Z \).
This means \( m - n \) is divisible by 7, so \( m - n = 7k \) for some integer \( k \).
We need to show that \( n R m \), which means \( n - m \) is divisible by 7.
From \( m - n = 7k \), we multiply by -1 to get \( -(m - n) = -7k \).
So, \( n - m = 7(-k) \).
Since \( k \) is an integer, \( -k \) is also an integer. Let \( k' = -k \).
Then \( n - m = 7k' \), which means \( n - m \) is divisible by 7.
Thus, \( n R m \). Therefore, R is symmetric.

(c) Transitive: Let \( m R n \) and \( n R r \), where \( m, n, r \in Z \).
\( m R n \) means \( m - n = 7k_1 \) for some integer \( k_1 \). (Equation 1)
\( n R r \) means \( n - r = 7k_2 \) for some integer \( k_2 \). (Equation 2)
We need to show that \( m R r \), which means \( m - r \) is divisible by 7.
Add Equation 1 and Equation 2:
\( (m - n) + (n - r) = 7k_1 + 7k_2 \)
\( m - r = 7(k_1 + k_2) \).
Since \( k_1 \) and \( k_2 \) are integers, \( k_1 + k_2 \) is also an integer. Let \( k_3 = k_1 + k_2 \).
Then \( m - r = 7k_3 \), which means \( m - r \) is divisible by 7.
Thus, \( m R r \). Therefore, R is transitive.

Since R is reflexive, symmetric, and transitive, R is an equivalence relation. This kind of relation is known as a congruence relation modulo 7.
In simple words: We are checking if "m minus n is divisible by 7" is an equivalence relation. It's reflexive because m - m is 0, and 0 is divisible by 7. It's symmetric because if m - n is divisible by 7, then n - m (which is just -(m - n)) is also divisible by 7. It's transitive because if m - n is divisible by 7 and n - r is divisible by 7, then if you add them up, m - r is also divisible by 7. Since all three are true, it's an equivalence relation.

🎯 Exam Tip: When proving relations of divisibility or congruence, clearly define what the relation means (e.g., \( m-n = 7k \)). Then use algebraic manipulation to demonstrate each property rigorously.