OP Malhotra Class 12 Maths Solutions Chapter 1 Relations Exercise 1

Question 1. Consider the following properties of relations : Symmetric (S), Transitive (T), Reflexive (R), Equivalence (E), None of these (N). State which property/properties are satisfied by each of the following relations: (Give the answers in terms of S, T, R, E and N).
(a) "is greater than" for the set of real numbers.
(b) "is the cube of" for the set of all real numbers.
(c) "is the sister of" for a set of children.
(d) "is similar to” for the set of triangles.
(e) "is perpendicular to” for the set of co-planar lines.
Answer:
(a) For the relation "is greater than" (\( a > b \)) on real numbers:
Reflexive: A number is not greater than itself (e.g., \( a \ngtr a \)). So, it is not reflexive.
Symmetric: If \( a > b \), it does not mean \( b > a \) (e.g., if \( 5 > 3 \), then \( 3 \ngtr 5 \)). So, it is not symmetric.
Transitive: If \( a > b \) and \( b > c \), then \( a > c \) is true (e.g., if \( 5 > 3 \) and \( 3 > 1 \), then \( 5 > 1 \)). So, it is transitive.
This relation only satisfies the Transitive property (T).
(b) For the relation "is the cube of" (\( a = b^3 \)) on real numbers:
Reflexive: A number is not always its own cube (e.g., \( 2 \neq 2^3 \)). So, it is not reflexive.
Symmetric: If \( a = b^3 \), it does not mean \( b = a^3 \) (e.g., \( 8 = 2^3 \) but \( 2 \neq 8^3 \)). So, it is not symmetric.
Transitive: If \( a = b^3 \) and \( b = c^3 \), it does not mean \( a = c^3 \) (e.g., \( 512 = 8^3 \) and \( 8 = 2^3 \), but \( 512 \neq 2^3 \)). So, it is not transitive.
This relation satisfies None of the options (N).
(c) For the relation "is the sister of" on a set of children:
Reflexive: A child cannot be their own sister. So, it is not reflexive.
Symmetric: If A is the sister of B, B might be a boy, so B would not be the sister of A. So, it is not symmetric.
Transitive: If A is the sister of B, and B is the sister of C, then A must also be the sister of C. So, it is transitive.
This relation only satisfies the Transitive property (T).
(d) For the relation "is similar to" on a set of triangles:
Reflexive: Every triangle is similar to itself. So, it is reflexive.
Symmetric: If triangle A is similar to triangle B, then triangle B is also similar to triangle A. So, it is symmetric.
Transitive: If triangle A is similar to triangle B, and triangle B is similar to triangle C, then triangle A is similar to triangle C. So, it is transitive.
Since it is reflexive, symmetric, and transitive, this relation is an Equivalence relation (E).
(e) For the relation "is perpendicular to" (\( a \perp b \)) on a set of co-planar lines:
Reflexive: A line cannot be perpendicular to itself. So, it is not reflexive.
Symmetric: If line A is perpendicular to line B, then line B is also perpendicular to line A. So, it is symmetric.
Transitive: If line A is perpendicular to line B, and line B is perpendicular to line C, then line A must be parallel to line C, not perpendicular to it. So, it is not transitive.
This relation only satisfies the Symmetric property (S).
In simple words: We check each relation against the definitions of reflexive, symmetric, and transitive properties. A relation is reflexive if every element is related to itself. It is symmetric if \( (a,b) \) being related means \( (b,a) \) is also related. It is transitive if \( (a,b) \) and \( (b,c) \) being related implies \( (a,c) \) is also related. If all three hold, it's an equivalence relation.

🎯 Exam Tip: Remember the clear definitions for each type of relation (reflexive, symmetric, transitive) and always test with counter-examples if a property does not hold. This helps prove non-satisfaction easily.

Question 2. Write down a relation which is
(a) Only transitive
(b) Only symmetric
(c) Only reflexive and transitive
(d) Only symmetric and reflexive.
Answer:
(a) A relation that is only transitive is "is greater than" on the set of real numbers. As shown in Question 1(a), it is not reflexive or symmetric but is transitive.
(b) A relation that is only symmetric is "is perpendicular to" on the set of co-planar lines. As shown in Question 1(e), it is not reflexive or transitive but is symmetric.
(c) A relation that is only reflexive and transitive (but not symmetric) is "is a multiple of" on the set of positive integers.
Reflexive: \( a \) is a multiple of \( a \) (e.g., \( a = a \times 1 \)). So, \( (a,a) \) is in the relation.
Transitive: If \( a \) is a multiple of \( b \) and \( b \) is a multiple of \( c \), then \( a \) is also a multiple of \( c \). For instance, if \( a=kb \) and \( b=mc \), then \( a=(km)c \).
Symmetric: If \( a \) is a multiple of \( b \), \( b \) is not necessarily a multiple of \( a \) (e.g., \( 6 \) is a multiple of \( 3 \), but \( 3 \) is not a multiple of \( 6 \)). So, it is not symmetric.
(d) A relation that is only symmetric and reflexive (but not transitive) is "is a friend of" on a set of people.
Reflexive: Every person is considered a friend of themselves in this context, or we can define it such that \( (a,a) \) is always true.
Symmetric: If A is a friend of B, then B is also a friend of A.
Transitive: If A is a friend of B, and B is a friend of C, it does not mean A is a friend of C. The friend-of-a-friend relationship is not always direct. For a relation to be an equivalence relation, all three properties must hold.
In simple words: We need to think of examples for each type of relation. For "only transitive", "greater than" works. For "only symmetric", "perpendicular to" works. For "reflexive and transitive", "is a multiple of" works because you can't reverse it. For "symmetric and reflexive", "is a friend of" works because your friend's friend isn't always your friend.

🎯 Exam Tip: When asked to provide an example of a specific type of relation, choose simple, intuitive mathematical or real-world scenarios and clearly demonstrate why it satisfies some properties and fails others.

Question 3. Prove that if A is the set of the members of a family and R means “is brother of” then it is a transitive relation.
Answer: Let A be the set of family members and R be the relation "is brother of".
We need to prove that R is transitive. For a relation to be transitive, if \( (a,b) \in R \) and \( (b,c) \in R \), then \( (a,c) \in R \) must hold.
Let \( a, b, c \in A \).
1. Assume \( (a,b) \in R \). This means \( a \) is the brother of \( b \). (This implies \( a \) is male.)
2. Assume \( (b,c) \in R \). This means \( b \) is the brother of \( c \). (This implies \( b \) is male.)
From these two statements, if \( a \) is the brother of \( b \), and \( b \) is the brother of \( c \), it logically follows that \( a \) is also the brother of \( c \). All three share the same parents, making \( a \) and \( c \) siblings, with \( a \) being male.

\( \implies (a,c) \in R \).
Therefore, the relation "is brother of" is a transitive relation.

Let's also quickly check other properties for context, though not asked:
Reflexive: A person cannot be their own brother. So, R is not reflexive.
Symmetric: If \( a \) is the brother of \( b \), \( b \) might be female, so \( b \) is not necessarily the brother of \( a \). So, R is not symmetric.
In simple words: If person A is the brother of person B, and person B is the brother of person C, then A must also be the brother of C. They share the same parents, so A is a male sibling to C. This makes the relation "is brother of" transitive.

🎯 Exam Tip: When proving transitivity for a specific relationship, ensure that the conditions for the first two pairs logically lead to the third pair being true. Always consider the definition of the relationship carefully.

Question 4. Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.
Answer: Let the set be \( A = \{1, 2, 3\} \) and the relation be \( R = \{(1, 2), (2, 1)\} \).
We need to examine its properties:
1. Reflexive property:
For R to be reflexive, every element in A must be related to itself. This means \( (1,1), (2,2), (3,3) \) must be in R. However, these pairs are not in R.

\( \implies \) R is not reflexive.
2. Symmetric property:
For R to be symmetric, if \( (a,b) \in R \), then \( (b,a) \in R \) must also be true. In our relation, we have \( (1,2) \in R \), and its reverse \( (2,1) \) is also in R. There are no other pairs to check.

\( \implies \) R is symmetric.
3. Transitive property:
For R to be transitive, if \( (a,b) \in R \) and \( (b,c) \in R \), then \( (a,c) \in R \) must be true.
Consider the pairs \( (1,2) \in R \) and \( (2,1) \in R \). For transitivity, \( (1,1) \) should be in R. However, \( (1,1) \notin R \).

\( \implies \) R is not transitive.
Thus, we have shown that R is symmetric but neither reflexive nor transitive.
In simple words: For the set \( \{1, 2, 3\} \), the relation \( \{(1, 2), (2, 1)\} \) is symmetric because if 1 is related to 2, then 2 is also related to 1. But it's not reflexive because numbers like 1, 2, or 3 are not related to themselves. It's also not transitive because even though 1 relates to 2 and 2 relates to 1, 1 is not related to 1, which it should be for transitivity.

🎯 Exam Tip: When analyzing small, finite sets, explicitly list out all required pairs for each property (reflexive, symmetric, transitive) and check if they are present in the given relation. A single missing pair is enough to disprove a property.

Question 5. Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}. Choose the correct answer:
(a) R is reflexive and symmetric but not transitive.
(b) R is reflexive and transitive but not symmetric.
(c) R is symmetric and transitive but not reflexive.
(d) R is an equivalence relation.
Answer: (b) R is reflexive and transitive but not symmetric.
In simple words: We check if the relation R has all three properties: reflexive (each number related to itself), symmetric (if (a,b) is in R, then (b,a) is too), and transitive (if (a,b) and (b,c) are in R, then (a,c) is too). We found it is reflexive and transitive, but not symmetric.

🎯 Exam Tip: For MCQ questions involving relation properties, quickly test each property (reflexive, symmetric, transitive) with the given set and relation. A single counterexample is sufficient to rule out a property, helping to efficiently select the correct option.

Question 6. Show that the relation R in the set A of real numbers defined as R = {(a, b) : a ≤ b} is reflexive and transitive but not symmetric.
Answer: Let A be the set of real numbers and R be the relation defined as \( R = \{(a, b) : a \le b\} \).
We need to show if it is reflexive, symmetric, and transitive.
1. Reflexive property:
For any real number \( a \in A \), \( a \le a \) is always true.

\( \implies (a,a) \in R \).
Therefore, R is reflexive.
2. Symmetric property:
For R to be symmetric, if \( (a,b) \in R \), then \( (b,a) \in R \) must be true.
Let \( (a,b) \in R \). This means \( a \le b \).
However, if \( a \le b \) (e.g., \( 2 \le 5 \)), it does not mean \( b \le a \) (e.g., \( 5 \not\le 2 \)).

\( \implies (b,a) \notin R \) for all \( a, b \).
Therefore, R is not symmetric.
3. Transitive property:
For R to be transitive, if \( (a,b) \in R \) and \( (b,c) \in R \), then \( (a,c) \in R \) must be true.
Let \( (a,b) \in R \) and \( (b,c) \in R \). This means \( a \le b \) and \( b \le c \).
From these inequalities, it logically follows that \( a \le c \).

\( \implies (a,c) \in R \).
Therefore, R is transitive.
Thus, the relation R = \( \{(a, b) : a \le b\} \) is reflexive and transitive, but not symmetric.
In simple words: We are checking the "less than or equal to" relation on all real numbers. It's reflexive because any number is less than or equal to itself. It's transitive because if \( a \le b \) and \( b \le c \), then \( a \) must be \( \le c \). But it's not symmetric because if \( a \le b \), it's not always true that \( b \le a \) (e.g., \( 2 \le 5 \) but \( 5 \not\le 2 \)).

🎯 Exam Tip: The "less than or equal to" relation (\( \le \)) is a classic example of an ordering relation that is reflexive and transitive but not symmetric. Always remember this common example.

Question 7. Show that the relation R in the set A of real numbers defined as R = {(a, b): a < b²} is neither reflexive nor symmetric nor transitive.
Answer: Let A be the set of real numbers and R be the relation defined as \( R = \{(a, b) : a < b^2\} \).
We need to check its properties:
1. Reflexive property:
For R to be reflexive, \( (a,a) \in R \) must be true for all \( a \in A \). This means \( a < a^2 \).
Consider \( a = \frac{1}{2} \). Then \( \frac{1}{2} < (\frac{1}{2})^2 \implies \frac{1}{2} < \frac{1}{4} \), which is false.
Also, for \( a = 1 \), \( 1 < 1^2 \implies 1 < 1 \), which is false.

\( \implies \) R is not reflexive.
2. Symmetric property:
For R to be symmetric, if \( (a,b) \in R \), then \( (b,a) \in R \) must be true.
Let \( (a,b) \in R \). This means \( a < b^2 \).
Consider \( a = 1, b = 2 \). Then \( 1 < 2^2 \implies 1 < 4 \), which is true, so \( (1,2) \in R \).
Now check \( (b,a) \): \( 2 < 1^2 \implies 2 < 1 \), which is false. So \( (2,1) \notin R \).

\( \implies \) R is not symmetric.
3. Transitive property:
For R to be transitive, if \( (a,b) \in R \) and \( (b,c) \in R \), then \( (a,c) \in R \) must be true.
Let \( (a,b) \in R \) and \( (b,c) \in R \). This means \( a < b^2 \) and \( b < c^2 \).
Consider \( a = 40, b = 7, c = 3 \).
\( (40,7) \in R \) because \( 40 < 7^2 \implies 40 < 49 \), which is true.
\( (7,3) \in R \) because \( 7 < 3^2 \implies 7 < 9 \), which is true.
Now check \( (a,c) \): \( 40 < 3^2 \implies 40 < 9 \), which is false. So \( (40,3) \notin R \).

\( \implies \) R is not transitive.
Therefore, the relation R is neither reflexive nor symmetric nor transitive.
In simple words: The relation is "a is less than b squared". For reflexive, a number is not always less than its own square (e.g., \( 0.5 \) is not less than \( 0.5^2 \)). For symmetric, if \( 1 < 2^2 \), it doesn't mean \( 2 < 1^2 \). For transitive, if \( 40 < 7^2 \) and \( 7 < 3^2 \), it doesn't mean \( 40 < 3^2 \). So, it fails all three tests.

🎯 Exam Tip: When dealing with inequalities or conditions involving powers, always test fractional values between 0 and 1, zero, negative numbers, and common positive integers. These diverse test cases help uncover counter-examples for properties like reflexivity and transitivity.

Question 8. In the set of all triangles in a plane, show that the relation of similarity is an equivalence relation.
Answer: Let S be the set of all triangles in a plane and R be the relation "is similar to". We need to show that R is an equivalence relation, which means it must be reflexive, symmetric, and transitive.
1. Reflexive property:
For any triangle \( T \in S \), every triangle is similar to itself.

\( \implies (T,T) \in R \).
Therefore, R is reflexive.
2. Symmetric property:
For R to be symmetric, if \( (T_1,T_2) \in R \), then \( (T_2,T_1) \in R \) must be true.
If triangle \( T_1 \) is similar to triangle \( T_2 \), then it naturally follows that triangle \( T_2 \) is also similar to triangle \( T_1 \). The order of comparison does not change similarity.

\( \implies (T_2,T_1) \in R \).
Therefore, R is symmetric.
3. Transitive property:
For R to be transitive, if \( (T_1,T_2) \in R \) and \( (T_2,T_3) \in R \), then \( (T_1,T_3) \in R \) must be true.
If triangle \( T_1 \) is similar to triangle \( T_2 \), and triangle \( T_2 \) is similar to triangle \( T_3 \), then by the definition of similarity (corresponding angles are equal and corresponding sides are proportional), it means that triangle \( T_1 \) is similar to triangle \( T_3 \).

\( \implies (T_1,T_3) \in R \).
Therefore, R is transitive.
Since the relation "is similar to" satisfies the reflexive, symmetric, and transitive properties, it is an equivalence relation.
In simple words: For triangles, "similarity" means they have the same shape. Any triangle is similar to itself (reflexive). If triangle A is similar to B, then B is similar to A (symmetric). If A is similar to B, and B is similar to C, then A is also similar to C (transitive). Because all three rules are met, being "similar to" is an equivalence relation.

🎯 Exam Tip: Equivalence relations are fundamental in mathematics. "Similarity" and "congruence" for geometric figures, and "equality" for numbers, are classic examples. Clearly stating each property and its justification is key to a complete proof.

Question 9. Is the relation 'is the square of' for the set of natural numbers N an equivalence relation?
Answer: Let N be the set of natural numbers and R be the relation "is the square of", so \( R = \{(a, b) : a = b^2, \text{ for } a, b \in N\} \).
We need to check if R is an equivalence relation.
1. Reflexive property:
For R to be reflexive, \( (a,a) \in R \) must be true for all \( a \in N \). This means \( a = a^2 \).
This is only true for \( a = 1 \) (since \( 1 = 1^2 \)). For any other natural number like \( 2 \), \( 2 \neq 2^2 \).

\( \implies \) R is not reflexive for all \( a \in N \).
Since R is not reflexive, it cannot be an equivalence relation. An equivalence relation must satisfy all three properties: reflexive, symmetric, and transitive.

(For completeness, let's briefly check the other properties as well, although not strictly needed to answer the main question):
2. Symmetric property:
If \( (a,b) \in R \), then \( a = b^2 \). For R to be symmetric, \( (b,a) \in R \) must be true, meaning \( b = a^2 \).
Consider \( (4,2) \in R \) since \( 4 = 2^2 \). But \( (2,4) \notin R \) because \( 2 \neq 4^2 \).

\( \implies \) R is not symmetric.
3. Transitive property:
If \( (a,b) \in R \) and \( (b,c) \in R \), then \( a = b^2 \) and \( b = c^2 \). For R to be transitive, \( (a,c) \in R \) must be true, meaning \( a = c^2 \).
If \( a = b^2 \) and \( b = c^2 \), then \( a = (c^2)^2 = c^4 \). For \( a \) to be \( c^2 \), it would require \( c^4 = c^2 \), which implies \( c^2 = 1 \), so \( c=1 \). This is not true for all natural numbers.
Consider \( a=16, b=4, c=2 \). \( (16,4) \in R \) since \( 16=4^2 \). \( (4,2) \in R \) since \( 4=2^2 \). But \( (16,2) \notin R \) since \( 16 \neq 2^2 \).

\( \implies \) R is not transitive.
Therefore, the relation "is the square of" is not an equivalence relation on the set of natural numbers.
In simple words: No, the relation "is the square of" is not an equivalence relation for natural numbers. For a relation to be an equivalence relation, it must be reflexive (a number is its own square), symmetric (if A is square of B, B is square of A), and transitive. This relation fails all three tests, for example, 2 is not its own square, if 4 is the square of 2, 2 is not the square of 4, and if 16 is the square of 4 and 4 is the square of 2, 16 is not the square of 2.

🎯 Exam Tip: To prove a relation is NOT an equivalence relation, you only need to show that it fails one of the three properties (reflexive, symmetric, transitive). Pick the easiest property to disprove and provide a clear counter-example.

Question 10. Consider the following properties of relations : Symmetric (S), Transitive (T), Reflexive (R) and Equivalence (E). A relation may have all, some or none of these properties. For each part, state whether the relation has some or all of the properties mentioned by writing the letters S, T, R, E in the space provided. Write N, if none of the properties satisfy.
(a) 'is smaller than',
(b) 'is the father of',
(c) 'is parallel to' for set of straight lines.
(d) 'is a multiple of for a set of positive integers'
(e) 'is congruent to'.
Answer:
(a) For the relation "is smaller than" (\( a < b \)) on real numbers (Let R1 = \( \{(a, b) : a < b\} \)):
Reflexive: A number is not smaller than itself (\( a \nless a \)). So, not reflexive.
Symmetric: If \( a < b \), it does not mean \( b < a \) (e.g., \( 1 < 2 \) but \( 2 \nless 1 \)). So, not symmetric.
Transitive: If \( a < b \) and \( b < c \), then \( a < c \) is true. So, transitive.
Properties satisfied: T
(b) For the relation "is the father of" on a set of people (Let R1 = \( \{(a, b) : a \text{ is the father of } b\} \)):
Reflexive: A person cannot be their own father. So, not reflexive.
Symmetric: If \( a \) is the father of \( b \), then \( b \) is the son or daughter of \( a \), not the father of \( a \). So, not symmetric.
Transitive: If \( a \) is the father of \( b \), and \( b \) is the father of \( c \), then \( a \) is the grandfather of \( c \), not the father of \( c \). So, not transitive.
Properties satisfied: N
(c) For the relation "is parallel to" on a set of straight lines (Let R1 = \( \{(a, b) : a \parallel b\} \)):
Reflexive: Every line is parallel to itself. So, reflexive.
Symmetric: If line \( a \) is parallel to line \( b \), then line \( b \) is also parallel to line \( a \). So, symmetric.
Transitive: If line \( a \) is parallel to line \( b \), and line \( b \) is parallel to line \( c \), then line \( a \) is parallel to line \( c \). So, transitive.
Since it is reflexive, symmetric, and transitive, this relation is an Equivalence relation.
Properties satisfied: R, S, T, E
(d) For the relation "is a multiple of" for a set of positive integers (Let R1 = \( \{(a, b) : a \text{ is a multiple of } b\} \)):
Reflexive: Every positive integer is a multiple of itself (\( a = a \times 1 \)). So, reflexive.
Symmetric: If \( a \) is a multiple of \( b \), \( b \) is not necessarily a multiple of \( a \) (e.g., \( 6 \) is a multiple of \( 3 \), but \( 3 \) is not a multiple of \( 6 \)). So, not symmetric.
Transitive: If \( a \) is a multiple of \( b \), and \( b \) is a multiple of \( c \), then \( a \) is a multiple of \( c \) (e.g., if \( a=kb \) and \( b=mc \), then \( a=kmc \)). So, transitive.
Properties satisfied: R, T
(e) For the relation "is congruent to" on a set of triangles (Let R1 = \( \{(a, b) : a \text{ is congruent to } b\} \)):
Reflexive: Every triangle is congruent to itself. So, reflexive.
Symmetric: If triangle \( a \) is congruent to triangle \( b \), then triangle \( b \) is also congruent to triangle \( a \). So, symmetric.
Transitive: If triangle \( a \) is congruent to triangle \( b \), and triangle \( b \) is congruent to triangle \( c \), then triangle \( a \) is congruent to triangle \( c \). So, transitive.
Since it is reflexive, symmetric, and transitive, this relation is an Equivalence relation.
Properties satisfied: R, S, T, E
In simple words: We check each everyday relationship to see if it fits the rules of reflexive (related to self), symmetric (can switch order), and transitive (chain reaction). "Smaller than" is only transitive. "Father of" has none. "Parallel to" is all three, making it an equivalence relation. "Multiple of" is reflexive and transitive. "Congruent to" is also all three, making it an equivalence relation.

🎯 Exam Tip: For each relation, systematically check for reflexivity, symmetry, and transitivity. If all three hold, then it's an equivalence relation. Provide a concise reason or counter-example for each check.

Question 11. Answer true or false: The relation "is congruent to” in a set of triangles is a transitive relation.
Answer: True
In simple words: If one triangle is congruent (exactly the same size and shape) to a second triangle, and that second triangle is congruent to a third triangle, then the first triangle must also be congruent to the third. This "chain reaction" means the relation is transitive.

🎯 Exam Tip: Congruence for geometric figures (like triangles) is a fundamental equivalence relation, meaning it satisfies all three properties: reflexive, symmetric, and transitive. Recognizing these common examples can save time.

Question 12. If is a relation in N x N defined by (a, b) R (c, d) if and only if a + d = b + c, show that R is an equivalence relation.
Answer: Let N be the set of natural numbers, and the relation R on \( N \times N \) is defined as \( (a,b) R (c,d) \) if and only if \( a+d = b+c \). We need to show that R is an equivalence relation.
1. Reflexive property:
For R to be reflexive, \( (a,b) R (a,b) \) must be true for all \( (a,b) \in N \times N \).
Using the definition, this means \( a+b = b+a \).
This statement is always true due to the commutative law of addition for natural numbers.

\( \implies (a,b) R (a,b) \).
Therefore, R is reflexive.
2. Symmetric property:
For R to be symmetric, if \( (a,b) R (c,d) \), then \( (c,d) R (a,b) \) must be true.
Assume \( (a,b) R (c,d) \). By definition, \( a+d = b+c \).
We can rearrange this equation to \( c+b = d+a \).
By the definition of R, \( c+b = d+a \) means \( (c,d) R (a,b) \).

\( \implies (c,d) R (a,b) \).
Therefore, R is symmetric.
3. Transitive property:
For R to be transitive, if \( (a,b) R (c,d) \) and \( (c,d) R (e,f) \), then \( (a,b) R (e,f) \) must be true.
Assume \( (a,b) R (c,d) \). This means \( a+d = b+c \) ... (1)
Assume \( (c,d) R (e,f) \). This means \( c+f = d+e \) ... (2)
Now, add equation (1) and equation (2):
\( (a+d) + (c+f) = (b+c) + (d+e) \)
Rearrange the terms:
\( a+d+c+f = b+c+d+e \)
Subtract \( c \) and \( d \) from both sides:
\( a+f = b+e \)
By the definition of R, \( a+f = b+e \) means \( (a,b) R (e,f) \).

\( \implies (a,b) R (e,f) \).
Therefore, R is transitive.
Since R is reflexive, symmetric, and transitive, it is an equivalence relation on \( N \times N \).
In simple words: This relation checks if the sum of the first number of the first pair and the second number of the second pair equals the sum of the second number of the first pair and the first number of the second pair. It's reflexive because \( a+b \) always equals \( b+a \). It's symmetric because if \( a+d=b+c \), then you can swap the pairs to get \( c+b=d+a \). It's transitive because if the first pair relates to the second, and the second relates to the third, you can add their rules and simplify to show the first pair relates to the third. All three properties are true, so it's an equivalence relation.

🎯 Exam Tip: When proving equivalence relations on ordered pairs or more complex structures, break down the definition for each property, substitute the relation's rule, and use algebraic manipulation (like addition and subtraction of equations) to show the property holds.

Question 13. Prove that the relation R in the set \( A = \{1, 2, 3, 4, 5\} \) given by \( R = \{(a, b): | a - b | \) is even} is an equivalence relation.
Answer:
Given that the set is \( A = \{1, 2, 3, 4, 5\} \) and the relation is \( R = \{(a, b) : |a - b| \) is an even number}. We need to prove this is an equivalence relation, meaning it must be reflexive, symmetric, and transitive. **Reflexive:** For any element \( a \in A \), we check if \( (a, a) \in R \). \( |a - a| = 0 \). Since 0 is an even number, \( (a, a) \in R \) is true.
\( \implies \) Therefore, R is reflexive on A. **Symmetric:** For any \( (a, b) \in R \), we check if \( (b, a) \in R \). If \( (a, b) \in R \), then \( |a - b| \) is an even number. This means \( |a - b| = 2m \) for some integer \( m \). We know that \( |a - b| = |-(b - a)| = |b - a| \). So, \( |b - a| = 2m \), which means \( |b - a| \) is also an even number.
\( \implies \) Therefore, \( (b, a) \in R \), and R is symmetric on A. **Transitive:** For any \( (a, b) \in R \) and \( (b, c) \in R \), we check if \( (a, c) \in R \). If \( (a, b) \in R \), then \( |a - b| \) is even, so \( a - b = \pm 2m \) for some integer \( m \) (Equation 1). If \( (b, c) \in R \), then \( |b - c| \) is even, so \( b - c = \pm 2n \) for some integer \( n \) (Equation 2). Adding Equation 1 and Equation 2: \( (a - b) + (b - c) = \pm 2m \pm 2n \) \( a - c = \pm 2(m \pm n) \) Let \( m' = m \pm n \). Then \( a - c = \pm 2m' \). This shows that \( a - c \) is an even number.
\( \implies \) Thus, \( |a - c| \) is an even number, so \( (a, c) \in R \).
\( \implies \) Therefore, R is transitive on A. Since R is reflexive, symmetric, and transitive, it is an equivalence relation on A. A relation being an equivalence relation helps categorize or partition a set into disjoint subsets called equivalence classes.
In simple words: This relation checks if the difference between two numbers is an even number. It works for all numbers (reflexive), works backward (symmetric), and if it works for A to B and B to C, it also works for A to C (transitive). Because of these three rules, it's called an equivalence relation.

🎯 Exam Tip: To prove a relation is an equivalence relation, you must explicitly demonstrate its reflexivity, symmetry, and transitivity with clear mathematical steps and reasoning for each property. Don't skip any step.

Question 14. Let I be the set of all integers and R be the relation on I defined by a R b iff \( (a + b) \) is an even integer for all \( a, b \in I \). Prove that R is an equivalence relation.
Answer:
Given that I is the set of all integers, and the relation R is defined such that \( a R b \) if and only if \( (a + b) \) is an even integer. We need to prove this is an equivalence relation. **Reflexive:** For any integer \( a \in I \), we check if \( a R a \). This means we check if \( (a + a) \) is an even integer. \( a + a = 2a \). Since \( 2a \) is always an even integer (as it's a multiple of 2), \( a R a \) is true for all \( a \in I \).
\( \implies \) Therefore, R is reflexive on I. **Symmetric:** For any integers \( a, b \in I \) such that \( a R b \), we check if \( b R a \). If \( a R b \), then \( (a + b) \) is an even integer. We know that \( a + b = b + a \) (due to the commutative property of addition for integers). Since \( (a + b) \) is even, \( (b + a) \) must also be an even integer.
\( \implies \) Therefore, \( b R a \), and R is symmetric on I. **Transitive:** For any integers \( a, b, c \in I \) such that \( a R b \) and \( b R c \), we check if \( a R c \). If \( a R b \), then \( (a + b) \) is an even integer. So, \( a + b = 2m \) for some integer \( m \) (Equation 1). If \( b R c \), then \( (b + c) \) is an even integer. So, \( b + c = 2n \) for some integer \( n \) (Equation 2). Adding Equation 1 and Equation 2: \( (a + b) + (b + c) = 2m + 2n \) \( a + 2b + c = 2(m + n) \) Subtract \( 2b \) from both sides: \( a + c = 2(m + n) - 2b \) \( a + c = 2(m + n - b) \) Let \( m' = m + n - b \). Since \( m, n, b \) are integers, \( m' \) is also an integer. So, \( a + c = 2m' \), which means \( (a + c) \) is an even integer.
\( \implies \) Therefore, \( a R c \), and R is transitive on I. Since the relation R is reflexive, symmetric, and transitive, R is an equivalence relation on I. This type of relation helps to group integers into categories based on whether their sum is even.
In simple words: This relation means two numbers are "related" if their sum is an even number. It is reflexive because any number added to itself is even (like 2+2=4). It is symmetric because if A+B is even, then B+A is also even. It is transitive because if A+B is even and B+C is even, then A+C must also be even. Because it follows all these three rules, it is an equivalence relation.

🎯 Exam Tip: When proving relations with properties like "even" or "multiple of N", remember to use algebraic expressions (e.g., \( 2m \), \( Nm \)) to clearly show the property holds for all cases, not just specific examples.

Question 15. Let I be the set of all integers and R be the relation on I defined by \( R = \{(x, y): x, y \in I, x - y \) is divisible by 11}. Prove that R is an equivalence relation.
Answer:
Given that I is the set of all integers, and the relation R is defined such that \( (x, y) \in R \) if and only if \( (x - y) \) is divisible by 11. We need to prove this is an equivalence relation. **Reflexive:** For any integer \( x \in I \), we check if \( (x, x) \in R \). This means we check if \( (x - x) \) is divisible by 11. \( x - x = 0 \). Since 0 is divisible by any non-zero integer (0 divided by 11 is 0), \( (x - x) \) is divisible by 11.
\( \implies \) Therefore, \( (x, x) \in R \), and R is reflexive on I. **Symmetric:** For any integers \( x, y \in I \) such that \( (x, y) \in R \), we check if \( (y, x) \in R \). If \( (x, y) \in R \), then \( (x - y) \) is divisible by 11. This means \( x - y = 11K \) for some integer \( K \). Then, \( y - x = -(x - y) = -(11K) = 11(-K) \). Let \( K' = -K \). Since \( K \) is an integer, \( K' \) is also an integer. So, \( y - x = 11K' \). This means \( (y - x) \) is divisible by 11.
\( \implies \) Therefore, \( (y, x) \in R \), and R is symmetric on I. **Transitive:** For any integers \( x, y, z \in I \) such that \( (x, y) \in R \) and \( (y, z) \in R \), we check if \( (x, z) \in R \). If \( (x, y) \in R \), then \( (x - y) \) is divisible by 11. So, \( x - y = 11K_1 \) for some integer \( K_1 \) (Equation 1). If \( (y, z) \in R \), then \( (y - z) \) is divisible by 11. So, \( y - z = 11K_2 \) for some integer \( K_2 \) (Equation 2). Adding Equation 1 and Equation 2: \( (x - y) + (y - z) = 11K_1 + 11K_2 \) \( x - z = 11(K_1 + K_2) \) Let \( K' = K_1 + K_2 \). Since \( K_1 \) and \( K_2 \) are integers, \( K' \) is also an integer. So, \( x - z = 11K' \), which means \( (x - z) \) is divisible by 11.
\( \implies \) Therefore, \( (x, z) \in R \), and R is transitive on I. Since the relation R is reflexive, symmetric, and transitive, R is an equivalence relation on I. This concept groups numbers that have the same remainder when divided by 11.
In simple words: This relation connects two numbers if their difference can be divided by 11 without any remainder. It is reflexive because any number minus itself is zero, and zero can be divided by 11. It is symmetric because if (A-B) can be divided by 11, then (B-A) can also be divided by 11. It is transitive because if (A-B) is divisible by 11 and (B-C) is divisible by 11, then (A-C) must also be divisible by 11. All these properties make it an equivalence relation.

🎯 Exam Tip: When dealing with divisibility relations, remember that 0 is divisible by any non-zero integer, which is key for proving reflexivity. Always show the algebraic steps clearly for symmetry and transitivity.