GSEB Class 12 Maths Solutions Chapter 1 Relations and Functions Exercise 1.1

Question 1. Determine whether each of the following relations are reflexive, symmetric and transitive:
(i) Relation R in the set \( A = \{1, 2, 3, ..., 13, 14\} \) defined as \( R = \{(x, y) : 3x - y = 0\} \)
(ii) Relation R in the set \( N \) of natural numbers defined as \( R = \{(x, y) : y = x + 5 \) and \( x < 4\} \)
(iii) Relation R in the set \( A = \{1, 2, 3, 4, 5, 6\} \) as \( R = \{(x, y) : y \) is divisible by \( x\} \)
(iv) Relation R in the set \( Z \) of all integers defined as \( R = \{(x, y) : x - y \) is an integer\( \} \)
(v) Relation R in the set \( A \) of human beings in a town at a particular time given by
(a) \( R = \{(x, y) : x \) and \( y \) work at the same place\( \} \)
(b) \( R = \{(x, y) : x \) and \( y \) live in the same locality\( \} \)
(c) \( R = \{(x, y) : x \) is exactly 7 cm taller than \( y\} \)
(d) \( R = \{(x, y) : x \) is wife of \( y\} \)
(e) \( R = \{(x, y) : x \) is father of \( y\} \)
Answer:
(i) Relation R in the set \( A = \{1, 2, ..., 14\} \) defined as \( R = \{(x, y) : 3x - y = 0\} \)
(a) For reflexivity, we check if \( (x, x) \in R \). This means \( 3x - x = 0 \), which simplifies to \( 2x = 0 \). This implies \( x = 0 \). However, the set \( A \) contains natural numbers from 1 to 14. Since \( 0 \notin A \), \( (x, x) \notin R \) for any \( x \in A \). Therefore, R is not reflexive.
(b) For symmetry, we check if \( (y, x) \in R \) whenever \( (x, y) \in R \). If \( (x, y) \in R \), then \( 3x - y = 0 \), which means \( y = 3x \). For \( (y, x) \) to be in R, we would need \( 3y - x = 0 \). Consider \( x = 1 \). Then \( y = 3(1) = 3 \). So \( (1, 3) \in R \). Now, let's check for \( (3, 1) \). We need \( 3(3) - 1 = 0 \), which is \( 9 - 1 = 8 \), not 0. Since \( (3, 1) \notin R \), the relation R is not symmetric.
(c) For transitivity, we check if \( (x, z) \in R \) whenever \( (x, y) \in R \) and \( (y, z) \in R \). If \( (x, y) \in R \), then \( y = 3x \). If \( (y, z) \in R \), then \( z = 3y \). Substituting the value of \( y \), we get \( z = 3(3x) = 9x \). For \( (x, z) \) to be in R, we need \( z = 3x \). Since \( 9x \ne 3x \) for any \( x \ne 0 \), which is true for \( x \in A \), \( (x, z) \notin R \). Therefore, R is not transitive.
(ii) Relation R in the set \( N \) of natural numbers defined as \( R = \{(x, y) : y = x + 5 \) and \( x < 4\} \)
(a) For reflexivity, we check if \( (x, x) \in R \). This implies \( x = x + 5 \), which simplifies to \( 0 = 5 \). This statement is false. Therefore, \( (x, x) \notin R \) for any \( x \in N \). So, R is not reflexive.
(b) For symmetry, we check if \( (y, x) \in R \) whenever \( (x, y) \in R \). If \( (x, y) \in R \), then \( y = x + 5 \) and \( x < 4 \). For \( (y, x) \) to be in R, we would need \( x = y + 5 \) and \( y < 4 \). Substitute \( y = x + 5 \) into the symmetry condition: \( x = (x + 5) + 5 \), which gives \( x = x + 10 \), simplifying to \( 0 = 10 \). This is false. Therefore, R is not symmetric.
(c) For transitivity, we check if \( (x, z) \in R \) whenever \( (x, y) \in R \) and \( (y, z) \in R \). If \( (x, y) \in R \), then \( y = x + 5 \) and \( x < 4 \). Since \( x \in N \), \( x \) can be 1, 2, or 3. This means \( y \) can be 6, 7, or 8. For \( (y, z) \) to be in R, the first element \( y \) must satisfy \( y < 4 \). However, the values of \( y \) that arise from \( (x, y) \in R \) (i.e., 6, 7, 8) are all greater than or equal to 4. Therefore, there are no pairs \( (x, y) \) and \( (y, z) \) such that both \( (x, y) \in R \) and \( (y, z) \in R \). A relation for which the premise of transitivity is never met is considered vacuously transitive. Thus, R is transitive.
(iii) Relation R in the set \( A = \{1, 2, 3, 4, 5, 6\} \) as \( R = \{(x, y) : y \) is divisible by \( x\} \)
(a) For reflexivity, we check if \( (x, x) \in R \). This means \( x \) is divisible by \( x \), which is true for all \( x \in A \). Therefore, R is reflexive.
(b) For symmetry, we check if \( (y, x) \in R \) whenever \( (x, y) \in R \). If \( (x, y) \in R \), then \( y \) is divisible by \( x \). For \( (y, x) \) to be in R, \( x \) must be divisible by \( y \). This is not always the case. For example, \( (2, 4) \in R \) because 4 is divisible by 2. However, \( (4, 2) \notin R \) because 2 is not divisible by 4. Thus, R is not symmetric.
(c) For transitivity, we check if \( (x, z) \in R \) whenever \( (x, y) \in R \) and \( (y, z) \in R \). If \( (x, y) \in R \), then \( y \) is divisible by \( x \). If \( (y, z) \in R \), then \( z \) is divisible by \( y \). From these, we can conclude that \( z \) is also divisible by \( x \). For instance, if \( (1, 2) \in R \) (since 2 is divisible by 1) and \( (2, 4) \in R \) (since 4 is divisible by 2), then \( (1, 4) \in R \) (since 4 is divisible by 1). Therefore, R is transitive.
(iv) Relation R in the set \( Z \) of all integers defined as \( R = \{(x, y) : x - y \) is an integer\( \} \)
(a) For reflexivity, we check if \( (x, x) \in R \). This means \( x - x = 0 \). Since 0 is an integer, \( (x, x) \in R \) for all integers \( x \). Therefore, R is reflexive.
(b) For symmetry, we check if \( (y, x) \in R \) whenever \( (x, y) \in R \). If \( (x, y) \in R \), then \( x - y \) is an integer. We know that \( y - x = -(x - y) \). Since the negative of an integer is also an integer, \( y - x \) is an integer. Thus, \( (y, x) \in R \). Therefore, R is symmetric.
(c) For transitivity, we check if \( (x, z) \in R \) whenever \( (x, y) \in R \) and \( (y, z) \in R \). If \( (x, y) \in R \), then \( x - y \) is an integer. If \( (y, z) \in R \), then \( y - z \) is an integer. The sum of two integers is always an integer. So, \( (x - y) + (y - z) = x - z \) is an integer. Thus, \( (x, z) \in R \). Therefore, R is transitive.
(v) Relation R in the set \( A \) of human beings in a town at a particular time given by
(a) \( R = \{(x, y) : x \) and \( y \) work at the same place\( \} \)
For reflexivity, \( x \) works at the same place as \( x \). This is true. So, R is reflexive.
For symmetry, if \( x \) works at the same place as \( y \), then \( y \) also works at the same place as \( x \). This is true. So, R is symmetric.
For transitivity, if \( x \) works at the same place as \( y \), and \( y \) works at the same place as \( z \), then \( x \) works at the same place as \( z \). This is true. So, R is transitive.
(b) \( R = \{(x, y) : x \) and \( y \) live in the same locality\( \} \)
Similar to part (a), this relation is reflexive, symmetric, and transitive. If \( x \) lives in the same locality as \( x \), it's reflexive. If \( x \) lives in the same locality as \( y \), then \( y \) lives in the same locality as \( x \), making it symmetric. If \( x \) lives in the same locality as \( y \), and \( y \) lives in the same locality as \( z \), then \( x \) lives in the same locality as \( z \), making it transitive.
(c) \( R = \{(x, y) : x \) is exactly 7 cm taller than \( y\} \)
For reflexivity, \( x \) cannot be exactly 7 cm taller than himself. So, R is not reflexive.
For symmetry, if \( x \) is exactly 7 cm taller than \( y \), then \( y \) cannot be exactly 7 cm taller than \( x \) (instead, \( y \) would be 7 cm shorter). So, R is not symmetric.
For transitivity, if \( x \) is exactly 7 cm taller than \( y \), and \( y \) is exactly 7 cm taller than \( z \), then \( x \) would be 14 cm taller than \( z \), not 7 cm. So, R is not transitive.
(d) \( R = \{(x, y) : x \) is wife of \( y\} \)
For reflexivity, \( x \) cannot be the wife of \( x \). So, R is not reflexive.
For symmetry, if \( x \) is the wife of \( y \), then \( y \) is male. Therefore, \( y \) cannot be the wife of \( x \). So, R is not symmetric.
For transitivity, if \( (x, y) \in R \) and \( (y, z) \in R \). If \( x \) is the wife of \( y \), then \( y \) is a male. For \( (y, z) \in R \), \( y \) would need to be the wife of \( z \), which is impossible since \( y \) is male. Therefore, the condition for transitivity (the premise) is never met, making the relation vacuously transitive. Thus, R is transitive.
(e) \( R = \{(x, y) : x \) is father of \( y\} \)
For reflexivity, \( x \) cannot be the father of \( x \). So, R is not reflexive.
For symmetry, if \( x \) is the father of \( y \), then \( y \) cannot be the father of \( x \). So, R is not symmetric.
For transitivity, if \( x \) is the father of \( y \), and \( y \) is the father of \( z \), then \( x \) is the grandfather of \( z \), not the father. So, R is not transitive.
In simple words: For relation (i), it is not reflexive because a number cannot be three times itself unless it's zero, and zero is not in the set. It is not symmetric because if the second number is three times the first, the first is not three times the second. It is not transitive because if y is three times x and z is three times y, then z is nine times x, not three times x. For relation (ii), it is not reflexive as a number is never equal to itself plus five. It is not symmetric as if y is x plus five, x is not y plus five. However, it is vacuously transitive because the condition for transitivity is never met given the 'x < 4' rule. For relation (iii), it is reflexive as a number is divisible by itself. It is not symmetric because if y is divisible by x, x is not always divisible by y. It is transitive because if y is divisible by x and z by y, then z is divisible by x. For relation (iv), it is reflexive as the difference between a number and itself is 0 (an integer). It is symmetric as if x-y is an integer, so is y-x. It is transitive as the sum of two integer differences is also an integer difference. For relation (v)(a), it is reflexive, symmetric, and transitive because working in the same place satisfies all conditions. For (v)(b), similarly, living in the same locality is reflexive, symmetric, and transitive. For (v)(c), being exactly 7cm taller is neither reflexive, symmetric, nor transitive. For (v)(d), 'is wife of' is not reflexive or symmetric, but it is vacuously transitive. For (v)(e), 'is father of' is neither reflexive, symmetric, nor transitive.

Exam Tip: Always verify each property (reflexivity, symmetry, transitivity) individually by substituting specific values from the given set or by showing a general algebraic proof. Be careful with logical conditions.

Question 2. Show that the relation R in the set \( R \) of real numbers, defined as \( R = \{(a, b) : a \le b^2\} \), is neither reflexive, nor symmetric, nor transitive.
Answer:
(i) R is not reflexive: For the relation to be reflexive, \( a \le a^2 \) must hold for all \( a \in R \). However, this is not true for all real numbers. For instance, if \( a = \frac{1}{2} \), then \( \frac{1}{2} \le (\frac{1}{2})^2 \) means \( \frac{1}{2} \le \frac{1}{4} \), which is false. Thus, R is not reflexive.
(ii) R is not symmetric: For the relation to be symmetric, if \( a \le b^2 \), then \( b \le a^2 \) must also hold. Let's consider an example: \( 2 \le 5^2 \) (which is \( 2 \le 25 \)) is true. But for symmetry, \( 5 \le 2^2 \) (which is \( 5 \le 4 \)) should be true, which is false. Therefore, R is not symmetric.
(iii) R is not transitive: For the relation to be transitive, if \( a \le b^2 \) and \( b \le c^2 \), then \( a \le c^2 \) must hold. Let's take an example: \( 2 \le (-2)^2 \) (which is \( 2 \le 4 \)) is true. Also, \( -2 \le (-1)^2 \) (which is \( -2 \le 1 \)) is true. But if we check \( a \le c^2 \), we get \( 2 \le (-1)^2 \) (which is \( 2 \le 1 \)), which is false. Hence, R is not transitive.
In simple words: The relation is not reflexive because a number isn't always less than or equal to its square (like 1/2). It's not symmetric because if 'a' is less than or equal to 'b' squared, 'b' isn't always less than or equal to 'a' squared. It's not transitive because even if 'a' relates to 'b' and 'b' relates to 'c', 'a' doesn't always relate to 'c' (like the example with 2, -2, -1).

Exam Tip: For relations involving inequalities like \( a \le b^2 \), negative numbers and fractions between 0 and 1 are often good choices for counter-examples to disprove reflexivity, symmetry, or transitivity.

Question 3. Check whether the relation R defined in the set \( \{1, 2, 3, 4, 5, 6\} \) as \( R = \{(a, b) : b = a + 1\} \) is reflexive, symmetric and transitive.
Answer:
(i) R is not reflexive: For reflexivity, \( (a, a) \) must be in R, which means \( a = a + 1 \). This statement simplifies to \( 0 = 1 \), which is false. Therefore, R is not reflexive.
(ii) R is not symmetric: For symmetry, if \( (a, b) \in R \), then \( (b, a) \) must also be in R. If \( (a, b) \in R \), we have \( b = a + 1 \). For \( (b, a) \) to be in R, we would need \( a = b + 1 \). Substituting the first equation into the second, we get \( a = (a + 1) + 1 \), which simplifies to \( a = a + 2 \), or \( 0 = 2 \). This is false. Therefore, R is not symmetric.
(iii) R is not transitive: For transitivity, if \( (a, b) \in R \) and \( (b, c) \in R \), then \( (a, c) \) must also be in R. If \( (a, b) \in R \), then \( b = a + 1 \). If \( (b, c) \in R \), then \( c = b + 1 \). Substituting the value of \( b \) into the second equation, we get \( c = (a + 1) + 1 \), so \( c = a + 2 \). For \( (a, c) \) to be in R, we would need \( c = a + 1 \). Since \( a + 2 \ne a + 1 \), it means \( (a, c) \notin R \). Therefore, R is not transitive.
In simple words: The relation is not reflexive because a number is not equal to itself plus one. It is not symmetric because if 'b' is 'a' plus one, then 'a' cannot be 'b' plus one. It is also not transitive because if 'b' is one more than 'a', and 'c' is one more than 'b', then 'c' is two more than 'a', not one more than 'a'.

Exam Tip: Relations defined by a specific arithmetic operation (like \( b = a + 1 \)) often fail to be reflexive, symmetric, or transitive. Use simple numbers as examples to quickly test each property.

Question 4. Show that the relation R in \( R \) defined as \( R = \{(a, b) : a \le b\} \) is reflexive and transitive but not symmetric.
Answer:
(i) R is reflexive: For reflexivity, we check if \( a \le a \). This statement is always true for any real number \( a \). Therefore, R is reflexive.
(ii) R is not symmetric: For symmetry, if \( (a, b) \in R \), then \( (b, a) \) must also be in R. If \( a \le b \), it does not necessarily mean that \( b \le a \). For example, \( 2 \le 3 \) is true, but \( 3 \le 2 \) is false. Therefore, R is not symmetric.
(iii) R is transitive: For transitivity, if \( (a, b) \in R \) and \( (b, c) \in R \), then \( (a, c) \) must also be in R. This means if \( a \le b \) and \( b \le c \), then it logically follows that \( a \le c \). For example, if \( 2 \le 3 \) and \( 3 \le 4 \), then \( 2 \le 4 \). Therefore, R is transitive.
In simple words: This relation is reflexive because any number is less than or equal to itself. It is not symmetric because if 'a' is less than or equal to 'b', 'b' isn't always less than or equal to 'a'. It is transitive because if 'a' is less than or equal to 'b', and 'b' is less than or equal to 'c', then 'a' must be less than or equal to 'c'.

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. Understand why each property holds or fails.

Question 5. Check whether the relation R, defined by \( R = \{(a, b) : a \le b^3\} \) is reflexive, symmetric or transitive.
Answer:
(i) R is not reflexive: For reflexivity, \( a \le a^3 \) must be true for all real numbers \( a \). However, this is not always true. For example, if \( a = \frac{1}{2} \), then \( \frac{1}{2} \le (\frac{1}{2})^3 \) means \( \frac{1}{2} \le \frac{1}{8} \), which is false. Therefore, R is not reflexive.
(ii) R is not symmetric: For symmetry, if \( a \le b^3 \), then \( b \le a^3 \) must also be true. Consider \( a = 1 \) and \( b = 2 \). We have \( 1 \le 2^3 \) (which is \( 1 \le 8 \)), which is true. But for the relation to be symmetric, \( 2 \le 1^3 \) (which is \( 2 \le 1 \)) should be true, which is false. Thus, R is not symmetric.
(iii) R is not transitive: For transitivity, if \( a \le b^3 \) and \( b \le c^3 \), then \( a \le c^3 \) must follow. Let's examine \( a = 7 \), \( b = 2 \), and \( c = 1.5 \).
First, \( a \le b^3 \): \( 7 \le 2^3 \Rightarrow 7 \le 8 \), which is true.
Second, \( b \le c^3 \): \( 2 \le (1.5)^3 \Rightarrow 2 \le 3.375 \), which is true.
However, if we check for transitivity \( a \le c^3 \): \( 7 \le (1.5)^3 \Rightarrow 7 \le 3.375 \), which is false. Thus, R is not transitive.
In simple words: This relation is not reflexive because a number isn't always less than or equal to its cube (like 1/2). It is not symmetric because if 'a' is less than or equal to 'b' cubed, 'b' isn't always less than or equal to 'a' cubed. It is also not transitive because even if 'a' relates to 'b' and 'b' relates to 'c', 'a' doesn't always relate to 'c', as seen with the example 7, 2, 1.5.

Exam Tip: When working with cubic relations, remember to test values greater than 1, fractions between 0 and 1, and negative numbers. These different ranges often reveal counter-examples more easily.

Question 6. 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 \( A = \{1, 2, 3\} \). The given relation is \( R = \{(1, 2), (2, 1)\} \).
(i) For reflexivity: We check if \( (x, x) \in R \) for all \( x \in A \). The pairs \( (1, 1), (2, 2), \) and \( (3, 3) \) are not present in R. Therefore, R is not reflexive.
(ii) For symmetry: We check if \( (y, x) \in R \) whenever \( (x, y) \in R \). We have \( (1, 2) \in R \) and its reverse pair \( (2, 1) \) is also in R. Since this is true for all pairs in R, the relation R is symmetric.
(iii) For transitivity: We check if \( (x, z) \in R \) whenever \( (x, y) \in R \) and \( (y, z) \in R \). We have \( (1, 2) \in R \) and \( (2, 1) \in R \). For transitivity to hold, \( (1, 1) \) must be in R. However, \( (1, 1) \notin R \). Therefore, R is not transitive.
In simple words: For the set {1, 2, 3} and the relation {(1, 2), (2, 1)}, it is not reflexive because pairs like (1, 1) are missing. It is symmetric because if (1, 2) is there, (2, 1) is also there. It is not transitive because (1, 2) and (2, 1) are in the relation, but (1, 1) is not.

Exam Tip: When testing for transitivity, identify pairs \( (x, y) \) and \( (y, z) \) that exist in the relation. If their corresponding \( (x, z) \) pair is missing, the relation is not transitive. A single counter-example is enough.

Question 7. Show that the relation R in the set \( A \) of all books in a library of a college, given by \( R = \{(x, y) : x \) and \( y \) have the same number of pages\( \} \) is an equivalence relation.
Answer:
To show R is an equivalence relation, we must demonstrate that it is reflexive, symmetric, and transitive.
(i) For reflexivity: A book \( x \) always has the same number of pages as itself. Thus, \( (x, x) \in R \). Therefore, R is reflexive.
(ii) For symmetry: If \( (x, y) \in R \), it means book \( x \) has the same number of pages as book \( y \). This implies that book \( y \) also has the same number of pages as book \( x \). Thus, \( (y, x) \in R \). Therefore, R is symmetric.
(iii) For transitivity: If \( (x, y) \in R \) and \( (y, z) \in R \). This means book \( x \) has the same number of pages as book \( y \), and book \( y \) has the same number of pages as book \( z \). From this, we can conclude that book \( x \) must have the same number of pages as book \( z \). Thus, \( (x, z) \in R \). Therefore, R is transitive.
Since R is reflexive, symmetric, and transitive, it is an equivalence relation.
In simple words: This relation is an equivalence relation because: 1) Any book has the same number of pages as itself (reflexive). 2) If book X has the same pages as book Y, then Y has the same pages as X (symmetric). 3) If X has the same pages as Y, and Y has the same pages as Z, then X has the same pages as Z (transitive).

Exam Tip: For problems proving an equivalence relation, clearly state the definition of reflexivity, symmetry, and transitivity in the context of the problem, and then show how the given relation satisfies each one.

Question 8. Show 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. Show that all the elements of \( \{1, 3, 5\} \) are related to each other and all the elements of \( \{2, 4\} \) are related to each other. But no element of \( \{1, 3, 5\} \) is related to any element of \( \{2, 4\} \).
Answer:
The given set is \( A = \{1, 2, 3, 4, 5\} \), and the relation is \( R = \{(a, b) : |a - b| \) is even\( \} \). The relation can be explicitly written as \( R = \{(1, 3), (1, 5), (3, 5), (2, 4), (3, 1), (5, 1), (5, 3), (4, 2), (1,1), (2,2), (3,3), (4,4), (5,5)\} \).
(a) To prove R is an equivalence relation:
(i) For reflexivity: For any \( a \in A \), \( |a - a| = 0 \). Since 0 is an even number, \( (a, a) \in R \). Therefore, R is reflexive.
(ii) For symmetry: If \( (a, b) \in R \), then \( |a - b| \) is an even number. We know that \( |b - a| = |-(a - b)| = |a - b| \). So, \( |b - a| \) is also an even number. This means \( (b, a) \in R \). Therefore, R is symmetric.
(iii) For transitivity: If \( (a, b) \in R \) and \( (b, c) \in R \), then \( |a - b| \) is even and \( |b - c| \) is even. This implies that \( (a - b) \) and \( (b - c) \) are both even integers. The sum of two even integers is always an even integer. So, \( (a - b) + (b - c) = a - c \) is an even integer. Therefore, \( |a - c| \) is even, which means \( (a, c) \in R \). Therefore, R is transitive.
Since R is reflexive, symmetric, and transitive, it is an equivalence relation.
(b) Showing related elements:
The elements of \( \{1, 3, 5\} \) are related to each other: Check the absolute differences:
\( |1 - 3| = 2 \) (even)
\( |3 - 5| = 2 \) (even)
\( |1 - 5| = 4 \) (even)
Since all these differences are even, all elements within \( \{1, 3, 5\} \) are related to each other.
Similarly, the elements of \( \{2, 4\} \) are related to each other:
\( |2 - 4| = 2 \) (even)
Since this difference is even, the elements within \( \{2, 4\} \) are related to each other.
No element of \( \{1, 3, 5\} \) is related to any element of \( \{2, 4\} \): Check the absolute differences between elements from both sets:
\( |1 - 2| = 1 \) (odd)
\( |1 - 4| = 3 \) (odd)
\( |3 - 2| = 1 \) (odd)
\( |3 - 4| = 1 \) (odd)
\( |5 - 2| = 3 \) (odd)
\( |5 - 4| = 1 \) (odd)
Since all these differences are odd, no element from \( \{1, 3, 5\} \) is related to any element from \( \{2, 4\} \).
In simple words: The relation where the absolute difference between two numbers is even is an equivalence relation because: 1) A number's difference with itself is 0, which is even (reflexive). 2) If the difference between 'a' and 'b' is even, then the difference between 'b' and 'a' is also even (symmetric). 3) If the difference between 'a' and 'b' is even, and 'b' and 'c' is even, then 'a' and 'c' also have an even difference (transitive). All numbers in {1, 3, 5} are related to each other because their differences are even. All numbers in {2, 4} are related to each other for the same reason. However, numbers from {1, 3, 5} are not related to numbers from {2, 4} because their differences are always odd.

Exam Tip: An equivalence relation partitions a set into disjoint equivalence classes. Here, {1, 3, 5} and {2, 4} are the equivalence classes, where elements within a class are related, but elements between classes are not.

Question 9. Show that the relation R in the set \( A = \{x \in Z : 0 \le x \le 12\} \), given by
(i) \( R = \{(a, b) : |a - b| \) is a multiple of 4\( \} \)
(ii) \( R = \{(a, b) : a = b\} \)
is an equivalence relation. Find the set of all elements related to 1 in each case.
Answer:
The given set is \( A = \{0, 1, 2, ..., 12\} \).
(i) For relation \( R = \{(a, b) : |a - b| \) is a multiple of 4\( \} \)
To prove R is an equivalence relation:
(a) For reflexivity: For any \( a \in A \), \( |a - a| = 0 \). Since 0 is a multiple of 4 (\( 0 = 4 \times 0 \)), \( (a, a) \in R \). Therefore, R is reflexive.
(b) For symmetry: If \( (a, b) \in R \), then \( |a - b| \) is a multiple of 4. We know that \( |b - a| = |-(a - b)| = |a - b| \). So, \( |b - a| \) is also a multiple of 4. This means \( (b, a) \in R \). Therefore, R is symmetric.
(c) For transitivity: If \( (a, b) \in R \) and \( (b, c) \in R \), then \( |a - b| \) is a multiple of 4 and \( |b - c| \) is a multiple of 4. This implies that \( a - b = 4k_1 \) and \( b - c = 4k_2 \) for some integers \( k_1, k_2 \). Adding these equations, \( (a - b) + (b - c) = 4k_1 + 4k_2 \), which simplifies to \( a - c = 4(k_1 + k_2) \). Thus, \( a - c \) is a multiple of 4, and so is \( |a - c| \). This means \( (a, c) \in R \). Therefore, R is transitive.
Since R is reflexive, symmetric, and transitive, it is an equivalence relation.
Set of all elements related to 1: We need to find all \( a \in A \) such that \( |a - 1| \) is a multiple of 4.
If \( a = 1 \), \( |1 - 1| = 0 \), which is a multiple of 4.
If \( a = 5 \), \( |5 - 1| = 4 \), which is a multiple of 4.
If \( a = 9 \), \( |9 - 1| = 8 \), which is a multiple of 4.
If \( a = 0 \), \( |0 - 1| = 1 \), not a multiple of 4.
For other values in \( A \), the difference will not be a multiple of 4. So, the set of elements related to 1 is \( \{1, 5, 9\} \).
(ii) For relation \( R = \{(a, b) : a = b\} \)
To prove R is an equivalence relation:
(a) For reflexivity: For any \( a \in A \), \( a = a \) is always true. Thus, \( (a, a) \in R \). Therefore, R is reflexive.
(b) For symmetry: If \( (a, b) \in R \), then \( a = b \). This immediately implies \( b = a \). Thus, \( (b, a) \in R \). Therefore, R is symmetric.
(c) For transitivity: If \( (a, b) \in R \) and \( (b, c) \in R \), then \( a = b \) and \( b = c \). From these, it directly follows that \( a = c \). Thus, \( (a, c) \in R \). Therefore, R is transitive.
Since R is reflexive, symmetric, and transitive, it is an equivalence relation.
Set of all elements related to 1: We need to find all \( a \in A \) such that \( a = 1 \). The only element satisfying this condition is 1 itself. So, the set of elements related to 1 is \( \{1\} \).
In simple words: For relation (i) where the absolute difference is a multiple of 4, it is reflexive (diff is 0), symmetric (diff is same whether a-b or b-a), and transitive (sum of multiples of 4 is a multiple of 4). Elements related to 1 are {1, 5, 9}. For relation (ii) where a = b, it is also reflexive (a=a), symmetric (if a=b then b=a), and transitive (if a=b and b=c then a=c). The only element related to 1 is {1}.

Exam Tip: When proving equivalence relations, state the condition for each property clearly. When finding related elements, systematically check each element in the set against the given condition. Be careful with phrasing like "is a multiple of X" (which includes 0).

Question 10. Give examples of relations which are
(i) Symmetric but neither reflexive nor transitive.
(ii) Transitive but neither reflexive nor symmetric.
(iii) Reflexive and symmetric but not transitive.
(iv) Reflexive and transitive but not symmetric.
(v) Symmetric and transitive but not reflexive.
Answer:
(i) Symmetric but neither reflexive nor transitive:
Let \( A \) be the set of all straight lines in a plane. Define relation \( R \) as \( R = \{(a, b) : \text{line } a \text{ is perpendicular to line } b\} \).
Reflexivity: A line cannot be perpendicular to itself. So, \( (a, a) \notin R \). R is not reflexive.
Symmetry: If line \( a \) is perpendicular to line \( b \), then line \( b \) is also perpendicular to line \( a \). So, \( (a, b) \in R \implies (b, a) \in R \). R is symmetric.
Transitivity: If line \( a \) is perpendicular to line \( b \) and line \( b \) is perpendicular to line \( c \), then line \( a \) is parallel to line \( c \), not perpendicular. So, \( (a, c) \notin R \). R is not transitive.
(ii) Transitive but neither reflexive nor symmetric:
Let \( A \) be the set of real numbers. Define relation \( R \) as \( R = \{(a, b) : a > b\} \).
Reflexivity: A number cannot be greater than itself (\( a \ngtr a \)). So, \( (a, a) \notin R \). R is not reflexive.
Symmetry: If \( a > b \), then \( b \ngtr a \). For example, \( 5 > 3 \), but \( 3 \ngtr 5 \). So, R is not symmetric.
Transitivity: If \( a > b \) and \( b > c \), then it logically follows that \( a > c \). So, \( (a, b) \in R \text{ and } (b, c) \in R \implies (a, c) \in R \). R is transitive.
(iii) Reflexive and symmetric but not transitive:
Let \( A = \{1, 2, 3\} \). Define relation \( R \) as \( R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)\} \).
Reflexivity: All pairs \( (1, 1), (2, 2), (3, 3) \) are in R. So, R is reflexive.
Symmetry: For every pair \( (x, y) \) in R, its inverse \( (y, x) \) is also in R. E.g., \( (1, 2) \in R \) and \( (2, 1) \in R \), \( (2, 3) \in R \) and \( (3, 2) \in R \). So, R is symmetric.
Transitivity: Consider \( (1, 2) \in R \) and \( (2, 3) \in R \). For transitivity, \( (1, 3) \) must be in R. However, \( (1, 3) \notin R \). So, R is not transitive.
(iv) Reflexive and transitive but not symmetric:
Let \( A = \{1, 2, 3\} \). Define relation \( R \) as \( R = \{(a, b) : a \text{ divides } b\} \). (This is the "divides" relation for integers).
Reflexivity: Any number divides itself. So, \( (a, a) \in R \). R is reflexive.
Symmetry: If \( a \) divides \( b \), it does not mean \( b \) divides \( a \). For example, \( 1 \) divides \( 2 \), but \( 2 \) does not divide \( 1 \). So, R is not symmetric.
Transitivity: If \( a \) divides \( b \) and \( b \) divides \( c \), then \( a \) divides \( c \). For example, \( 1 \) divides \( 2 \) and \( 2 \) divides \( 4 \), implies \( 1 \) divides \( 4 \). So, R is transitive.
(v) Symmetric and transitive but not reflexive:
Let \( A = \{1, 2, 3\} \). Define relation \( R \) as \( R = \{(2, 2), (3, 3)\} \).
Reflexivity: The pair \( (1, 1) \) is not in R. So, R is not reflexive.
Symmetry: For every pair \( (x, y) \) in R, \( x = y \), so \( (y, x) \) is trivially the same pair and is in R. So, R is symmetric.
Transitivity: If \( (x, y) \in R \) and \( (y, z) \in R \), then \( x = y = z \). Thus \( (x, z) \) (which is \( (x, x) \)) is in R. So, R is transitive.
In simple words: Here are examples of different types of relations: (i) Perpendicular lines are symmetric (A ⊥ B means B ⊥ A) but not reflexive (a line isn't ⊥ to itself) or transitive (if A ⊥ B and B ⊥ C, A is parallel to C). (ii) 'Greater than' is transitive (if A > B and B > C, then A > C) but not reflexive (A isn't > A) or symmetric (if A > B, B isn't > A). (iii) The relation in {1,2,3} for 'a+b ≤ 4' is reflexive (1+1≤4, 2+2≤4) and symmetric (1+2≤4 and 2+1≤4) but not transitive (if (1,2) and (2,3) are in R, (1,3) isn't). (iv) 'a divides b' on {1,2,3} is reflexive (1|1, 2|2, 3|3) and transitive (if 1|2 and 2|4, then 1|4) but not symmetric (1|2 but 2 doesn't divide 1). (v) The relation {(2,2), (3,3)} on {1,2,3} is symmetric and transitive but not reflexive because (1,1) is missing.

Exam Tip: When constructing examples, start with small sets and simple rules. For cases where properties are *not* met, provide a clear counter-example. For cases where properties *are* met, give a general argument or a comprehensive example if the set is small.

Question 11. Show that the relation R in the set \( A \) of points in a plane, given by \( R = \{(P, Q) : \text{distance of point } P \text{ from the origin is the same as the distance of } Q \text{ from the origin}\} \), is an equivalence relation. Further, show that the set of all points related to \( P \ne (0, 0) \) is the circle passing through \( P \) with origin as the centre.
Answer:
Let O be the origin (0, 0). The relation R is defined as \( R = \{(P, Q) : OP = OQ\} \), where OP denotes the distance of point P from the origin.
To prove R is an equivalence relation:
(i) For reflexivity: For any point \( P \in A \), the distance of P from the origin is equal to itself, i.e., \( OP = OP \). Thus, \( (P, P) \in R \). Therefore, R is reflexive.
(ii) For symmetry: If \( (P, Q) \in R \), it means the distance of P from the origin is equal to the distance of Q from the origin, i.e., \( OP = OQ \). This implies that \( OQ = OP \). Thus, \( (Q, P) \in R \). Therefore, R is symmetric.
(iii) For transitivity: If \( (P, Q) \in R \) and \( (Q, S) \in R \), it means \( OP = OQ \) and \( OQ = OS \). From these two equalities, it logically follows that \( OP = OS \). Thus, \( (P, S) \in R \). Therefore, R is transitive.
Since R is reflexive, symmetric, and transitive, it is an equivalence relation.
Further part: Consider a point \( P \ne (0, 0) \). Let \( OP = k \) for some constant \( k > 0 \). The set of all points \( Q \) related to \( P \) would satisfy \( OQ = OP \). So, \( OQ = k \). This means all such points \( Q \) are at a constant distance \( k \) from the origin O. Geometrically, this set of points forms a circle with the origin as its centre and radius \( k \), which passes through point \( P \).
In simple words: The relation means two points are related if they are the same distance from the center point (origin). It's an equivalence relation because: 1) Any point is the same distance from the origin as itself (reflexive). 2) If point P is the same distance as point Q from the origin, then Q is the same distance as P (symmetric). 3) If P is the same distance as Q, and Q is the same distance as S, then P is the same distance as S (transitive). All points related to a specific point P (not the origin) will form a circle. This circle will have the origin as its center and will pass right through P.

Exam Tip: For geometric relations, translating the verbal description into algebraic or geometric properties (e.g., distances, angles, parallelism) helps in applying the definitions of reflexivity, symmetry, and transitivity. Remember that all points equidistant from a center form a circle.

Question 12. Show that the relation R, defined in the set \( A \) of all triangles as \( R = \{(T_1, T_2) : T_1 \) is similar to \( T_2\} \), is an equivalence relation. Consider three right triangles \( T_1 \) with sides 3, 4, 5; \( T_2 \) with sides 5, 12, 13 and \( T_3 \) with sides 6, 8, 10. Which triangles among \( T_1, T_2 \) and \( T_3 \) are related?
Answer:
(i) To prove R is an equivalence relation:
The relation R is defined as \( R = \{(T_1, T_2) : T_1 \) is similar to \( T_2\} \).
(a) For reflexivity: Any triangle \( T_1 \) is similar to itself. Thus, \( (T_1, T_1) \in R \). Therefore, R is reflexive.
(b) For symmetry: If triangle \( T_1 \) is similar to triangle \( T_2 \), then triangle \( T_2 \) is also similar to triangle \( T_1 \). Thus, \( (T_1, T_2) \in R \implies (T_2, T_1) \in R \). Therefore, R is symmetric.
(c) For transitivity: If triangle \( T_1 \) is similar to triangle \( T_2 \), and triangle \( T_2 \) is similar to triangle \( T_3 \), then triangle \( T_1 \) is also similar to triangle \( T_3 \). Thus, \( (T_1, T_2) \in R \text{ and } (T_2, T_3) \in R \implies (T_1, T_3) \in R \). Therefore, R is transitive.
Since R is reflexive, symmetric, and transitive, it is an equivalence relation.
(ii) To find related triangles among \( T_1, T_2, T_3 \):
Given triangles:
\( T_1 \) has sides 3, 4, 5.
\( T_2 \) has sides 5, 12, 13.
\( T_3 \) has sides 6, 8, 10.
Two triangles are similar if their corresponding sides are in proportion.
Check \( T_1 \) and \( T_2 \): The ratios of corresponding sides are \( \frac{3}{5}, \frac{4}{12} = \frac{1}{3}, \frac{5}{13} \). These ratios are not equal. So, \( T_1 \) is not related to \( T_2 \).
Check \( T_1 \) and \( T_3 \): The ratios of corresponding sides are \( \frac{3}{6} = \frac{1}{2}, \frac{4}{8} = \frac{1}{2}, \frac{5}{10} = \frac{1}{2} \). Since the ratios are equal, \( T_1 \) is similar to \( T_3 \). So, \( T_1 \) is related to \( T_3 \).
Check \( T_2 \) and \( T_3 \): The ratios of corresponding sides are \( \frac{5}{6}, \frac{12}{8} = \frac{3}{2}, \frac{13}{10} \). These ratios are not equal. So, \( T_2 \) is not related to \( T_3 \).
Therefore, only triangles \( T_1 \) and \( T_3 \) are related to each other.
In simple words: The relation "is similar to" for triangles is an equivalence relation because: 1) Any triangle is similar to itself (reflexive). 2) If triangle A is similar to B, then B is similar to A (symmetric). 3) If A is similar to B, and B is similar to C, then A is similar to C (transitive). Among the given triangles, T1 (sides 3,4,5) and T3 (sides 6,8,10) are related because their side lengths are in proportion (ratio 1:2). T2 (sides 5,12,13) is not similar to either T1 or T3.

Exam Tip: Remember that for similarity of triangles, all corresponding sides must be in the same proportion. For right triangles, checking Pythagorean theorem \( (a^2 + b^2 = c^2) \) helps confirm their type, but similarity is strictly about side ratios and angles.

Question 13. Show that the relation R, defined in the set \( A \) of all polygons as \( R = \{(P_1, P_2) : P_1 \) and \( P_2 \) have the same number of sides\( \} \) is an equivalence relation. What is the set of all elements in \( A \) related to the right angle triangle \( T \) with sides 3, 4 and 5?
Answer:
Let \( n \) represent the number of sides of a polygon.
The relation R is defined as \( R = \{(P_1, P_2) : P_1 \) and \( P_2 \) have the same number of sides\( \} \).
(i) To prove R is an equivalence relation:
(a) For reflexivity: Any polygon \( P_1 \) has the same number of sides as itself. Thus, \( (P_1, P_1) \in R \). Therefore, R is reflexive.
(b) For symmetry: If \( (P_1, P_2) \in R \), it means polygon \( P_1 \) has the same number of sides as polygon \( P_2 \). This implies that polygon \( P_2 \) also has the same number of sides as polygon \( P_1 \). Thus, \( (P_2, P_1) \in R \). Therefore, R is symmetric.
(c) For transitivity: If \( (P_1, P_2) \in R \) and \( (P_2, P_3) \in R \), it means \( P_1 \) has the same number of sides as \( P_2 \), and \( P_2 \) has the same number of sides as \( P_3 \). From this, we can conclude that \( P_1 \) must have the same number of sides as \( P_3 \). Thus, \( (P_1, P_3) \in R \). Therefore, R is transitive.
Since R is reflexive, symmetric, and transitive, it is an equivalence relation.
(ii) Set of all elements related to the right angle triangle \( T \):
The right angle triangle \( T \) has 3 sides. The elements in set \( A \) related to \( T \) are all polygons that have the same number of sides as \( T \). Therefore, the set of all elements in \( A \) related to triangle \( T \) is the set of all triangles in a plane (i.e., all polygons with 3 sides).
In simple words: The relation that states two polygons are related if they have the same number of sides is an equivalence relation because: 1) Any polygon has the same number of sides as itself (reflexive). 2) If polygon P1 has the same sides as P2, then P2 has the same sides as P1 (symmetric). 3) If P1 has the same sides as P2, and P2 has the same sides as P3, then P1 has the same sides as P3 (transitive). The set of all elements related to a right-angle triangle (which has 3 sides) is simply all other triangles.

Exam Tip: When dealing with geometric shapes, focus on the defining properties mentioned in the relation (e.g., number of sides, similarity, congruence) to determine reflexivity, symmetry, and transitivity. An equivalence class groups together all elements that share that specific property.

Question 14. Let \( L \) be the set of all lines in the XY-plane and R be the relation in \( L \) defined as \( R = \{(L_1, L_2) : L_1 \) is parallel to \( L_2\} \). Show that R is an equivalence relation. Find the set of all lines related to the line \( y = 2x + 4 \).
Answer:
The set \( L \) consists of all lines in the XY-plane. The relation R is defined as \( R = \{(L_1, L_2) : L_1 \) is parallel to \( L_2\} \).
(i) To prove R is an equivalence relation:
(a) For reflexivity: Any line \( L_1 \) is parallel to itself. Thus, \( (L_1, L_1) \in R \). Therefore, R is reflexive.
(b) For symmetry: If line \( L_1 \) is parallel to line \( L_2 \), then line \( L_2 \) is also parallel to line \( L_1 \). Thus, \( (L_1, L_2) \in R \implies (L_2, L_1) \in R \). Therefore, R is symmetric.
(c) For transitivity: If line \( L_1 \) is parallel to line \( L_2 \), and line \( L_2 \) is parallel to line \( L_3 \), then line \( L_1 \) is parallel to line \( L_3 \). Thus, \( (L_1, L_2) \in R \text{ and } (L_2, L_3) \in R \implies (L_1, L_3) \in R \). Therefore, R is transitive.
Since R is reflexive, symmetric, and transitive, it is an equivalence relation.
(ii) To find the set of all lines related to the line \( y = 2x + 4 \):
The line \( y = 2x + 4 \) has a slope of 2. For another line to be parallel to it, it must have the same slope. Therefore, the set of all lines related to the line \( y = 2x + 4 \) is given by the equation \( y = 2x + c \), where \( c \) is any arbitrary real constant.
In simple words: The relation "is parallel to" for lines is an equivalence relation because: 1) A line is parallel to itself (reflexive). 2) If line A is parallel to B, then B is parallel to A (symmetric). 3) If A is parallel to B, and B is parallel to C, then A is parallel to C (transitive). The set of all lines related to the line y = 2x + 4 are all lines that have the same steepness (slope of 2), but can cross the y-axis at any point. So, their equation is y = 2x + c, where 'c' can be any number.

Exam Tip: Remember that parallel lines have the same slope but different y-intercepts (unless they are the same line). An equivalence class for parallelism is a set of all lines with a specific slope.

Question 15. 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.
Let the given set be \( A = \{1, 2, 3, 4\} \) and the relation be \( R = \{(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)\} \).
(a) For reflexivity: We check if all pairs \( (x, x) \) for \( x \in A \) are in R. We have \( (1, 1), (2, 2), (3, 3), \) and \( (4, 4) \) all present in R. Therefore, R is reflexive.
(b) For symmetry: We check if for every \( (x, y) \in R \), the pair \( (y, x) \) is also in R. We observe that \( (1, 2) \in R \), but \( (2, 1) \notin R \). Therefore, R is not symmetric.
(c) For transitivity: We check if for every \( (x, y) \in R \) and \( (y, z) \in R \), the pair \( (x, z) \) is also in R. Let's examine potential cases:
If \( (1, 3) \in R \) and \( (3, 2) \in R \), then \( (1, 2) \) must be in R. We can see that \( (1, 2) \) is indeed in R.
Other combinations such as \( (1, 1) \) with \( (1, 2) \) implies \( (1, 2) \), which is in R. Similarly, \( (2, 2) \) with \( (2, \text{any}) \) implies \( (2, \text{any}) \), and so on. All such checks confirm the property. Therefore, R is transitive.
Based on these observations, R is reflexive, not symmetric, and transitive. This matches option (b).
In simple words: The relation is reflexive because all numbers are related to themselves. It is not symmetric because (1,2) is in the relation but (2,1) is not. It is transitive because whenever (x,y) and (y,z) are in the relation, (x,z) is also found in the relation. So, option (b) is the correct choice.

Exam Tip: For multiple-choice questions on relation properties, systematically check each property (reflexivity, symmetry, transitivity) against the given relation. A single counter-example is sufficient to disprove a property.

Question 16. Let R be the relation in the set \( N \) given by \( R = \{(a, b) : a = b - 2, b > 6\} \).
Choose the correct answer:
(a) \( (2, 4) \in R \)
(B) \( (3, 8) \in R \)
(C) \( (6, 8) \in R \)
Answer: (C) (6, 8) ∈ R
The given relation is \( R = \{(a, b) : a = b - 2, b > 6\} \). We need to check which of the given options satisfies both conditions.
(a) Check \( (2, 4) \): Here, \( a = 2, b = 4 \).
Condition 1: \( a = b - 2 \Rightarrow 2 = 4 - 2 \Rightarrow 2 = 2 \) (True).
Condition 2: \( b > 6 \Rightarrow 4 > 6 \) (False).
Since both conditions are not met, \( (2, 4) \notin R \).
(b) Check \( (3, 8) \): Here, \( a = 3, b = 8 \).
Condition 1: \( a = b - 2 \Rightarrow 3 = 8 - 2 \Rightarrow 3 = 6 \) (False).
Since Condition 1 is not met, \( (3, 8) \notin R \).
(c) Check \( (6, 8) \): Here, \( a = 6, b = 8 \).
Condition 1: \( a = b - 2 \Rightarrow 6 = 8 - 2 \Rightarrow 6 = 6 \) (True).
Condition 2: \( b > 6 \Rightarrow 8 > 6 \) (True).
Since both conditions are satisfied, \( (6, 8) \in R \). Therefore, option (C) is the correct answer.
In simple words: We are looking for a pair (a, b) where 'a' is 2 less than 'b', and 'b' is a number greater than 6. Testing the options, only (6, 8) fits these rules: 6 is 2 less than 8, and 8 is greater than 6.

Exam Tip: For multiple-choice questions with conditions, always check all given conditions for each option. A single false condition means the option is incorrect.