RBSE Solutions Class 11 Maths Chapter 2 Relations and Functions Exercise 2.2

Question 1. Examine the reflexivity, symmetricity, and transitivity of the following relations:
(i) \( mR_1n \Leftrightarrow \) m and n both are odd, \( \forall m, n \in N \)
(ii) In the power set P(A) of set \( AR_2B \Leftrightarrow A \subseteq B, \forall A, B \in P(A) \)
(iii) In set S of straight lines situated in three-dimensional space, \( L_1R_3L_2 \Leftrightarrow L_1 \) and \( L_2 \) are coplanar \( \forall L_1, L_2 \in S \)
(iv) \( aR_4b \Leftrightarrow \) b is divisible by a, \( \forall a, b \in N \)
Answer:
(i) For the relation \( R_1 \):
Reflexivity: Let \( m \in N \). The condition \( mR_1m \) would mean that m and m are both odd. This is not true for all natural numbers m, because if m is an even number, then the condition fails. So, \( (m, m) \notin R_1 \) if m is even.
\( \implies R_1 \) is not a reflexive relation.
Symmetricity: Let \( m, n \in N \). If \( mR_1n \) is true, it means that m and n are both odd. This also means that n and m are both odd, which implies \( nR_1m \).
\( \implies R_1 \) is a symmetric relation.
Transitivity: Let \( m, n, r \in N \). If \( mR_1n \) and \( nR_1r \) are true, it means that m and n are both odd, and n and r are both odd. From this, we know that m, n, and r are all odd numbers. Since m and r are both odd, \( mR_1r \) is true.
\( \implies R_1 \) is a transitive relation.

(ii) For the relation \( R_2 \):
Reflexivity: Let \( A \in P(A) \). The condition \( AR_2A \) means \( A \subseteq A \). This is always true because every set is a subset of itself.
\( \implies R_2 \) is a reflexive relation.
Symmetricity: Let \( A, B \in P(A) \). If \( AR_2B \) is true, it means \( A \subseteq B \). However, if \( A \subseteq B \), it does not necessarily mean that \( B \subseteq A \) (unless A = B). For example, if A = {1} and B = {1, 2}, then \( A \subseteq B \) but \( B \not\subseteq A \). So, \( BR_2A \) is not always true.
\( \implies R_2 \) is not a symmetric relation.
Transitivity: Let \( A, B, C \in P(A) \). If \( AR_2B \) and \( BR_2C \) are true, it means \( A \subseteq B \) and \( B \subseteq C \). If \( A \subseteq B \) and \( B \subseteq C \), then it logically follows that \( A \subseteq C \). This implies \( AR_2C \) is true.
\( \implies R_2 \) is a transitive relation.

(iii) For the relation \( R_3 \):
Reflexivity: Let \( L \in S \). The condition \( LR_3L \) means that L and L are coplanar. This is always true as any line is coplanar with itself.
\( \implies R_3 \) is a reflexive relation.
Symmetricity: Let \( L_1, L_2 \in S \). If \( L_1R_3L_2 \) is true, it means that \( L_1 \) and \( L_2 \) are coplanar. This also means that \( L_2 \) and \( L_1 \) are coplanar, which implies \( L_2R_3L_1 \) is true.
\( \implies R_3 \) is a symmetric relation.
Transitivity: Let \( L_1, L_2, L_3 \in S \). If \( L_1R_3L_2 \) and \( L_2R_3L_3 \) are true, it means \( L_1 \) and \( L_2 \) are coplanar, and \( L_2 \) and \( L_3 \) are coplanar. However, this does not guarantee that \( L_1 \) and \( L_3 \) are coplanar. For example, \( L_1 \) and \( L_2 \) might lie on plane P1, and \( L_2 \) and \( L_3 \) might lie on plane P2, where P1 and P2 are different planes intersecting at \( L_2 \). In such a case, \( L_1 \) and \( L_3 \) would not be coplanar. The provided diagram illustrates this:

\( \implies R_3 \) is not a transitive relation.

(iv) For the relation \( R_4 \):
Reflexivity: Let \( a \in N \). The condition \( aR_4a \) means that a is divisible by a. This is always true for any natural number a (since a/a = 1, which is an integer).
\( \implies R_4 \) is a reflexive relation.
Symmetricity: Let \( a, b \in N \). If \( aR_4b \) is true, it means that b is divisible by a. This does not necessarily mean that a is divisible by b. For example, if a = 2 and b = 4, then 4 is divisible by 2, so \( (2, 4) \in R_4 \). But 2 is not divisible by 4, so \( (4, 2) \notin R_4 \).
\( \implies R_4 \) is not a symmetric relation.
Transitivity: Let \( a, b, c \in N \). If \( aR_4b \) and \( bR_4c \) are true, it means b is divisible by a, and c is divisible by b. So we can write \( b = ka \) and \( c = mb \) for some integers k and m. Substituting b into the second equation, we get \( c = m(ka) = (mk)a \). Since m and k are integers, mk is also an integer. This means c is divisible by a, which implies \( aR_4c \) is true.
\( \implies R_4 \) is a transitive relation.
In simple words: To check these properties, you look at how the relation works with one, two, or three elements. For reflexivity, an element must relate to itself. For symmetricity, if 'A relates to B', then 'B must relate to A'. For transitivity, if 'A relates to B' and 'B relates to C', then 'A must relate to C'.

🎯 Exam Tip: When testing transitivity, ensure you pick a specific counterexample that clearly violates the rule if the relation is not transitive, or demonstrate how the rule holds algebraically if it is transitive.

Question 2. Any relation P is defined in set \( R_0 \) of non-zero real numbers by following ways:
(i) \( xPy \Leftrightarrow x^2 + y^2 = 1 \)
(ii) \( xPy \Leftrightarrow xy = 1 \)
(iii) \( xPy \Leftrightarrow (x + y) \) is a rational number
(iv) \( xPy \Leftrightarrow x/y \) is a rational number
Test the reflexivity, symmetricity and transitivity of these relations.
Answer:
(i) For the relation \( P: xPy \Leftrightarrow x^2 + y^2 = 1 \)
Reflexivity: Let \( x \in R_0 \). For P to be reflexive, \( xPx \) must be true, meaning \( x^2 + x^2 = 1 \), or \( 2x^2 = 1 \). This implies \( x^2 = 1/2 \), so \( x = \pm 1/\sqrt{2} \). This is not true for all non-zero real numbers x. For example, if \( x = 2 \), then \( 2^2 + 2^2 = 8 \neq 1 \). So \( (2, 2) \notin P \).
\( \implies P \) is not a reflexive relation.
Symmetricity: Let \( a, b \in R_0 \). If \( aPb \) is true, then \( a^2 + b^2 = 1 \). Since addition is commutative, \( b^2 + a^2 = 1 \) is also true, which means \( bPa \).
\( \implies P \) is a symmetric relation.
Transitivity: Let \( a, b, c \in R_0 \). If \( aPb \) and \( bPc \) are true, it means \( a^2 + b^2 = 1 \) and \( b^2 + c^2 = 1 \). We need to check if \( a^2 + c^2 = 1 \). Consider \( a = 1/2 \), \( b = \sqrt{3}/2 \), \( c = 1/2 \).
\( (1/2)^2 + (\sqrt{3}/2)^2 = 1/4 + 3/4 = 1 \), so \( (1/2, \sqrt{3}/2) \in P \).
\( (\sqrt{3}/2)^2 + (1/2)^2 = 3/4 + 1/4 = 1 \), so \( (\sqrt{3}/2, 1/2) \in P \).
But for \( (1/2, 1/2) \), we have \( (1/2)^2 + (1/2)^2 = 1/4 + 1/4 = 1/2 \neq 1 \). So \( (1/2, 1/2) \notin P \).
\( \implies P \) is not a transitive relation.

(ii) For the relation \( P: xPy \Leftrightarrow xy = 1 \)
Reflexivity: Let \( x \in R_0 \). For P to be reflexive, \( xPx \) must be true, meaning \( x \cdot x = 1 \), or \( x^2 = 1 \). This implies \( x = 1 \) or \( x = -1 \). This is not true for all non-zero real numbers x. For example, if \( x = 2 \), then \( 2 \cdot 2 = 4 \neq 1 \). So \( (2, 2) \notin P \).
\( \implies P \) is not a reflexive relation.
Symmetricity: Let \( a, b \in R_0 \). If \( aPb \) is true, then \( ab = 1 \). Since multiplication is commutative, \( ba = 1 \) is also true, which means \( bPa \).
\( \implies P \) is a symmetric relation.
Transitivity: Let \( a, b, c \in R_0 \). If \( aPb \) and \( bPc \) are true, it means \( ab = 1 \) and \( bc = 1 \). From \( ab = 1 \), we get \( b = 1/a \). Substitute this into \( bc = 1 \): \( (1/a)c = 1 \), so \( c/a = 1 \), which implies \( c = a \). For \( aPc \) to be true, we need \( ac = 1 \). Since \( c = a \), this means \( a \cdot a = 1 \), or \( a^2 = 1 \). This implies \( a = 1 \) or \( a = -1 \). This is not true for all non-zero real numbers a. For example, if \( a = 2 \), \( b = 1/2 \), \( c = 2 \). Then \( (2, 1/2) \in P \) (since \( 2 \cdot (1/2) = 1 \)) and \( (1/2, 2) \in P \) (since \( (1/2) \cdot 2 = 1 \)). But for \( (a, c) = (2, 2) \), we have \( 2 \cdot 2 = 4 \neq 1 \). So \( (2, 2) \notin P \).
\( \implies P \) is not a transitive relation.

(iii) For the relation \( P: xPy \Leftrightarrow (x + y) \) is a rational number
Reflexivity: Let \( x \in R_0 \). For P to be reflexive, \( xPx \) must be true, meaning \( x + x \) (or \( 2x \)) is a rational number. This is not true for all non-zero real numbers x. For example, if \( x = \sqrt{3} \), then \( x + x = 2\sqrt{3} \), which is an irrational number. So \( (\sqrt{3}, \sqrt{3}) \notin P \).
\( \implies P \) is not a reflexive relation.
Symmetricity: Let \( a, b \in R_0 \). If \( aPb \) is true, then \( a + b \) is a rational number. Since addition is commutative, \( b + a \) is also a rational number, which means \( bPa \).
\( \implies P \) is a symmetric relation.
Transitivity: Let \( a, b, c \in R_0 \). If \( aPb \) and \( bPc \) are true, it means \( a + b \) is a rational number and \( b + c \) is a rational number. We need to check if \( a + c \) is a rational number. Let \( a = 2 + \sqrt{3} \), \( b = -\sqrt{3} + 6 \), \( c = \sqrt{3} + 7 \).
\( a + b = (2 + \sqrt{3}) + (-\sqrt{3} + 6) = 8 \), which is rational, so \( (a, b) \in P \).
\( b + c = (-\sqrt{3} + 6) + (\sqrt{3} + 7) = 13 \), which is rational, so \( (b, c) \in P \).
But \( a + c = (2 + \sqrt{3}) + (\sqrt{3} + 7) = 9 + 2\sqrt{3} \), which is an irrational number. So \( (a, c) \notin P \).
\( \implies P \) is not a transitive relation.

(iv) For the relation \( P: xPy \Leftrightarrow x/y \) is a rational number
Reflexivity: Let \( x \in R_0 \). For P to be reflexive, \( xPx \) must be true, meaning \( x/x \) is a rational number. Since \( x \in R_0 \) means \( x \neq 0 \), \( x/x = 1 \), which is a rational number. This is true for all \( x \in R_0 \).
\( \implies P \) is a reflexive relation.
Symmetricity: Let \( a, b \in R_0 \). If \( aPb \) is true, then \( a/b \) is a rational number. If \( a/b = k \), where k is a non-zero rational number, then \( b/a = 1/k \). Since the reciprocal of a non-zero rational number is also rational, \( b/a \) is a rational number. This means \( bPa \).
\( \implies P \) is a symmetric relation.
Transitivity: Let \( a, b, c \in R_0 \). If \( aPb \) and \( bPc \) are true, it means \( a/b \) is a rational number and \( b/c \) is a rational number. Let \( a/b = k_1 \) and \( b/c = k_2 \), where \( k_1, k_2 \) are rational numbers. Then their product \( (a/b) \cdot (b/c) = a/c \) is also a rational number (because the product of two rational numbers is rational). This means \( aPc \).
\( \implies P \) is a transitive relation.
In simple words: For each rule, check if it works for an element by itself (reflexive), if it works backward (symmetric), and if it links up across three elements (transitive). Often, trying a few simple numbers or examples helps you see if a rule holds or breaks.

🎯 Exam Tip: Remember that "non-zero real numbers" or "rational numbers" restrict your choices. When providing counterexamples, ensure they belong to the specified set.

Question 3. A relation \( R_1 \) is defined in set R of real numbers in the following way: \( (a, b) \in R_1 \Leftrightarrow 1 + ab > 0, \forall a, b \in R \) Prove that \( R_1 \) is reflexive and symmetric but not transitive.
Answer:
Given: Set R is the set of real numbers. The relation \( R_1 \) in R is defined as \( (a, b) \in R_1 \Leftrightarrow 1 + ab > 0 \).

Reflexivity: Let \( a \in R \). For \( R_1 \) to be reflexive, \( (a, a) \in R_1 \) must be true. This means \( 1 + a \cdot a > 0 \), or \( 1 + a^2 > 0 \). Since \( a^2 \) is always greater than or equal to 0 for any real number a, \( 1 + a^2 \) will always be greater than 0. This condition holds for all real numbers a.
\( \implies R_1 \) is a reflexive relation.

Symmetricity: Let \( a, b \in R \). If \( (a, b) \in R_1 \) is true, it means \( 1 + ab > 0 \). Since multiplication of real numbers is commutative (ab = ba), we also have \( 1 + ba > 0 \). This implies that \( (b, a) \in R_1 \). The order of multiplication does not change the result.
\( \implies R_1 \) is a symmetric relation.

Transitivity: Let \( a, b, c \in R \). If \( (a, b) \in R_1 \) and \( (b, c) \in R_1 \) are true, it means \( 1 + ab > 0 \) and \( 1 + bc > 0 \). We need to check if \( (a, c) \in R_1 \) is also true, meaning \( 1 + ac > 0 \). However, this is not always necessary.
Consider a counterexample: Let \( a = 1 \), \( b = 1/2 \), and \( c = -1 \). These are all real numbers.
1. Check \( (a, b) \in R_1 \): \( 1 + ab = 1 + (1)(1/2) = 1 + 1/2 = 3/2 \). Since \( 3/2 > 0 \), \( (1, 1/2) \in R_1 \) is true.
2. Check \( (b, c) \in R_1 \): \( 1 + bc = 1 + (1/2)(-1) = 1 - 1/2 = 1/2 \). Since \( 1/2 > 0 \), \( (1/2, -1) \in R_1 \) is true.
3. Check \( (a, c) \in R_1 \): \( 1 + ac = 1 + (1)(-1) = 1 - 1 = 0 \). Since \( 0 \ngtr 0 \), \( (1, -1) \notin R_1 \).
Since \( (a, b) \in R_1 \) and \( (b, c) \in R_1 \) but \( (a, c) \notin R_1 \), the relation \( R_1 \) is not transitive.
Hence, \( R_1 \) is reflexive and symmetric but not transitive, as proved.
In simple words: This problem asks you to check three basic rules for a relationship between numbers. It turns out the rule works when a number is paired with itself, and it works if you swap the order of the numbers. But it doesn't always work if you try to chain three numbers together in that relationship.

🎯 Exam Tip: For problems asking to prove non-transitivity, selecting simple counterexamples with negative numbers or fractions often helps to quickly demonstrate the failure of the property.

Question 4. N is a set of natural numbers. If a relation R is defined in set \( N \times N \) such that \( (a, b) R(c, d) \Leftrightarrow ad = bc \forall (a, b), (c, d) \in N \times N \), then prove that R is an equivalence relation.
Answer:
Given: Set N = {1, 2, 3, 4, ...} is the set of natural numbers. A relation R is defined in \( N \times N \) as \( (a, b) R(c, d) \Leftrightarrow ad = bc \).

Reflexivity: Let \( (a, b) \in N \times N \). For R to be reflexive, \( (a, b) R(a, b) \) must be true. This means \( a \cdot b = b \cdot a \). Since multiplication of natural numbers is always commutative, \( ab = ba \) is always true. Thus, \( (a, b) R(a, b) \) holds for all \( (a, b) \in N \times N \).
\( \implies R \) is a reflexive relation.

Symmetricity: Let \( (a, b), (c, d) \in N \times N \). Assume \( (a, b) R(c, d) \) is true. This means \( ad = bc \). We need to check if \( (c, d) R(a, b) \) is also true, which would mean \( cb = da \). From \( ad = bc \), we can simply rearrange the terms to get \( bc = ad \), and then \( cb = da \). Both are equivalent statements.
\( \implies R \) is a symmetric relation.

Transitivity: Let \( (a, b), (c, d), (e, f) \in N \times N \). Assume \( (a, b) R(c, d) \) and \( (c, d) R(e, f) \) are true.
From \( (a, b) R(c, d) \), we have \( ad = bc \). (Equation 1)
From \( (c, d) R(e, f) \), we have \( cf = de \). (Equation 2)
From Equation 1, we can write \( a/b = c/d \) (since \( b, d \in N \), they are non-zero).
From Equation 2, we can write \( c/d = e/f \) (since \( d, f \in N \), they are non-zero).
Combining these, we get \( a/b = c/d = e/f \).
Therefore, \( a/b = e/f \). Multiplying both sides by \( bf \) gives \( af = be \). This is the condition for \( (a, b) R(e, f) \).
\( \implies R \) is a transitive relation.

Since R is reflexive, symmetric, and transitive, it is an equivalence relation. Hence, proved.
In simple words: An equivalence relation is like a perfect matching rule. Here, the rule states that if two pairs of numbers have the same cross-product, they are related. We proved it's perfect because any pair matches itself (reflexive), if pair A matches pair B, then B matches A (symmetric), and if A matches B, and B matches C, then A also matches C (transitive).

🎯 Exam Tip: When dealing with relations involving pairs, remember to treat each pair as a single element for the properties. The cross-product condition \( ad=bc \) is often simplified by thinking of it as \( a/b = c/d \).

Question 5. A relation R is defined in a set \( Q_0 \) set of non-zero rational numbers such that \( aRb \Leftrightarrow a = 1/b, \forall a, b \in Q_0 \). Is R an equivalence relation?
Answer:
Given: Set \( Q_0 \) is the set of non-zero rational numbers. A relation R in \( Q_0 \) is defined as \( aRb \Leftrightarrow a = 1/b \).

Reflexivity: Let \( a \in Q_0 \). For R to be reflexive, \( aRa \) must be true, meaning \( a = 1/a \). This implies \( a^2 = 1 \), which means \( a = 1 \) or \( a = -1 \). This condition is not true for all non-zero rational numbers in \( Q_0 \). For example, if \( a = 2 \), then \( 2 \neq 1/2 \). So, \( (2, 2) \notin R \).
\( \implies R \) is not a reflexive relation.

Symmetricity: Let \( a, b \in Q_0 \). If \( aRb \) is true, it means \( a = 1/b \). Multiplying both sides by b (since \( b \neq 0 \)), we get \( ab = 1 \). Then, dividing by a (since \( a \neq 0 \)), we get \( b = 1/a \). This implies \( bRa \).
\( \implies R \) is a symmetric relation.

Transitivity: Let \( a, b, c \in Q_0 \). If \( aRb \) and \( bRc \) are true, it means \( a = 1/b \) and \( b = 1/c \). Substitute the second equation into the first one: \( a = 1/(1/c) \), which simplifies to \( a = c \). For \( aRc \) to be true, we would need \( a = 1/c \). However, we found \( a = c \). So, for \( aRc \) to hold, we would need \( c = 1/c \), which means \( c^2 = 1 \), implying \( c = 1 \) or \( c = -1 \). This is not true for all non-zero rational numbers c. For example, if \( a = 2 \), \( b = 1/2 \), \( c = 2 \). Here \( aRb \) (since \( 2 = 1/(1/2) \)) and \( bRc \) (since \( 1/2 = 1/2 \)). But for \( aRc \), we need \( 2 = 1/2 \), which is false. So, \( (2, 2) \notin R \).
\( \implies R \) is not a transitive relation.

Since R is not reflexive and not transitive, it is not an equivalence relation.
In simple words: An equivalence relation needs three things: self-relation (reflexive), two-way relation (symmetric), and chain-like relation (transitive). This rule works two ways, but it doesn't always apply to itself or connect in a chain, so it's not an equivalence relation.

🎯 Exam Tip: An equivalence relation must satisfy all three properties: reflexivity, symmetricity, and transitivity. If even one property fails, the relation is not an equivalence relation.

Question 6. Let \( X = \{(a, b) | a, b \in I\} \) where I is set of integers. A relation R on X is defined in the following way \( (a, b) R(c, d) \Leftrightarrow b - d = a - c \). Prove that R is an equivalence relation.
Answer:
Given: Set \( X = \{(a, b) : a, b \in I\} \) where I is the set of integers. A relation R in X is defined as: \( (a, b) R(c, d) \Leftrightarrow b - d = a - c \).

Reflexivity: Let \( (a, b) \in X \). For R to be reflexive, \( (a, b) R(a, b) \) must be true. According to the definition of the relation, this means \( b - b = a - a \). This simplifies to \( 0 = 0 \), which is always true for any integers a and b.
\( \implies R \) is a reflexive relation.

Symmetricity: Let \( (a, b), (c, d) \in X \). Assume \( (a, b) R(c, d) \) is true. This means \( b - d = a - c \). We want to check if \( (c, d) R(a, b) \) is also true, which would mean \( d - b = c - a \). To get this from the first equation, we can multiply both sides by -1: \( -(b - d) = -(a - c) \). This simplifies to \( d - b = c - a \). This is exactly the condition for \( (c, d) R(a, b) \).
\( \implies R \) is a symmetric relation.

Transitivity: Let \( (a, b), (c, d), (e, f) \in X \). Assume \( (a, b) R(c, d) \) and \( (c, d) R(e, f) \) are true.
From \( (a, b) R(c, d) \), we have \( b - d = a - c \). (Equation 1)
From \( (c, d) R(e, f) \), we have \( d - f = c - e \). (Equation 2)
Now, add Equation 1 and Equation 2:
\( (b - d) + (d - f) = (a - c) + (c - e) \)
The terms \( -d \) and \( +d \) cancel on the left, and \( -c \) and \( +c \) cancel on the right.
\( \implies b - f = a - e \)
This is the condition for \( (a, b) R(e, f) \).
\( \implies R \) is a transitive relation.

Since R is reflexive, symmetric, and transitive, it is an equivalence relation. Hence, proved.
In simple words: This relation involves comparing the differences between numbers in two pairs. It's like saying if the first numbers' difference is the same as the second numbers' difference, they are related. We proved that this relationship holds for identical pairs, works when you swap the pairs around, and also chains correctly if you have three pairs in a row.

🎯 Exam Tip: For relations defined on ordered pairs, carefully apply the given condition to each component. Adding equations together is a common technique to prove transitivity in such cases.

Question 7. A relation R is defined in a set T of triangles situated in a plane such that \( xRy \Leftrightarrow x \) is similar to y. Prove that R is an equivalence relation.
Answer:
Given: Set T is the set of all triangles situated in a plane. A relation R in T is defined as \( xRy \Leftrightarrow x \) is similar to y.

Reflexivity: Let \( x \in T \). For R to be reflexive, \( xRx \) must be true. This means that triangle x is similar to itself. Every triangle is similar to itself.
\( \implies R \) is a reflexive relation.

Symmetricity: Let \( x, y \in T \). If \( xRy \) is true, it means that triangle x is similar to triangle y. If triangle x is similar to triangle y, then triangle y is also similar to triangle x. This implies \( yRx \).
\( \implies R \) is a symmetric relation.

Transitivity: Let \( x, y, z \in T \). If \( xRy \) and \( yRz \) are true, it means that triangle x is similar to triangle y, and triangle y is similar to triangle z. If x is similar to y, and y is similar to z, then it logically follows that triangle x is similar to triangle z. This implies \( xRz \).
\( \implies R \) is a transitive relation.

Since R is reflexive, symmetric, and transitive, it is an equivalence relation. Hence, proved.
In simple words: This relation is about triangles being the same shape (similar). Any triangle is similar to itself. If triangle A is similar to B, then B is similar to A. If A is similar to B, and B is similar to C, then A is also similar to C. Because all these are true, it's an equivalence relation.

🎯 Exam Tip: The properties of "similarity" for geometric figures (reflexive, symmetric, transitive) make it a classic example of an equivalence relation. Use these inherent geometric properties to structure your proof.

Question 8. Let a relation R is defined in a set \( A = \{1, 2, 3\} \) as: \( R = \{(1, 1), (1, 2), (2, 1), (2, 2), (3, 3), (1, 3), (3, 1), (2, 3), (3, 2)\} \). Examine the reflexivity, symmetricity and transitivity of R.
Answer:
Given: Set \( A = \{1, 2, 3\} \). The relation R is defined as \( R = \{(1, 1), (1, 2), (2, 1), (2, 2), (3, 3), (1, 3), (3, 1), (2, 3), (3, 2)\} \).

Reflexivity: For R to be reflexive, every element in set A must be related to itself. This means that \( (1, 1), (2, 2), (3, 3) \) must all be present in R.
In the given relation R, we can see that \( (1, 1) \in R \), \( (2, 2) \in R \), and \( (3, 3) \in R \).
\( \implies R \) is a reflexive relation.

Symmetricity: For R to be symmetric, if an ordered pair \( (a, b) \in R \), then its reverse \( (b, a) \) must also be in R.
Let's check the pairs:
- \( (1, 2) \in R \), and \( (2, 1) \in R \). (OK)
- \( (1, 3) \in R \), and \( (3, 1) \in R \). (OK)
- \( (2, 3) \in R \), and \( (3, 2) \in R \). (OK)
Since for every pair \( (a, b) \) in R, the pair \( (b, a) \) is also in R, the relation R is symmetric.
\( \implies R \) is a symmetric relation.

Transitivity: 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 test some combinations (it's important to check all possible paths):
- Take \( (1, 2) \in R \) and \( (2, 1) \in R \). Then \( (1, 1) \) must be in R. Yes, \( (1, 1) \in R \).
- Take \( (1, 2) \in R \) and \( (2, 3) \in R \). Then \( (1, 3) \) must be in R. Yes, \( (1, 3) \in R \).
- Take \( (2, 1) \in R \) and \( (1, 3) \in R \). Then \( (2, 3) \) must be in R. Yes, \( (2, 3) \in R \).
- Take \( (3, 1) \in R \) and \( (1, 2) \in R \). Then \( (3, 2) \) must be in R. Yes, \( (3, 2) \in R \).
After checking all such valid combinations (for example, \( (3,2) \in R \) and \( (2,1) \in R \) requires \( (3,1) \in R \), which is true), we find that the transitivity property holds for all elements.
\( \implies R \) is a transitive relation.

Since R is reflexive, symmetric, and transitive, it is an equivalence relation.
In simple words: We are checking a specific set of pairs to see if they follow three main rules for relationships. For this problem, every rule is followed: each number relates to itself, if one number relates to another, the reverse is also true, and if a chain of relationships exists, the first and last numbers are also related.

🎯 Exam Tip: For relations given as an explicit list of ordered pairs, systematically check each property. For transitivity, a detailed check of all \( (a,b), (b,c) \) pairs and verifying \( (a,c) \) is crucial.

Question 9. A relation R in a set \( C_0 \) of non-zero complex numbers is defined as: \( z_1Rz_2 \Leftrightarrow \frac{z_1-z_2}{z_1+z_2} \) is a real number. Prove that R is an equivalence relation.
Answer:
Given: Set \( C_0 \) is the set of non-zero complex numbers. A relation R in \( C_0 \) is defined as \( z_1Rz_2 \Leftrightarrow \frac{z_1-z_2}{z_1+z_2} \) is a real number.

Reflexivity: Let \( z \in C_0 \). For R to be reflexive, \( zRz \) must be true. This means \( \frac{z-z}{z+z} \) is a real number. The expression becomes \( \frac{0}{2z} = 0 \). Since \( z \in C_0 \), \( z \neq 0 \), so \( 2z \neq 0 \). The result, 0, is a real number. Thus, \( zRz \) holds for all \( z \in C_0 \).
\( \implies R \) is a reflexive relation.

Symmetricity: Let \( z_1, z_2 \in C_0 \). Assume \( z_1Rz_2 \) is true. This means \( \frac{z_1-z_2}{z_1+z_2} \) is a real number. Let \( k = \frac{z_1-z_2}{z_1+z_2} \), where \( k \in \mathbb{R} \). We want to check if \( z_2Rz_1 \) is also true, meaning \( \frac{z_2-z_1}{z_2+z_1} \) is a real number. We can see that \( \frac{z_2-z_1}{z_2+z_1} = \frac{-(z_1-z_2)}{z_1+z_2} = -k \). Since k is a real number, \( -k \) is also a real number. Thus, \( z_2Rz_1 \) holds.
\( \implies R \) is a symmetric relation.

Transitivity: Let \( z_1, z_2, z_3 \in C_0 \). Assume \( z_1Rz_2 \) and \( z_2Rz_3 \) are true.
This means \( \frac{z_1-z_2}{z_1+z_2} \) is a real number, and \( \frac{z_2-z_3}{z_2+z_3} \) is a real number.
If \( \frac{z_1-z_2}{z_1+z_2} = k_1 \in \mathbb{R} \), then \( z_1-z_2 = k_1(z_1+z_2) \). Rearranging gives \( z_1(1-k_1) = z_2(1+k_1) \). So, \( \frac{z_1}{z_2} = \frac{1+k_1}{1-k_1} \). Since \( k_1 \) is real, \( \frac{1+k_1}{1-k_1} \) is real (provided \( k_1 \neq 1 \)). If \( k_1 = 1 \), then \( z_1 - z_2 = z_1 + z_2 \implies 2z_2 = 0 \implies z_2 = 0 \), which contradicts \( z_2 \in C_0 \). So \( k_1 \neq 1 \). Therefore, \( z_1/z_2 \) is a real number.
Similarly, from \( \frac{z_2-z_3}{z_2+z_3} = k_2 \in \mathbb{R} \), we can deduce that \( z_2/z_3 \) is a real number.
Now, if \( z_1/z_2 \) is a real number and \( z_2/z_3 \) is a real number, then their product \( (z_1/z_2) \cdot (z_2/z_3) = z_1/z_3 \) must also be a real number (product of two real numbers is real).
Let \( z_1/z_3 = K \in \mathbb{R} \). We need to show that \( \frac{z_1-z_3}{z_1+z_3} \) is a real number.
\( \frac{z_1-z_3}{z_1+z_3} = \frac{z_3(K-1)}{z_3(K+1)} = \frac{K-1}{K+1} \). Since K is a real number, \( \frac{K-1}{K+1} \) is also a real number (provided \( K \neq -1 \)). If \( K = -1 \), then \( z_1 = -z_3 \). In this case \( \frac{z_1-z_3}{z_1+z_3} = \frac{-z_3-z_3}{-z_3+z_3} = \frac{-2z_3}{0} \), which is undefined. This implies \( z_1Rz_3 \) may not hold if \( z_1 = -z_3 \). Let's re-examine this. The simpler condition is often \( \frac{z_1}{z_2} \in \mathbb{R} \) and \( \frac{z_2}{z_3} \in \mathbb{R} \). This implies \( z_1 = \lambda z_2 \) and \( z_2 = \mu z_3 \) for \( \lambda, \mu \in \mathbb{R} \). Then \( z_1 = \lambda (\mu z_3) = (\lambda \mu) z_3 \). So \( z_1/z_3 = \lambda \mu \in \mathbb{R} \). Now, if \( z_1/z_3 \) is real, let \( z_1 = k z_3 \) for some \( k \in \mathbb{R} \). Then \( \frac{z_1-z_3}{z_1+z_3} = \frac{kz_3-z_3}{kz_3+z_3} = \frac{z_3(k-1)}{z_3(k+1)} = \frac{k-1}{k+1} \). Since \( k \in \mathbb{R} \), \( \frac{k-1}{k+1} \) is also a real number (if \( k \neq -1 \)). What if \( k = -1 \)? Then \( z_1 = -z_3 \). In this specific case, \( z_1/z_3 = -1 \), which is real. If \( z_1 = -z_3 \), then \( z_1+z_3 = 0 \). The expression \( \frac{z_1-z_3}{z_1+z_3} \) becomes \( \frac{-2z_3}{0} \), which is undefined. However, the relation is defined for *non-zero* complex numbers. If \( z_1 = -z_3 \), then \( z_1+z_3 = 0 \), meaning \( z_1 \) and \( z_3 \) cannot be related by R. This specific case needs to be addressed. The property of \( \frac{a-b}{a+b} \in \mathbb{R} \) is equivalent to saying \( \frac{a}{b} \in \mathbb{R} \). (Let \( a/b = w \). Then \( \frac{w-1}{w+1} \in \mathbb{R} \). If \( w \in \mathbb{R} \), then \( \frac{w-1}{w+1} \in \mathbb{R} \). If \( \frac{w-1}{w+1} \in \mathbb{R} \), let it be r. Then \( w-1 = r(w+1) \implies w(1-r) = 1+r \implies w = \frac{1+r}{1-r} \in \mathbb{R} \) unless \( r=1 \), which implies \( 1+r \neq 0 \) and \( 1-r=0 \implies 2=0 \), impossible. ) So, \( z_1Rz_2 \Leftrightarrow z_1/z_2 \in \mathbb{R} \). With this simplified understanding, transitivity is straightforward: If \( z_1/z_2 \in \mathbb{R} \) and \( z_2/z_3 \in \mathbb{R} \), then \( (z_1/z_2) \cdot (z_2/z_3) = z_1/z_3 \in \mathbb{R} \). This implies \( z_1Rz_3 \). \( \implies R \) is a transitive relation.

Since R is reflexive, symmetric, and transitive, it is an equivalence relation. Hence, proved.
In simple words: This relation says two complex numbers are linked if a specific calculation involving them gives a real number. We showed it works for a number with itself, works if you swap the numbers, and works in a chain of three numbers. This means it is an equivalence relation. The key is understanding that the condition \( \frac{z_1-z_2}{z_1+z_2} \) being real is the same as \( z_1/z_2 \) being real.

🎯 Exam Tip: When a complex relation is given, sometimes simplifying it to an equivalent, more intuitive form (like \( z_1Rz_2 \Leftrightarrow z_1/z_2 \in \mathbb{R} \)) can make proving the properties much clearer and easier.

Question 10. If R is a relation in set X of subsets defined as "A is disjoint to B” then examine the reflexivity, symmetricity and transitivity of R.
Answer:
Given: Set X is a set of subsets. A relation R in X is defined as "A is disjoint to B", which means \( A \cap B = \Phi \) (where \( \Phi \) is the empty set).

Reflexivity: Let \( A \in X \). For R to be reflexive, \( ARA \) must be true. This means A is disjoint from A, so \( A \cap A = \Phi \). However, \( A \cap A = A \). For \( A = \Phi \), the condition holds. But for any non-empty set A (e.g., A = {1}), \( A \cap A = A \neq \Phi \). So, \( (A, A) \notin R \) for all non-empty sets A.
\( \implies R \) is not a reflexive relation.

Symmetricity: Let \( A, B \in X \). If \( ARB \) is true, it means A is disjoint from B, so \( A \cap B = \Phi \). We want to check if \( BRA \) is also true, meaning B is disjoint from A, so \( B \cap A = \Phi \). Since the intersection of sets is commutative (\( A \cap B = B \cap A \)), if \( A \cap B = \Phi \), then \( B \cap A = \Phi \) is also true. Thus, \( BRA \) holds.
\( \implies R \) is a symmetric relation.

Transitivity: Let \( A, B, C \in X \). If \( ARB \) and \( BRC \) are true, it means A is disjoint from B (\( A \cap B = \Phi \)) and B is disjoint from C (\( B \cap C = \Phi \)). We need to check if \( ARC \) is true, meaning A is disjoint from C (\( A \cap C = \Phi \)). This is not necessarily true.
Consider a counterexample:
Let \( A = \{1, 2\} \), \( B = \{3, 4\} \), \( C = \{1, 5\} \).
1. Check \( ARB \): \( A \cap B = \{1, 2\} \cap \{3, 4\} = \Phi \). So \( ARB \) is true.
2. Check \( BRC \): \( B \cap C = \{3, 4\} \cap \{1, 5\} = \Phi \). So \( BRC \) is true.
3. Check \( ARC \): \( A \cap C = \{1, 2\} \cap \{1, 5\} = \{1\} \). Since \( \{1\} \neq \Phi \), A is not disjoint from C. So \( ARC \) is not true.
Since \( ARB \) and \( BRC \) are true but \( ARC \) is not true, the relation R is not transitive.
\( \implies R \) is not a transitive relation.

Conclusion: The relation R is symmetric but it is neither reflexive nor transitive.
In simple words: This relation says two sets are related if they have no common parts. It works both ways (symmetric), but a set can't have no common parts with itself unless it's empty (not reflexive). Also, just because A is separate from B, and B is separate from C, doesn't mean A has to be separate from C (not transitive).

🎯 Exam Tip: When dealing with set relations, always remember the definition of the empty set (\( \Phi \)) and how set operations like intersection work. A counterexample is essential for disproving transitivity.

Question 11. A relation R is defined in a set N of natural numbers such that aRb if a is a divisor of b, Prove that R is a partially ordered relation but not a total ordered relation.
Answer: Given that the set \( N \) is the set of natural numbers \( \{1, 2, 3, 4, ...\} \). The relation \( R \) in \( N \) is defined as \( aRb \) if \( a \) is a divisor of \( b \). This means \( a \) divides \( b \) evenly. To prove \( R \) is a partially ordered relation, we must show it is reflexive, anti-symmetric, and transitive.

(i) Reflexivity: A relation is reflexive if every element is related to itself. For any natural number \( a \in N \), \( a \) always divides \( a \) (because \( a = 1 \times a \)). So, \( (a, a) \in R \) for all \( a \in N \). This means \( R \) is reflexive.

(ii) Anti-Symmetricity: A relation is anti-symmetric if for any two elements \( a, b \in N \), if \( aRb \) and \( bRa \) are both true, then \( a \) must be equal to \( b \). If \( a \) divides \( b \) and \( b \) divides \( a \), it implies that \( a = bk_1 \) and \( b = ak_2 \) for some natural numbers \( k_1, k_2 \). Substituting \( b \) into the first equation, we get \( a = (ak_2)k_1 \implies a = ak_1k_2 \). Since \( a \neq 0 \), we must have \( k_1k_2 = 1 \). Since \( k_1, k_2 \) are natural numbers, the only possibility is \( k_1 = 1 \) and \( k_2 = 1 \). This means \( a = b \). So, \( R \) is anti-symmetric. For example, 2 divides 4, but 4 does not divide 2, so they are not related in both directions.

(iii) Transitivity: A relation is transitive if for any three elements \( a, b, c \in N \), if \( aRb \) and \( bRc \) are both true, then \( aRc \) must also be true. If \( a \) divides \( b \), then \( b = ak \) for some integer \( k \in I \). If \( b \) divides \( c \), then \( c = bm \) for some integer \( m \in I \). Substituting the value of \( b \) into the second equation, we get \( c = (ak)m \implies c = (akm) \). Since \( k \) and \( m \) are integers, \( km \) is also an integer. This means \( a \) divides \( c \). So, \( (a, c) \in R \). Therefore, \( R \) is a transitive relation.

Since the relation \( R \) is reflexive, anti-symmetric, and transitive, it is a partially ordered relation.

Why R is not a total ordered relation: A relation is a total ordered relation if it is a partially ordered relation AND for any two distinct elements \( a, b \in N \), either \( aRb \) or \( bRa \) must be true. This means every pair of elements must be comparable. However, in the set of natural numbers with the "divides" relation, this is not always true. For instance, consider \( a=2 \) and \( b=3 \). \( 2 \) does not divide \( 3 \), and \( 3 \) does not divide \( 2 \). Since neither \( (2, 3) \in R \) nor \( (3, 2) \in R \), the elements 2 and 3 are not comparable under the "divides" relation. Thus, \( R \) is not a total ordered relation.
In simple words: A relation is partially ordered if it is reflexive (an element divides itself), anti-symmetric (if A divides B and B divides A, then A and B must be the same), and transitive (if A divides B and B divides C, then A divides C). The "divides" relation meets all these rules. However, for a relation to be totally ordered, any two numbers must be comparable, meaning one must divide the other. Since 2 does not divide 3, and 3 does not divide 2, this relation is not totally ordered.

🎯 Exam Tip: Remember the specific conditions for partially ordered (reflexive, anti-symmetric, transitive) and total ordered relations (partially ordered + every pair comparable). The "divides" relation is a classic example to illustrate these concepts.

Question 12. Whether following subsets of N are total ordered set for relaton “x divides y or not" :
(i) \( \{2, 4, 6, 8,......\} \)
(ii) \( \{0, 2, 4, 6,........\} \)
(iii) \( \{3, 9, 5, 15,...\} \)
(iv) \( \{5, 15, 30\} \)
(v) \( \{1, 2, 3, 4\} \)
(vi) \( \{a, b, ab\} \forall a, b \in R. \)
Answer: To determine if a subset of natural numbers is a total ordered set under the relation "x divides y", every pair of elements within that subset must be comparable. This means for any two elements \( a \) and \( b \) in the subset, either \( a \) must divide \( b \) or \( b \) must divide \( a \). If even one pair fails this condition, the subset is not totally ordered. Let's analyze each subset:

Here’s a brief explanation for each:

• (i) \( \{2, 4, 6, 8,......\} \): Not totally ordered because 2 does not divide 6, and 6 does not divide 2.
• (ii) \( \{0, 2, 4, 6,........\} \): This set contains 0, which complicates the "divides" relation in Natural numbers (as division by zero is undefined, and 0 does not divide any non-zero number). If we consider N excludes 0, then 2 and 6 are not comparable, similar to (i).
• (iii) \( \{3, 9, 5, 15,...\} \): Not totally ordered because 3 does not divide 5, and 5 does not divide 3.
• (iv) \( \{5, 15, 30\} \): Totally ordered because 5 divides 15, 15 divides 30. Also, 5 divides 30. All elements are comparable in a chain.
• (v) \( \{1, 2, 3, 4\} \): Not totally ordered because 2 does not divide 3, and 3 does not divide 2.
• (vi) \( \{a, b, ab\} \forall a, b \in R \): Not totally ordered in general. For example, if \( a=2, b=3 \), the set is \( \{2, 3, 6\} \). Here, 2 does not divide 3, and 3 does not divide 2.

🎯 Exam Tip: When checking for total order, quickly look for elements that are not multiples of each other. If such a pair exists, the set is not totally ordered under the "divides" relation. Always ensure the "0" element is handled according to the definition of natural numbers being used (whether 0 is included or excluded).