CBSE Class 12 Mathematics Relations And Functions Class Test Set 03

Question. Define Equivalence relation
Answer: A relation \( R \) on a set \( A \) is said to be an equivalence relation if and only if it satisfies the following three properties:
1. Reflexive: \( (a, a) \in R \) for all \( a \in A \).
2. Symmetric: If \( (a, b) \in R \), then \( (b, a) \in R \) for all \( a, b \in A \).
3. Transitive: If \( (a, b) \in R \) and \( (b, c) \in R \), then \( (a, c) \in R \) for all \( a, b, c \in A \).

Question. Show that the relation \( R \) in the Set \( \{1,2,3\} \) given by \( R = \{(1,1), (2,2), (3,3), (1,2), (2,3)\} \) is reflexive but neither symmetric nor transitive.
Answer: Let \( A = \{1, 2, 3\} \).
The given relation is \( R = \{(1,1), (2,2), (3,3), (1,2), (2,3)\} \).
1. Reflexivity: Since \( (1,1) \in R \), \( (2,2) \in R \), and \( (3,3) \in R \), every element in \( A \) is related to itself. Hence, \( R \) is reflexive.
2. Symmetry: We observe that \( (1,2) \in R \), but \( (2,1) \notin R \). Thus, \( R \) is not symmetric.
3. Transitivity: We observe that \( (1,2) \in R \) and \( (2,3) \in R \), but \( (1,3) \notin R \). Thus, \( R \) is not transitive.
Therefore, \( R \) is reflexive but neither symmetric nor transitive.

Question. Prove that the relation \( R \) in the set \( A = \{1,2,3,4,5\} \) given by \( R = \{(a,b) : |a-b| \text{ is even }, a, b \in A\} \) is an equivalence relation.
Answer: Let \( A = \{1,2,3,4,5\} \) and \( R = \{(a,b) : |a-b| \text{ is even}\} \).
1. Reflexivity: For any \( a \in A \), \( |a-a| = 0 \), which is an even number. Therefore, \( (a,a) \in R \) for all \( a \in A \). Thus, \( R \) is reflexive.
2. Symmetry: Let \( (a,b) \in R \). Then \( |a-b| \) is even. Since \( |b-a| = |-(a-b)| = |a-b| \), \( |b-a| \) is also even. Thus, \( (b,a) \in R \). Hence, \( R \) is symmetric.
3. Transitivity: Let \( (a,b) \in R \) and \( (b,c) \in R \). Then \( |a-b| \) is even and \( |b-c| \) is even. This implies that \( a-b \) is even and \( b-c \) is even. Since the sum of two even integers is also an even integer, \( (a-b) + (b-c) = a-c \) is even. Consequently, \( |a-c| \) is even, which means \( (a,c) \in R \). Hence, \( R \) is transitive.
Since \( R \) is reflexive, symmetric, and transitive, it is an equivalence relation.

Question. Determine whether each of the following relations are Ref, Sym, Trans.
(i) \( R = \{(x,y) : x \text{ is the wife of } y\} \)
(ii) \( R = \{(x,y) : x \text{ is exactly 8cm taller than } y\} \)
Answer:
(i) \( R = \{(x,y) : x \text{ is the wife of } y\} \)
- Reflexive: A person \( x \) cannot be the wife of themselves. Thus, \( (x,x) \notin R \). It is not reflexive.
- Symmetric: If \( (x,y) \in R \), then \( x \) is the wife of \( y \). This implies \( y \) is the husband of \( x \), so \( y \) cannot be the wife of \( x \). Thus, \( (y,x) \notin R \). It is not symmetric.
- Transitive: If \( (x,y) \in R \), then \( x \) is the wife of \( y \), meaning \( y \) is male. Since \( y \) is male, \( y \) cannot be the wife of any person \( z \). Consequently, the condition for transitivity "if \( (x,y) \in R \) and \( (y,z) \in R \)" is vacuously satisfied. It is transitive.
Conclusion: The relation is transitive but neither reflexive nor symmetric.

(ii) \( R = \{(x,y) : x \text{ is exactly 8cm taller than } y\} \)
- Reflexive: A person \( x \) cannot be exactly 8 cm taller than themselves. Thus, \( (x,x) \notin R \). It is not reflexive.
- Symmetric: If \( (x,y) \in R \), then \( x \) is exactly 8 cm taller than \( y \). This means \( y \) is exactly 8 cm shorter than \( x \), so \( y \) cannot be exactly 8 cm taller than \( x \). Thus, \( (y,x) \notin R \). It is not symmetric.
- Transitive: If \( (x,y) \in R \) and \( (y,z) \in R \), then \( x \) is exactly 8 cm taller than \( y \), and \( y \) is exactly 8 cm taller than \( z \). This implies \( x \) is exactly 16 cm taller than \( z \). Thus, \( (x,z) \notin R \). It is not transitive.
Conclusion: The relation is neither reflexive, symmetric, nor transitive.