CBSE Class 12 Mathematics Relations And Functions Class Test Set 06

Question. Show that the relation \( R \) in the Set \( Z \) of integers given by \( R = \{(a,b) : 2 \text{ divides } (a-b) \ \forall\ a,b \in Z\} \) is an equivalence relation.
Answer: To prove that \( R \) is an equivalence relation, we must show that it is reflexive, symmetric, and transitive.
1. Reflexive: For any \( a \in Z \), \( a - a = 0 \), which is divisible by \( 2 \). Therefore, \( (a, a) \in R \) for all \( a \in Z \). Thus, the relation is reflexive.
2. Symmetric: Let \( (a, b) \in R \). This implies \( 2 \) divides \( (a - b) \), so \( a - b = 2k \) for some integer \( k \). We can write \( b - a = -(a - b) = -2k = 2(-k) \). Since \( -k \) is also an integer, \( 2 \) divides \( (b - a) \), meaning \( (b, a) \in R \). Thus, the relation is symmetric.
3. Transitive: Let \( (a, b) \in R \) and \( (b, c) \in R \). This implies \( 2 \) divides \( (a - b) \) and \( 2 \) divides \( (b - c) \). So, \( a - b = 2k_1 \) and \( b - c = 2k_2 \) for some integers \( k_1, k_2 \). Adding these equations gives \( (a - b) + (b - c) = 2k_1 + 2k_2 \implies a - c = 2(k_1 + k_2) \). Since \( k_1 + k_2 \) is an integer, \( 2 \) divides \( (a - c) \), meaning \( (a, c) \in R \). Thus, the relation is transitive.
Since the relation \( R \) is reflexive, symmetric, and transitive, it is an equivalence relation.

Question. \( L \) is the set of all lines in a plane and \( R \) is the relation in \( L \) : \( R = \{(l_1, l_2) : l_1 \perp l_2, \ \forall\ l_1, l_2 \in L\} \). Is \( R \) an equivalence relation.
Answer: To determine if \( R \) is an equivalence relation, let us check its properties:
1. Reflexive: A line \( l_1 \) cannot be perpendicular to itself, so \( (l_1, l_1) \notin R \). Thus, \( R \) is not reflexive.
2. Symmetric: Let \( (l_1, l_2) \in R \). This means \( l_1 \perp l_2 \). Since \( l_1 \) is perpendicular to \( l_2 \), it follows that \( l_2 \) is also perpendicular to \( l_1 \), so \( l_2 \perp l_1 \). This implies \( (l_2, l_1) \in R \). Thus, \( R \) is symmetric.
3. Transitive: Let \( (l_1, l_2) \in R \) and \( (l_2, l_3) \in R \). This means \( l_1 \perp l_2 \) and \( l_2 \perp l_3 \). If line \( l_1 \) is perpendicular to \( l_2 \) and line \( l_2 \) is perpendicular to \( l_3 \), then \( l_1 \) is parallel to \( l_3 \) (not perpendicular). Thus, \( (l_1, l_3) \notin R \). So, \( R \) is not transitive.
Since the relation \( R \) is not reflexive and not transitive, it is not an equivalence relation.

Question. State the reason for the relation \( R \) in the Set \( \{1,2,3\} \) given by \( \{(1,2), (2,1)\} \) not to be Transitive.
Answer: For a relation \( R \) to be transitive, if \( (a, b) \in R \) and \( (b, c) \in R \), then \( (a, c) \) must also belong to \( R \).
For the given relation, we have \( (1, 2) \in R \) and \( (2, 1) \in R \). If \( R \) were transitive, then \( (1, 1) \) must be an element of \( R \). However, \( (1, 1) \notin R \). Therefore, the relation is not transitive.

Question. let \( A = \{1,2,3,4,5\} \) and \( R = \{(a,b) : b=a+1 \ \forall\ a,b\in A\} \) check whether the above relation is reflexive, symmetric or transitive.
Answer: Given \( A = \{1,2,3,4,5\} \) and \( R = \{(a,b) : b=a+1 \ \forall\ a,b\in A\} \).
Writing the relation \( R \) in roster form, we get:
\( R = \{(1,2), (2,3), (3,4), (4,5)\} \)
Let us check the three properties:
1. Reflexive: For \( R \) to be reflexive, \( (a, a) \in R \) for all \( a \in A \). Since \( (1, 1) \notin R \), the relation is not reflexive.
2. Symmetric: For \( R \) to be symmetric, if \( (a, b) \in R \), then \( (b, a) \in R \). Here, \( (1, 2) \in R \) but \( (2, 1) \notin R \). Thus, the relation is not symmetric.
3. Transitive: For \( R \) to be transitive, if \( (a, b) \in R \) and \( (b, c) \in R \), then \( (a, c) \in R \). Here, we have \( (1, 2) \in R \) and \( (2, 3) \in R \), but \( (1, 3) \notin R \). Thus, the relation is not transitive.
Consequently, the relation is neither reflexive, nor symmetric, nor transitive.