R is reflexive and transitive but not symmetric.
R is symmetric and transitive but not reflexive.
R is an equivalence relation.
R =\({(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}\)
It is seen that \((a, a) ∈ R\), for every \(a ∈{1, 2, 3, 4}\).
∴ R is reflexive. It is seen that \((1, 2) ∈\)R, but \((2, 1) ∉ R.\)
∴R is not symmetric. Also, it is observed that \((a, b), (b, c) ∈ R ⇒ (a, c) ∈ R\) for all \(a, b, c ∈ {1, 2, 3, 4}\).
∴ R is transitive. Hence, R is reflexive and transitive but not symmetric.
The correct answer is B.
A school is organizing a debate competition with participants as speakers and judges. $ S = \{S_1, S_2, S_3, S_4\} $ where $ S = \{S_1, S_2, S_3, S_4\} $ represents the set of speakers. The judges are represented by the set: $ J = \{J_1, J_2, J_3\} $ where $ J = \{J_1, J_2, J_3\} $ represents the set of judges. Each speaker can be assigned only one judge. Let $ R $ be a relation from set $ S $ to $ J $ defined as: $ R = \{(x, y) : \text{speaker } x \text{ is judged by judge } y, x \in S, y \in J\} $.
The correct IUPAC name of \([ \text{Pt}(\text{NH}_3)_2\text{Cl}_2 ]^{2+} \) is:
Relation is said to be empty relation if no element of set X is related or mapped to any element of X i.e, R = Φ.
A relation R in a set, say A is a universal relation if each element of A is related to every element of A.
R = A × A.
Every element of set A is related to itself only then the relation is identity relation.
Let R be a relation from set A to set B i.e., R ∈ A × B. The relation R-1 is said to be an Inverse relation if R-1 from set B to A is denoted by R-1
If every element of set A maps to itself, the relation is Reflexive Relation. For every a ∈ A, (a, a) ∈ R.
A relation R is said to be symmetric if (a, b) ∈ R then (b, a) ∈ R, for all a & b ∈ A.
A relation is said to be transitive if, (a, b) ∈ R, (b, c) ∈ R, then (a, c) ∈ R, for all a, b, c ∈ A
A relation is said to be equivalence if and only if it is Reflexive, Symmetric, and Transitive.