Question 51

Mathematics Function and Relation Hard

Let \( A=\{1,2,3\} \) and consider the relation \( R=\{(1,1),(2,2),(3,3),(1,2),(1,3)\} \) then \( R \) is

(A) Reflexive but not symmetric
(B) Reflexive but no transitive
(C) Symmetric and transitive
(D) Equivalence relation
View Dynamic Solution & Explanation
Correct Solution: Option A

Step-by-step Solution:

We are given a set \( A = \{1,2,3\} \) and a relation \( R \) defined as:
\[ R = \{(1,1), (2,2), (3,3), (1,2), (1,3)\} \] We need to analyze the properties of \( R \):
1. Reflexive Property:
A relation is reflexive if \( (a, a) \in R \) for all \( a \in A \).
We have \( (1,1), (2,2), (3,3) \) in \( R \), which means all elements have self-pairs.
\( R \) is reflexive.

2. Symmetric Property:
A relation is symmetric if whenever \( (a, b) \in R \), then \( (b, a) \in R \) must also be in \( R \).
We have \( (1,2) \in R \), but \( (2,1) \notin R \).
We have \( (1,3) \in R \), but \( (3,1) \notin R \).
\( R \) is not symmetric.

3. Transitive Property:
A relation is transitive if whenever \( (a, b) \in R \) and \( (b, c) \in R \), then \( (a, c) \in R \).
We have \( (1,2) \in R \), but no \( (2, x) \) pair exists, so no issue.
We have \( (1,3) \in R \), but no \( (3, x) \) pair exists, so no issue.
No violations, so \( R \) is transitive.

The correct option is:
(A) Reflexive but not symmetric.