GATE CS 1995 Q1.19 — Sets, Relations and Functions

A relation $R$ on a set $A$ is symmetric and transitive, but it is not necessarily reflexive. For $R$ to be reflexive, every element $a \in A$ must satisfy $(a, a) \in R$.

We can demonstrate this with a simple counterexample:

• Let the set be $A = \{1, 2\}$.
• Let the relation be $R = \{(1, 1)\}$.

Let's verify the properties of $R$ on $A$:

1. Symmetry: For any $x, y \in A$, if $(x, y) \in R$, then $(y, x) \in R$. Here, $(1, 1) \in R \implies (1, 1) \in R$, which is true. Thus, $R$ is symmetric.
2. Transitivity: For any $x, y, z \in A$, if $(x, y) \in R$ and $(y, z) \in R$, then $(x, z) \in R$. Here, $(1, 1) \in R$ and $(1, 1) \in R \implies (1, 1) \in R$, which is true. Thus, $R$ is transitive.
3. Reflexivity: For $R$ to be reflexive, we must have $(a, a) \in R$ for all $a \in A$. However, $(2, 2) \notin R$. Thus, $R$ is not reflexive.

Since $R$ is not guaranteed to be reflexive, statements A, B, and C are all false. Therefore, the correct option is D.