Give an example of a relation that is symmetric and transitive but not reflexive.

(N/A) Let $A = \{-5, -6\}$.
Define a relation $R$ on $A$ as $R = \{(-5, -6), (-6, -5), (-5, -5)\}$.
$1$. Reflexivity: For $R$ to be reflexive,$(-5, -5) \in R$ and $(-6, -6) \in R$ must hold. Since $(-6, -6) \notin R$,the relation $R$ is not reflexive.
$2$. Symmetry: For $R$ to be symmetric,if $(a, b) \in R$,then $(b, a) \in R$. Here,$(-5, -6) \in R$ and $(-6, -5) \in R$. Also,$(-5, -5) \in R$ and its reverse $(-5, -5) \in R$. Thus,$R$ is symmetric.
$3$. Transitivity: For $R$ to be transitive,if $(a, b) \in R$ and $(b, c) \in R$,then $(a, c) \in R$. Checking the pairs: $(-6, -5) \in R$ and $(-5, -6) \in R$ implies $(-6, -6) \in R$,which is false. Wait,let us re-evaluate the set. Let $A = \{1, 2\}$ and $R = \{(1, 2), (2, 1), (1, 1)\}$. Here $(1, 2) \in R$ and $(2, 1) \in R$,but $(2, 2) \notin R$. Thus,it is not transitive.
Correct Example: Let $A = \{1, 2, 3\}$ and $R = \{(1, 1), (1, 2), (2, 1), (2, 2)\}$.
- Not reflexive: $(3, 3) \notin R$.
- Symmetric: $(1, 2) \in R$ and $(2, 1) \in R$; $(1, 1)$ and $(2, 2)$ are symmetric.
- Transitive: All conditions $(a, b) \in R$ and $(b, c) \in R \implies (a, c) \in R$ are satisfied.