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Let $R$ be a symmetric and transitive relation on a set $A$. Then

1. $R$ is reflexive and hence an equivalence relation
2. $R$ is reflexive and hence a partial order
3. $R$ is reflexive and hence not an equivalence relation
4. None of the above
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Can we see this question from propositional logic point of view?

$\text{R be a Symmetric and Transitive relation on a set A }$ $\implies$ $\text{R is Reflexive & Equivalence relation}$

which is completely false.

It would be true if

$\text{R be a Reflexive, Symmetric and Transitive relation on a set A }$ $\implies$ $\text{Equivalence relation}$
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in implies R⟹E , the statement is true also if R is false...

so by that way, you mean ..if ( Refl. & Symm & Transitivity ) is false than also it'll be Equivalence

Answer is $D$.

Let $A={1,2,3}$ and relation $R={(1,2),(2,1),(1,1),(2,2)}$. $R$ is symmetric and transitive but not reflexive.( $(3,3)$ is not there.)
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Also, the Empty relation is symmetric and transitive by default but not reflexive.

Hence ans-(d)

Let A = {(1,2),(2,1),(1,1)}

A is symmetric and transitive but not reflexive as (2,2) is not there.
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According to the example you have assumed, thought the answer remains correct, but you must include (2,2) into the relation as well because of transitivity. and may be you can change the set A to , A={1,2,3}
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simply, take (1,1) as relation  which is symmetric and transitive, but for reflexive it should have other pair ( b,b) &(c,c) if i consider set {1,2,3}

hence option D
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The explanation is not right!
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@rajshree you must include (2,2) in your explanation
here ans should be D

explanation:

here the relation is symmetric and transitive. if relation is symmetric and transitive then it need not necessariy be reflexive;i.e. it may or may not be reflexive. therefore ans is D
We can take an empty set { } which is both symmetric and and transitive but not reflexive because diagonal elememts are not present in the set so not reflexive.

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