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Below are my attempts at two proofs.

Let $R$ and $S$ be relations on a set $A$. Assume $A$ has at least three elements.

If $R$ and $S$ are reflexive, then $R \cap S$ is reflexive

proof:

Suppose (x,y) ∈ R $\cap$ S, --->

(x,y) ∈ S and (x,y) ∈ R --->

Since R and S are reflexive, (x,x)(y,y) ∈ S and (x,x)(y,y) ∈ R --->

(x,x)(y,y) ∈ R $\cap$ S, therefore it is reflexive.

If $R$ and $S$ are reflexive, then $R \cup S$ is reflexive

proof:

Suppose (x,y) ∈ R $\cup$ S, --->

(x,y) ∈ S or (x,y) ∈ R --->

Since R and S are reflexive, (x,x)(y,y) ∈ S or (x,x)(y,y) ∈ R --->

(x,x)(y,y) ∈ R $\cup$ S, therefore it is reflexive.

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Your answers are a little confusing to me. To prove reflexivity, you need to show that for all $x\in A, (x,x)$ is in the relation.

For $R\cap S$:

Let $x \in A$. Since $R,S$ are reflexive, $(x,x) \in R$ and $(x,x) \in S$. Therefore, $(x,x) \in R\cap S$. Therefore $R\cap S$ is reflexive.

For $R\cup S$:

Let $x \in A$. Since $R$ is reflexive, $(x,x) \in R$. Therefore, $(x,x) \in R\cup S$. Therefore $R\cup S$ is reflexive.

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