Solution
Since \(R_1\) and \(R_2\) are equivalence relations, \((a, a) \in R_1\) and \((a, a) \in R_2\), \(\forall\, a \in A\).
This implies that \((a, a) \in R_1 \cap
R_2\), \(\forall\, a\), showing
\(R_1 \cap R_2\) is reflexive.
Further, \((a, b) \in R_1 \Rightarrow (a, b)
\in R_1\) and \((b, a) \in
R_1\), and \((a, b) \in R_2 \Rightarrow
(b, a) \in R_2\).
Hence, \((a, b) \in R_1 \cap R_2 \Rightarrow
(b, a) \in R_1 \cap R_2\), so \(R_1
\cap R_2\) is symmetric.
Similarly, \((a, b) \in R_1 \cap R_2\)
and \((b, c) \in R_1 \cap R_2 \Rightarrow (a,
c) \in R_1\) and \((a, c) \in
R_2\).
This shows that \((a, c) \in R_1 \cap
R_2\) is transitive.
Thus, \(R_1 \cap R_2\) is an
equivalence relation.
