Solution:
and is even (As given)
If is Reflexive, Symmetric and Transitive, then is an equivalence relation.
Reflexivity:
Suppose a be an arbitrary element of
As is even
Therefore, is reflexive.
Symmetric:
Suppose a and , such that
even.
Therefore, is symmetric.
Transitivity:
Suppose and , such that and (which is an even)
and (which is an even)
On adding both the above equations, we obtain
is an even number
Therefore, is transitive.
As a result, is an equivalence relation.