Let and is even . Show that is an equivalence relation on .

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.