Question:
Let $R=\{(a, b): a, b \in Z$ and $(a-b)$ is even $\}$
Then, show that R is an equivalence relation on Z
Solution:
(i) Reflexivity: Let $a \in Z, a-a=0 \in Z$ which is also even.
Thus, $(a, a) \in R$ for all $a \in Z$. Hence, it is reflexive
(ii) Symmetry: Let $(a, b) \in R$
$(a, b) \in R$ è $a-b$ is even
$-(b-a)$ is even
$(b-a)$ is even
$(b, a) \in R$
Thus, it is symmetric
(iii) Transitivity: Let $(a, b) \in R$ and $(b, c) \in R$
Then, $(a-b)$ is even and $(b-c)$ is even.
$[(a-b)+(b-c)]$ is even
$(a-c)$ is even.
Thus $(a, c) \in R$.
Hence, it is transitive
Since, the given relation possesses the properties of reflexivity, symmetry and transitivity, it is an equivalence relation.