Solve this


Let $R=\{(a, b): a, b \in Z$ and $(a-b)$ is even $\}$

Then, show that R is an equivalence relation on Z



(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.


Leave a comment