Show that the relation $R$ in the set $A=\{1,2,3,4,5\}$ given by $R =\{(a, b):|a-b| \text{ is even}\}$ is an equivalence relation. Show that all the elements of $\{1,3,5\}$ are related to each other and all the elements of $\{2,4\}$ are related to each other,but no element of $\{1,3,5\}$ is related to any element of $\{2,4\}$.

(N/A) $A = \{1, 2, 3, 4, 5\}$ and $R = \{(a, b) : |a - b| \text{ is even}\}$.
$1$. Reflexivity: For any $a \in A$,$|a - a| = 0$,which is even. Thus,$(a, a) \in R$. So,$R$ is reflexive.
$2$. Symmetry: Let $(a, b) \in R$. Then $|a - b|$ is even. Since $|a - b| = |b - a|$,$|b - a|$ is also even. Thus,$(b, a) \in R$. So,$R$ is symmetric.
$3$. Transitivity: Let $(a, b) \in R$ and $(b, c) \in R$. Then $|a - b|$ is even and $|b - c|$ is even. Let $|a - b| = 2k$ and $|b - c| = 2m$ for some integers $k, m$. Then $(a - b) = \pm 2k$ and $(b - c) = \pm 2m$. Adding these,$(a - c) = (a - b) + (b - c) = \pm 2k \pm 2m = 2(\pm k \pm m)$,which is even. Thus,$|a - c|$ is even,so $(a, c) \in R$. So,$R$ is transitive.
Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation.
For the subsets: All elements of $\{1, 3, 5\}$ are odd. The difference between any two odd numbers is even,so they are all related. All elements of $\{2, 4\}$ are even. The difference between any two even numbers is even,so they are all related. However,the difference between an odd number and an even number is always odd,so no element of $\{1, 3, 5\}$ is related to any element of $\{2, 4\}$.