Show that the relation $R$ in the set $A = \{x \in Z : 0 \leq x \leq 12\},$ given by $R = \{(a, b) : |a - b| \text{ is a multiple of } 4\},$ is an equivalence relation. Find the set of all elements related to $1$.

(A) Set $A = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}$.
$R = \{(a, b) : |a - b| \text{ is a multiple of } 4\}$.
$1$. Reflexive: For any $a \in A$,$|a - a| = 0$,which is a multiple of $4$. Thus,$(a, a) \in R$. So,$R$ is reflexive.
$2$. Symmetric: Let $(a, b) \in R$. Then $|a - b|$ is a multiple of $4$. Since $|a - b| = |-(b - a)| = |b - a|$,$|b - a|$ is also a multiple of $4$. Thus,$(b, a) \in R$. So,$R$ is symmetric.
$3$. Transitive: Let $(a, b) \in R$ and $(b, c) \in R$. Then $|a - b| = 4k_1$ and $|b - c| = 4k_2$ for some integers $k_1, k_2$. Then $(a - b) = \pm 4k_1$ and $(b - c) = \pm 4k_2$. Adding these,$(a - c) = (a - b) + (b - c) = \pm 4k_1 \pm 4k_2 = 4(\pm k_1 \pm k_2)$,which is a multiple of $4$. Thus,$(a, c) \in R$. So,$R$ is transitive.
Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation.
The set of elements related to $1$ is $\{x \in A : |x - 1| \text{ is a multiple of } 4\}$.
$|x - 1| \in \{0, 4, 8, 12, \dots\}$.
If $|x - 1| = 0$,$x = 1$.
If $|x - 1| = 4$,$x - 1 = 4$ or $x - 1 = -4$,so $x = 5$ or $x = -3$ (not in $A$).
If $|x - 1| = 8$,$x - 1 = 8$ or $x - 1 = -8$,so $x = 9$ or $x = -7$ (not in $A$).
Thus,the set of elements related to $1$ is $\{1, 5, 9\}$.