सिद्ध किजिए कि समुच्चय $A =\{x \in Z : 0 \leq x \leq 12\},$ में दिए गए निम्नलिखित संबंधों $R$ में से प्रत्येक एक तुल्यता संबंध है:

$R =\{(a, b): \mid a-b \mid, 4$ का एक गुणज है $\}$

Set $A=\{x \in Z: 0 \leq x \leq 12\}=\{0,1,2,3,4,5,6,7,8,9,10,11,12\}$

$R =\{( a , b ):| a - b | $ is a multiple of $4\}$

For any element, $a \in A$, we have $(a, a) \in R$ as $|a-a|=0$ is a multiple of $4.$

$\therefore R$ is reflexive.

Now, let $(a, b) \in R \Rightarrow|a-b|$ is a multiple of $4$

$\Rightarrow|-(a-b)|=|b-a|$ is a multiple of $4$

$\Rightarrow(b, a) \in R$

$\therefore R$ is symmetric.

Now, let $(a, b),\,(b, c) \in R$

$\Rightarrow|a-b|$ is a multiple of $4$ and $|b-c|$ is a multiple of $4$

$\Rightarrow(a-b)$ is a multiple of $4$ and $(b-c)$ is a multiple of $4$

$\Rightarrow(a-c)=(a-b)+(b-c)$ is a multiple of $4$

$\Rightarrow|a-c|$ is a multiple of $4$

$\Rightarrow(a, c) \in R$

$\therefore R$ is transitive.

Hence, $R$ is an equivalence relation.

The set of elements related to $1$ is $\{1,5,9\}$ as

$|1-1|=0$ is a multiple of $4$

$|5-1|=4$ is a multiple of $4$

$|9-1|=8$ is a multiple of $4$