Show that each of the relation R in the set

Solution:

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

(i) $\mathrm{R}=\{(a, b):|a-b|$ 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 .

∴R is reflexive.

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

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

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

$\Rightarrow(b, a) \in \mathrm{R}$

∴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 .

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

∴ R is transitive.

Hence, R is an equivalence relation.

The set of elements related to 1 is {1, 5, 9} since

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

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

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

(ii) $R=\{(a, b): a=b\}$

For any element $a \in A$, we have $(a, a) \in R$, since $a=a .$

∴R is reflexive.

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

$\Rightarrow a=b$

$\Rightarrow b=a$

$\Rightarrow(b, a) \in \mathrm{R}$

∴R is symmetric.

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

$\Rightarrow a=b$ and $b=c$

$\Rightarrow a=c$

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

∴ R is transitive.

Hence, R is an equivalence relation.

The elements in R that are related to 1 will be those elements from set A which are equal to 1.

Hence, the set of elements related to 1 is {1}.