*This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com.This question has appeared in model paper 2012.*

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- A relation R in a set A is an equivalence relation if R is reflexive, symmetric and transitive.
- A relation R in a set A is called reflexive. if $(a,a) \in R\;for\; all\; a\in A$
- A relation R in a set A is called symmetric. if $(a_1,a_2) \in R\;\Rightarrow \; (a_2,a_1)\in R \;$ for $\;a_1,a_2 \in A$
- A relation R in a set A is called transitive. if $(a_1,a_2) \in\; R$ and $(a_2,a_3)\in R \Rightarrow \;(a_1,a_3)\in R\; $for all $\; a_1,a_2,a_3 \in A$
- $|a-b|=|b-a|$

Given a set $A=\{x \in Z:0 \leq x\leq 12\}$ and relation $R=\{(a,b):|a-b| \;is\;a\;multiple\; of \;4\}$:

For every $a \in A (a,a) \in R$, $|a-a|=0$, which is a multiple of 4. Therefore R is reflexive.

If $(a,b) \in R$ then $|a-b|$ is a multiple of 4. Similarly, if $(b,a) \in R$, then $|b-a|$ is also a multiple of 4.

Since $|a-b|=|b-a|$, if $(a,b) \in R$, then $(b,a) is also \in R$. Hence R is symmetric.

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

$\Rightarrow$ $|a-b|$ and $|b-c|$ are both multiples of 4.

For $(a,c) \in R$, |a-c|$ must be a multiple of 4

Since $a-c=a-b+b-c \rightarrow a-c$ is also a multiple of 4, as the sum of two elements that are multiple of a number is also a multiple of the same number.

Therefore R is transitive

Hence $R$ is an equivalence relation as its reflexive, symmetric and transitive.

Given a set $A=\{x \in Z:0 \leq x\leq 12\}$, the set of elements in $A$ is $\{0,1,2,3,4,5,6,7,8,9,10,11,12\}$

Among these elements the ones that are related to 1 are those that satisfy the condition $|a-1|$ is a multiple of 4.

Substituing for the elements in $A$, we find that the set of elements related to 1 is {1,5,9} because:

$|(1-1|=0$ multiple of 4

$|(5-1)|=4$ multiple of 4

$|(9-1)|=8$ multiple of 4

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