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# Show that each of the relation R in the set A = {x belongs to Z : 0 is less than or equal to x is less than or equal to 12}, given by R = {(a, b) : |a – b| is a multiple of 4} is an equivalence relation. Find the set of all elements related to 1 in each

Q.9 Show that each of the relation R in the set $A = \{x \in Z : 0 \leq x \leq 12\}$, given by

(i) $R = \{(a, b) : |a - b|\; is\;a\;multiple \;of\; 4\}$ is an equivalence relation. Find the set of all elements related to 1 in each case.

Views

$A = \{x \in Z : 0 \leq x \leq 12\}$

$A=\left \{ 0,1,2,3,4,5,6,7,8,9,10,11,12 \right \}$

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

For $a\in A$ , $(a,a)\in R$  as $\left | a-a \right |=0$ which is multiple of 4.

Henec, it is reflexive.

Let, $(a,b)\in R$ i.e. $\left | a-b \right |$ is multiple of 4.

then  $\left | b-a \right |$ is also multiple of 4 because $\left | a-b \right |$  =  $\left | b-a \right |$  i.e.$(b,a)\in R$

Hence, it is symmetric.

Let, $(a,b)\in R$ i.e. $\left | a-b \right |$ is multiple of 4   and    $(b,c)\in R$  i.e. $\left | b-c \right |$ is multiple of 4 .

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

$(a-c)=(a-b)+(b-c)$  is multiple of 4

$\left | a-c \right |$ is multiple of 4 i.e. $(a,c)\in R$

Hence, it is transitive.

Thus, it is reflexive,symmetric and transitive i.e. it is an equivalence relation.

The set of all elements related to 1 is  $\left \{1,5,9 \right \}$

$\left | 1-1 \right |=0$ is multiple of 4.

$\left | 5-1 \right |=4$ is multiple of 4.

$\left | 9-1 \right |=8$ is multiple of 4.

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