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Let $A= \left \{ x\, \epsilon \, Z:0\leq x\leq 12 \right \}\cdot$ Show that $R= \left \{ \left ( a,b \right ):a,b\, \epsilon \, A,\left | a-b \, is \: divisible\: by\: 4 \right | \right \}$ is an equivalence relation.
Find the set of all elements related to 1. Also write the equivalence
class [2].

Answers (1)

given $R= \left \{ \left ( a,b \right ):a,b\, \epsilon \, A,\left | a-b \, is \: divisible\: by\: 4 \right | \right \}$
Reflexivily: For any $a\, \epsilon \, A$
$\left | a-a \right |= 0$ which is divisible by 4
$\left ( a,Q \right )\epsilon R\; so\:$ R is reflexive
Symmetricly: Let $\left ( a,b \right )\epsilon\, R$
$\Rightarrow \left | a-b \right |$ is divisible by 4
$\Rightarrow \left | b-a \right |$ is divisible by 4 $\left [ \because \left | a-b \right | = \left | b-a \right |\right ]$
$\Rightarrow \left ( b,a \right )\, \epsilon \, R$
So, R is symmetric.
Transitive: Let $\left ( a,b \right )\, \epsilon \, R$ and $\left ( b,c \right )\, \epsilon \, R$
$\Rightarrow \left | a-b \right |$ is divisible by 4
$\Rightarrow \left | a-b \right |$ = 4 k
$\therefore a-b= \pm 4k,k\,\, \epsilon \, z---(i)$
Also $\left | b-c \right |$ is divisible by 4
$\Rightarrow \left | b-c \right |= 4m$
$\therefore$ $\therefore b-c= \pm 4m.m\, \epsilon\, z---(ii)$
Adding equations (i) and (ii)
$a-b+b-c= \pm 4\left ( k+m \right )$
$\Rightarrow a-c= \pm 4\left ( k+m \right )$
$\left | a-c \right |$ is divisible by 4
$\Rightarrow \left ( a,c \right )\, \epsilon \, R$
so R is symmetric
$\Rightarrow$ R is reflexive, symmetric and transitive
$\therefore$ R is an equivalence relation
Let x be an element of R such that $\left ( x,1 \right )\, \epsilon \, R$
Then $\left | x-1 \right |$ is divisible by 4
$x-1= 0,4,8,12$
$\Rightarrow -x= 4,5,9\, \left ( \because x\leq 12 \right )$
$\therefore$ set of all elements of A which are related to {1,5,9}.
The equivalence class of 2 ie
$\left [ 2 \right ]= \left \{ \left ( 2,2 \right ),a\, \epsilon A,\left | a-2 \right |\, is\, divisible \, by\, 4 \right \}$
$\Rightarrow \left | a-2 \right |= 4k\left ( K\, is\, a\, whole \, number,\, k\geq 3 \right )$
$\Rightarrow a= 2,6,10$
$\therefore$ equivalance class [2] is {2,6,10}