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Show that the relation S in the set A= \left \{ x\: \epsilon\: Z:0\leq x\leq 12 \right \}  given by S= \left \{ \left ( a,b \right ):a,b\, \epsilon \, Z,\left | a-b \right | \right  is divisible by 3} is an equivalence relation.

 

 

 

 
 
 
 
 

Answers (1)
R Ravindra Pindel

Reflexive
      S= \left \{ \left ( a,b \right ):a,b\, \epsilon A,\left | a-b \right |is \; divisible\; by\; 3 \right \}
    is defined on the set A= \left \{ x\, \epsilon \, z:0\leq x\leq 12 \right \}
     clearly, S is reflexive as 3 divides \left | a-a \right |= 0
                                                                  \psi a\, \epsilon \, A
symmetric
        further if \left ( a,b \right )\epsilon \, S  then 3 divides \left | a-b \right |.  therefore 3 divides \left | -\left ( b-a \right ) \right |.= \left | b-a \right |   as well
        Hence \left ( b,a \right )\, \epsilon S  which follows that S is symmetric
Transitive
             similarly, if \left ( a,b \right )\, \epsilon S  and \left ( b,c \right )\, \epsilon S  then \left | a-b \right |  and \left | b-c \right |   are both divisible by 3.
that is \left | a-b \right |= 3m     i.e a-b= \pm 3m
and    \left | b-c \right |= 3n    i.e  b-c= \pm 3n
so \left ( a-b \right )+\left ( b-c \right )= a-c= \pm 3\left ( m+n \right )
    \Rightarrow \left | a-c \right |= 3\left ( m+n \right )
\therefore \left | a-c \right |  is divisible by 3
This shows that S is transitive as \left ( a,c \right )\, \epsilon \, S
Then S is an equivalence relation in the set A
       

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