Show that the relation R in the set Z of integers given by R = {( a, b ) : divides a – b } is an equivalence relation. Solution R is reflexive, as divides ( a – a ) for all a ∈ Z . Further, if ( a, b ) ∈ R, then divides a – b . Therefore, divides b – a .
Hence, ( b , a ) ∈ R, which shows that R is symmetric. Similarly, if ( a , b ) ∈ R and ( b , c ) ∈ R, then a – b and b – c are divisible by . Now, a – c = ( a – b ) + ( b – c ) is even (Why?). So, ( a – c ) is divisible by .
This shows that R is transitive. Thus, R is an equivalence relation in Z . Fig . In Example , note that all even integers are related to zero, as ( , ± ), ( , ± ) etc., lie in R and no odd integer is related to , as ( , ± ), ( , ± ) etc., do not lie in R.
Similarly, all odd integers are related to one and no even integer is related to one. Therefore, the set E of all even integers and the set O of all odd integers are subsets of Z satisfying following conditions: (i) All elements of E are related to each other and all elements of O are related to each other. (ii) No element of E is related to any element of O and vice-versa. (iii) E and O are disjoint and Z = E ∪ O.
The subset E is called the equivalence class containing zero and is denoted by [ ]. Similarly, O is the equivalence class containing and is denoted by [ ]. Note that [ ] ≠ [ ], [ ] = [ r ] and [ ] = [ r + ], r ∈ Z . Infact, what we have seen above is true for an arbitrary equivalence relation