A relation which is reflexive, symmetric and transitive is an Equivalence relation on set.Relation R, defined in a set A, is said to be an equivalence relation only on the following conditions:
(i) aRa for all a ∈ A, that is,R is reflexive.
(ii) aRb⇒bRa for all a, b ∈ AR, that is, is symmetric
(iii) aRb and bRc⇒aRc for all a, b, c ∈ A., that is R is transitive
The relation which is defined by “x is equal to y” in the set A of real numbers is called as an equivalence relation.
Solved example on equivalence relation on set:
1. A relation R is defined on the set Z by “a R b if a – b is divisible by 5” for a, b ∈ Z. Find out if R is an equivalence relation on Z.
Solution:
(i) Let a ∈ Z. Then a –a is divisible by 5. HenceaRawill hold for all a in Z and R is reflexive.
(ii) Let a, b ∈ Z and let aRb hold. Then a – b is divisible by 5 and therefore b – a is divisible by 5.
Thus, aRbequalsbRa and therefore R is symmetric.
(iii) Let a, b, c ∈ Z and let aRb, bRc both hold. Then a – b and b – c are both divisible by 5.
Therefore a – c = (a – b) + (b – c) is divisible by 5.
Thus, aRb and bRc equalsaRc and therefore R is transitive.
As R is reflexive, symmetric and transitive, R is an equivalence relation on Z.
2. Let m e a positive integer. R (Relation) is defined on the set Z by “aRb if and only if a – b is divisible by m” for a, b ∈ Z. Examine if R is an equivalence relation on set Z.
Solution:
(i) Let a ∈ Z. Then a – a = 0, which is divisible by m
Therefore, aRa holds for all a ∈ Z.
Hence, R is reflexive.
(ii) If a, b ∈ Z and aRb holds, then a – b is divisible by m and therefore,
b –ais also divisible by m.
Thus, aRb⇒bRa.
Hence, R is symmetric.
(iii) If a, b, c ∈ Z and aRb, bRc both hold, then a – b is divisible by m and b – c is also divisible by m. Therefore, a – c = (a – b) + (b – c) is divisible by m.
Thus, aRb and bRc⇒aRc
Therefore, R is transitive.
Hence as we have defined, R is reflexive, symmetric and transitive so, R is an equivalence relation on set Z.
