Equivalence relation on set is a relation which is reflexive, symmetric and transitive.
A relation R, defined in a set A, is said to be an equivalence relation if and only if
(i) R is reflexive, that is, aRa for all a ∈ A.
(ii) R is symmetric, that is, aRb ⇒ bRa for all a, b ∈ A.
(iii) R is transitive, that is aRb and bRc ⇒ aRc for all a, b, c ∈ A.
The relation defined by “x is equal to y” in the set A of real numbers is an equivalence relation.
Let A be a set of triangles in a plane. The relation R is defined as “x is similar to y, x, y ∈ A”.
We see
that R is;
(i) Reflexive, for, every triangle is similar to itself.
(ii) Symmetric, for, if x be similar to y, then y is also similar to x.
(iii) Transitive, for, if x be similar to y and y be similar to z, then x is also similar to z.
Hence R is an equivalence relation.
A relation R in a set S is called a partial order relation if it satisfies the following conditions:
(i) aRa for all a∈ A, [Reflexivity]
(ii) aRb and bRa ⇒ a = b, [Anti-symmetry]
(iii) aRb and bRc ⇒ aRc, [Transitivity]
In the set of natural numbers, the relation R defined by “aRb if a divides b” is a partial order relation, since here R is reflexive, anti-symmetric and transitive.
A set, in which a partial order relation is defined, is called a partially ordered set or a poset.
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. Examine if R is an equivalence relation on Z.
Solution:
(i) Let a ∈ Z. Then a – a is divisible by 5. Therefore aRa holds for all a in Z and R is reflexive.
(ii) Let a, b ∈ Z and aRb hold. Then a – b is divisible by 5 and therefore b – a is divisible by 5.
Thus, aRb ⇒ bRa and therefore R is symmetric.
(iii) Let a, b, c ∈ Z and 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 ⇒ aRc and therefore R is transitive.
Since R is reflexive, symmetric and transitive so, R is an equivalence relation on Z.
2. Let m e a positive integer. A relation R is defined on the set Z by “aRb if and only if a – b is divisible by m” for a, b ∈ Z. Show that 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) Let a, b ∈ Z and aRb holds. Then a – b is divisible by m and therefore, b – a is also divisible by m.
Thus, aRb ⇒ bRa.
Hence, R is symmetric.
(iii) Let 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.
Since, R is reflexive, symmetric and transitive so, R is an equivalence relation on set Z
3. Let S be the set of all lines in 3 dimensional space. A relation ρ is defined on S by “lρm if and only if l lies on the plane of m” for l, m ∈ S.
Examine if ρ is (i) reflexive, (ii) symmetric, (iii) transitive
Solution:
(i) Reflexive: Let l ∈ S. Then l is coplanar with itself.
Therefore, lρl holds for all l in S.
Hence, ρ is reflexive
(ii) Symmetric: Let l, m ∈ S and lρm holds. Then l lies on the plane of m.
Therefore, m lies on the plane of l. Thus, lρm ⇒ mρl and therefore ρ is symmetric.
(iii) Transitive: Let l, m, p ∈ S and lρm, mρp both hold. Then l lies on the plane of m and m lies on the plane of p. This does not always implies that l lies on the plane of p.
That is, lρm and mρp do not necessarily imply lρp.
Therefore, ρ is not transitive.
Since, R is reflexive and symmetric but not transitive so, R is not an equivalence relation on set Z
● Set Theory
● Sets
● Subset
● Practice Test on Sets and Subsets
● Problems on Operation on Sets
● Practice Test on Operations on Sets
● Venn Diagrams in Different Situations
● Relationship in Sets using Venn Diagram
● Practice Test on Venn Diagrams
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