Equivalence Relation:

A relation R in a set A is known as an equivalence relation if

(i) R is reflexive i.e., (a, a) ∈ R, ∀ a ∈ A       

(ii) R is symmetric i.e., (a, b) ∈ R => (b, a) ∈ R

(iii) R is transitive  i.e., (a, b), (b, c) ∈ R => (a, c) ∈ R

The equivalence relation is normally shown by the symbol ~.

 Equivalence Classes of an Equivalence Relation:

 Consider R be equivalence relation in A (≠ Φ). Let a ∈ A.

Then the equivalence class of a shown by [a] or {} is described as the set of all those points of A which are belongs to a under the relation R. Therefore [a] = {x : x ∈ A, x R a}

It is simple to see that

(i) b ∈ [a] => a ∈ [b]

(ii) b ∈ [a] => [a] = [b]

(iii) Two equivalence classes are either identical or disjoint as an example we take a very important relation

x ≡ y (mod n) iff n divides (x -y),  is fixed positive number. Let n = 5 then

[0] = {x : x ≡ 0(mod 5)} = {5p : p ∈ z} = {0, ±5, ±10, ±15,....}

[1] = {x : x ≡ 1(mod 5)} = {x : x -1 = 5k, k ∈ z} = {5k + 1: k ∈ z} = {1, 6, 11, ...., -4, -9,....}

one may easily see that there are only 5 different equivalence classes viz. [0], [1], [2], [3] and [4] when n = 5.

Problem: N is the set of natural integer numbers. The relation R is described on N x N as follows:

                        (a, b) R (c, d) <=> a + d = b + c

                        Show that R is equivalence relation.

Solution:       (i) (a, b) R (a, b) <=> a + b = b + a

                        ∴ R is reflexive.

                        (ii) (a, b) R (c, d) => a + d = b + c

                        Þ c + b = d + a

                        Þ (c, d) R (a, b)

                        ∴ R is symmetric.

                        Now (iii) (a, b) R (c, d) and (c, d) R (e, f) => a + d = b + c & c + f = d + e

                        => a + d + c + f = b + c + d + e

                        => a + f = b + e => (a, b) R (e, f)

                         R is transitive. Thus R is an equivalence relation on N x N.

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