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We all have learned about fractions in our childhood and if we have then it is not unknown to us that every fraction has many equivalent forms. Let us take an example,

1/2, 2/4, 3/6, -1/-2, -3/-6, 15/30

The fractions given above may all look different from each other or maybe referred by different names but actually they are all equal and the same number.

This unique idea of classifying them together that “look different but are actually the same” is the fundamental idea of equivalence relations. Distribution of a set S is either a finite or infinite collection of a nonempty and mutually disjoint subset whose union is S.

A relation R on a set A can be considered as an equivalence relation only if the relation R will be reflexive, along with being symmetric, and transitive. But what does reflexive, symmetric, and transitive mean?

Reflexive: A relation is supposed to be reflexive, if (a, a) ∈ R, for every a ∈ A.

Symmetric: A relation is supposed to be symmetric, if (a, b) ∈ R, then (b, a) ∈ R.

Transitive: A relation is supposed to be transitive if (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R.

Question 1: Let us consider that F is a relation on the set R real numbers that are defined by xFy on a condition if x-y is an integer. Prove F as an equivalence relation on R.

Solution: Reflexive property: Assume that x belongs to R, and, x – x = 0 which is an integer. Thus, xFx.

Symmetric Property: Assume that x and y belongs to R and xFy. And x – y is an integer. Therefore, y – x = – ( x – y), y – x is too an integer. Thus, yFx.

Transitive Property: Assume that x and y belongs to R, xFy, and yFz. And both x-y and y-z are integers. So, according to the transitive property, ( x – y ) + ( y – z ) = x – z is also an integer. Therefore, xFz.

Hence, R is an equivalence relation on R.

Question 2: How do we know that the relation R is an equivalence relation in the set A = { 1, 2, 3, 4, 5 } given by the relation R = { (a, b):|a-b| is even }.

Solution : Here, R = { (a, b):|a-b| is even }. And a, b belongs to A

Reflexive Property : From the given relation,

|a – a| = | 0 |=0

And 0 is always even.

Thus, |a-a| is even

Therefore, (a, a) belongs to R

Hence R is Reflexive

Symmetric Property : From the given relation,

|a – b| = |b – a|

We know that |a – b| = |-(b – a)|= |b – a|

Hence |a – b| is even,

Then |b – a| is also even.

Therefore, if (a, b) ∈ R, then (b, a) belongs to R

Hence R is symmetric

Transitive Property : If |a-b| is even, then (a-b) is even.

In the same way, if |b-c| is even, then (b-c) is also even.

Sum of even number is also even

So, we can write it as a-b+ b-c is even

Then, a – c is also even

So,

|a – b| and |b – c| is even , then |a-c| is even.

Therefore, if (a, b) ∈ R and (b, c) ∈ R, then (a, c) also belongs to R

Hence R is transitive.

FAQ (Frequently Asked Questions)

1. What are the examples of equivalence relations?

It will be much easier if we try to understand equivalence relations in terms of the examples:

Example 1) “=” sign on a set of numbers. For example, 1/3 = 3/9

Example 2) In the triangles, we compare two triangles using terms like ‘is similar to’ and ‘is congruent to’.

Example 3) In integers, the relation of ‘is congruent to, modulo n’ shows equivalence.

Example 4) The image and the domain under a function, are the same and thus show a relation of equivalence.

Example 5) The cosines in the set of all the angles are the same.

Example 6) In a set, all the real has the same absolute value.

2. How can an equivalence relation be proved?

Given below are examples of an equivalence relation to proving the properties.

Let us consider that R is a relation on the set of ordered pairs that are positive integers such that ((a,b), (c,d))∈ Ron a condition that if ad=bc. Iso the question is if R is an equivalence relation?

To prove that R is an equivalence relation, we have to show that R is reflexive, symmetric, and transitive.

As par the reflexive property, if (a, a) ∈ R, for every a∈A

For every pair of positive integers,

((a,b),(a,b))∈ R.

Certainly, we can say

ab = ab for all the positive integers.

Therefore, the reflexive property is proved.

As par the symmetric property,

if (a, b) ∈ R, we can say that (b, a) ∈ R

As for the given condition,

if ((a,b),(c,d)) ∈ R, then ((c,d),(a,b)) ∈ R.

If ((a,b),(c,d))∈ R, then ad = bc and cb = da

Now just because the multiplication is commutative.

Therefore ((c,d),(a,b)) ∈ R

Consequently, the symmetric property is also proven.

As par the transitive property,

if (a, b) ∈ R and (b, c) ∈ R, then (a, c) too belongs to R

As for the given set of ordered pairs of positive integers,

((a,b), (c,d))∈ R and ((c,d), (e,f))∈ R,

then ((a,b),(e,f) ∈ R.

Now, consider that ((a,b), (c,d))∈ R and ((c,d), (e,f)) ∈ R.

Therefore, we get, ad = cb and cf = de.

The above relation suggest that a/b = c/d and that c/d = e/f,

so a/b = e/f we get af = be.

Therefore ((a,b),(e,f))∈ R.

Consequently, we have also proved transitive property.