Equivalence Relation

A relation R on a set A is said to be an equivalence relation if and only if the relation R is reflexive, symmetric and transitive.

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

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

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

Equivalence Relation Proof

Here is an equivalence relation example to prove the properties.

Let us assume that R be a relation on the set of ordered pairs of positive integers such that ((a,b), (c,d))∈ R if and only if ad=bc. Is R an equivalence relation?

In order to prove that R is an equivalence relation, we must show that R is reflexive, symmetric and transitive.

Proof

The Proof for the given condition is given below :

Reflexive

According to the reflexive property, if (a, a) ∈ R, for every a∈A

For all pairs of positive integers,

((a,b),(a,b))∈ R.

Clearly, we can say

ab = ab for all positive integers.

Hence reflexive property is proved.

Symmetric

From the symmetric property,

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

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

since multiplication is commutative.

Therefore ((c,d),(a,b)) ∈ R

Hence symmetric property is proved.

Transitive

From the transitive property,

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

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

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

The above relation implies 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.

Hence transitive property is proved.

Sample Problem

Go through the equivalence relation examples and solutions provided here

Question 1 :

Let assume that R be a relation on the set \(\mathbb{R}\) real numbers defined by xRy if and only if x-y is an integer. Prove that R is an equivalence relation on \(\mathbb{R}\).

Solution :

Reflexive: Consider x belongs to \(\mathbb{R}\),then x – x = 0 which is an integer. Therefore xRx.

Symmetric: Consider x and y belongs to \(\mathbb{R}\) and xRy. Then x – y is an integer. Thus, y – x = – ( x – y), y – x is also an integer. Therefore yRx.

Transitive: Consider x and y belongs to \(\mathbb{R}\), xRy and yRz. Therefore x-y and y-z are integers. According to the transitive property, ( x – y ) + ( y – z ) = x – z is also an integer. So that xRz.

Thus, R is an equivalence relation on \(\mathbb{R}\).

Question 2:

Show 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 :

R = { (a, b):|a-b| is even }. Where 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.

Similarly, 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.

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