Equivalence Relation: Overview, Questions, Preparation

Equivalence relation, in mathematics, is a binary relation that is symmetric, transitive and reflexive. Therefore, the relation R on the set A will be termed as the equivalence relation, only when the relation R is transitive, reflexive, and symmetric. 

Proof of equivalence relation

Let R be a relation on a set, let's say, ordered pairs of positive integers in such that- (a, b), (c, d) ∈ R only when ad= bc.
The proof is given below:

Reflexive : 
According to the property:
When (a, a) ∈ R for each ∈ A
For every pair of positive integers- ((a, b), (a, b)) ∈ R. Thus, we can see that ab = ab for every positive integer. Thus, the reflexive property proved. 

Symmetric Property
According the property :
When (a, b) ∈ R then, (b, a) ∈ R
Therefore,  when ((a, b), (c, d)) ∈ R in that case, ((c,d), (a, b)) ∈ R
When ((a, b),(c, d))∈ R in that case cb = da and ad = bc. 
Now, ((c, d),(a, b)) ∈ R. 
Therefore, the symmetric property is proven. 

Transitive property:
According to the transitive property :
(a, b) ∈ R and (b, c) ∈ R, then (a, c) belongs to R. 
Therefore, ((a, b), (c, d))∈ R
((c, d), (e, f))∈ R in that case,  ((a, b),(e, f) ∈ R.
Now, let's assume ((c, d), (e, f)) ∈ R and ((a, b), (c, d))∈ R. 
Therefore, ad = cb and cf = de. 
Now, a/b = c/d similarly c/d = e/f. Therefore, a/b = e/f thus, we'll get, af = be.  Thus, ((a, b),(e, f))∈ R. Therefore, transitive property proved.

Details about the topic
In maths, relation and function are one of the most important concepts in classes 11 and 12. These topics are very important in fields of economics, engineering, etc. 

Illustrations

Find if the given relations is reflexive, symmetric, and transitive: Relation R in the set A = {1, 2, 3...13, 14} defined as
R = {(x, y): 3x − y = 0}
Solution: A = {1, 2, 3 ... 13, 14}
 R = {(x, y): 3x − y = 0}

Thus, R = {(1, 3), (2, 6), (3, 9), (4, 12)}
R isn't reflexive
Also, R is not symmetric as (1, 3) ∈ R, but (3, 1) ∉ R. Thus, R is also not an equivalence relation. 

Is the following relation symmetric, reflexive or transitive? Relation R in the set N of natural numbers:
R = {(x, y): y = x + 5 and x   R = {(x, y): y = x + 5 and x

It is evident that (1, 1) ∉ R.

Thus, R is not reflexive.

Now, (1, 6) ∈ R but (1, 6) ∉ R.

Thus, R isn't symmetric.

Because there isn't a pair in R such that (x, y) and (y, z) ∈ R, then (x, z) can't belong to R.
Thus, R isn't reflexive, symmetric, or transitive.

Is the given relation symmetric, reflexive or transitive? Relation R in the set A = {1, 2, 3, 4, 5, 6} as

 R = {(x, y): y is divisible by x}
A = {1, 2, 3, 4, 5, 6}

R = {(x, y): y is divisible by x}
Thus, (x, x) ∈ R therefore, R is reflexive.

Frequently Asked Questions

1. What do you mean by equivalence relation? 
The relation R on the set A is known as the equivalence relation, only when the relation R is transitive, reflexive, and symmetric.

2. What are the three types of properties of equivalence relation? 
Reflexive property 
Transitive property 
Symmetric Property 

3. Is it possible to say that an empty relation is equal to an equivalence relation? 
Yes, an empty relation can indeed be equal to the equivalence relation.  

4. Is every relation an equivalence relation? 
No, every relation is not an equivalence relation but every function is an equivalence relation. 

5. What is the symmetric property? 
Any relationship will be called Symmetric only when (a, b) ∈ R, in that case (b, a) ∈ R.