What are Equivalence Relations?
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 to 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 relation. 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.
Solved Examples of Equivalence Relation
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.
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.
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 }.
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.
Connection of Equivalence Relation to other relations
An incomplete order is a reciprocal, system can be classified, and linear relation.
Equality is a complete order as well as an equivalence relation.
Equality is also the only inductive, symmetric, and antisymmetric relation on a set.
Equal variables in algebraic expressions can be replaced for one another, a feature not accessible for equivalence-related variables.
Persons inside an equivalence relation's equivalence classes can replace each other, but not people within a class.
A rigorous incomplete order is asymmetric, irreflexive, and bidirectional.
A complete equivalence connection is equal and linear. Such a relationship is bidirectional if and only if it is serial, that is, if for every display style, there occurs some display style text such that asim b. As a result, an equivalence relation can be defined in a variety of ways, including symmetric, transitive, and serial.
The ternary equivalent of the normal equivalence relation is the ternary comparability relation.
A reliance relation or a tolerance relation is a reciprocal and symmetrical relation.
A sequence is both bidirectional and inductive.
An equivalence relation whose domain X is also the bottom set for an algebraic form and which satisfies the extra form is known as a congruence relation. In general, congruence relations serve as homomorphism kernels, and a structure's quotient can be created using a congruence relation. Congruence relations have an alternate representation as structural components of the form on which they are established in many crucial circumstances.
Although the opposite assertion holds exclusively in classical mathematics because it is identical to the law of excluded middle, any equivalence connection is the negative of an apartness relationship.
Every recursive and left or right Riemann connection is also an interval estimate.
A Few key points to remember
i) Equations with similar solutions or bases are known as equivalent equations.
ii) An analogous equation is created by adding or subtracting the identical number or phrase to both sides of an equation.
iii) An analogous equation is created by multiplying or dividing both sides of an equation by the same non-zero value.
Conclusion
The primary focus lies in conceptual understanding and one who has mastered that art is sure to succeed. Practice sums after going through the concept for a better understanding of the topic. Equivalence relations can be a tricky affair if not practiced again and again.
FAQs on Equivalence Relation
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. So, 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.
The Proof for the Following Condition is Given Below
Reflexive Property
As per 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.
Symmetric Property
As per 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.
Transitive Property
As per 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.
3. What is group theory?
Equivalence relations are rooted in divided collections, which are groups closed over bijections that maintain division structure, much as order relations are based in sorted sets, sets closed over bilateral supremum and infimum. Because all such bijections translate an equality class onto themselves, they're also called permutations. As a result, permutation groups, also known as transformation groups and the associated concept of orbit, illuminate the mathematical structure of equivalence relations. Let " be a non-empty set A and an equivalence relation across the planet or bottom group. The above theorems are valid if this is the scenario.
i) A is divided into equivalence classes. This is the above-mentioned Fundamental Theorem of Equivalence Relations.)
ii) Given a partition of A, G is a composition group whose orbits are the partition's cells;
Given a partition of A, G is a composition group whose orbits are the partition's cells;
iii) Generally, provided an equivalence relation across A, a changing group G over A exists, where orbits are the equivalence categories of A.
4. Elaborate about equivalence relations and mathematical logic.
Equivalence relations provide a wealth of examples and counterexamples. A simple example of a theory that is categorical but not categorical for any bigger ordinal value is an equivalency connection with precisely two indefinite similarity categories. The model theory implies that the characteristics characterizing a relation can be demonstrated distinct from each other and therefore necessary elements of the definition if and only if instances of relations not matching the given resource while fulfilling all the other characteristics can be discovered for each value.
5. Explain about Algebraic structure.
In mathematics, An algebraic structure is made up of a nonempty group A, also known as the underlying set, carrier set, or domain, a set of bounded functions on A that are usually binary operations, and a limited set of assumptions that these functions must fulfill. The research of analogies and order connections underpins much of mathematics. The mathematical framework of order relations is captured by lattice theory. Despite the fact that equivalence relations are as common in mathematics as order relations, their algebraic structure is less well understood. The former structure is based mostly on group theory and, to a smaller extent, lattice, category, and groupoids theory.