Equivalence Classes: Solve for x in A

In summary, the given problem involves a relation R on the set A = {0, 1, 2, 3} defined by xRy if there exists an integer k such that x - y = 3k. The equivalence classes of R can be described by the sets E_x = {y ∈ A : y - x = 3k, k ∈ Z} for each value of x in A, where k must be an integer to satisfy the relation. This can be simplified to E_x = {x + 3k : k ∈ Z} for each element in A. Therefore, the four equivalence classes of R are {0, 3}, {1}, {2}, and {3}.
  • #1
mliuzzolino
58
0

Homework Statement



Let A be the set, A = {0, 1, 2, 3}.

Consider the relation R on A give by xRy iff there exists [itex] k \in \mathbb{Z} [/itex] such that x - y = 3k.

Describe the equivalence classes of R.

Homework Equations





The Attempt at a Solution



[itex] E_x = \{y \in A: y - x = 3k, k \in \mathbb{Z} \} [/itex]

For y = 0, k = -x/3

y = 1, k = (1-x)/3

y = 2, k = (2 - x)/3

y = 3, k = (3 - x)/4

Therefore there are four equivalence classes?

I have tried and tried to understand equivalence classes to no avail. I understand the analogies, but then when it transfers to numbers it's just completely lost on me. Even if what I've written above is correct, I just can't wrap my mind around what it means. I'm at one of those points where math makes you feel like the most worthless human to ever live.
 
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  • #2
mliuzzolino said:

Homework Statement



Let A be the set, A = {0, 1, 2, 3}.

Consider the relation R on A give by xRy iff there exists [itex] k \in \mathbb{Z} [/itex] such that x - y = 3k.

Describe the equivalence classes of R.

Homework Equations





The Attempt at a Solution



[itex] E_x = \{y \in A: y - x = 3k, k \in \mathbb{Z} \} [/itex]

For y = 0, k = -x/3

y = 1, k = (1-x)/3

y = 2, k = (2 - x)/3

y = 3, k = (3 - x)/4

Therefore there are four equivalence classes?

I have tried and tried to understand equivalence classes to no avail. I understand the analogies, but then when it transfers to numbers it's just completely lost on me. Even if what I've written above is correct, I just can't wrap my mind around what it means. I'm at one of those points where math makes you feel like the most worthless human to ever live.

math makes you feel like the most worthless human to ever live.

"Being a mathematician is a bit like being a manic depressive. You spend your life alternating between giddy elation and black despair" - Steven Krantz

Not every day is a good day. There will be some days where things may not seem to make sense or it may be hard to formulate reasoning, but tomorrow is another day.

As for your problem. What is it you don't understand? I can also tell you re-wrote the rule so you could get everything in terms of x's.
 
  • #3
I guess I am lost in multiple concepts.

R is a relation on A given by xRy. Given that A only contains 0, 1, 2, and 3, does that mean that x and y can only take on these values? The relation x - y = 3k defines how x is related to y?

So say y = 0. Then the relation is saying that x is related to 0 when k = -x/3? or should it be written in terms of k: x = -3k?

I guess I'm really confused right here. x can only take on the value of 0, 1, 2, or 3. So that means k must be 0, -1/3, -2/3, or -1 to satisfy the relation? I don't really understand what this is saying. I understand that equivalence classes delineate different classes of the relation (maybe?), but I'm just not seeing this delineation. All I see is a giant mess of meaningless relationships.

If y = 0, then k = -x/3. So k = 0, -1/3, -2/3, -1. (for x = 0, 1, 2, 3; respectively)
If y = 1, then k = (1 - x)/3. So k = 1/3, 0, -1/3, -2/3. (for x = 0, 1, 2, 3; respectively)
If y = 2, then k = (2 - x)/3. So k = 2/3, 1/3, 0, -1/3. (for x = 0, 1, 2, 3; respectively)
If y = 3, then k = (3 - x)/3. So k = 1, 2/3, 1/3, 0. (for x = 0, 1, 2, 3; respectively)

The Krantz quote is great, by the way.
 
  • #4
mliuzzolino said:
I guess I am lost in multiple concepts.

R is a relation on A given by xRy. Given that A only contains 0, 1, 2, and 3, does that mean that x and y can only take on these values? The relation x - y = 3k defines how x is related to y?

So say y = 0. Then the relation is saying that x is related to 0 when k = -x/3? or should it be written in terms of k: x = -3k?

I guess I'm really confused right here. x can only take on the value of 0, 1, 2, or 3. So that means k must be 0, -1/3, -2/3, or -1 to satisfy the relation? I don't really understand what this is saying. I understand that equivalence classes delineate different classes of the relation (maybe?), but I'm just not seeing this delineation. All I see is a giant mess of meaningless relationships.

If y = 0, then k = -x/3. So k = 0, -1/3, -2/3, -1. (for x = 0, 1, 2, 3; respectively)
If y = 1, then k = (1 - x)/3. So k = 1/3, 0, -1/3, -2/3. (for x = 0, 1, 2, 3; respectively)
If y = 2, then k = (2 - x)/3. So k = 2/3, 1/3, 0, -1/3. (for x = 0, 1, 2, 3; respectively)
If y = 3, then k = (3 - x)/3. So k = 1, 2/3, 1/3, 0. (for x = 0, 1, 2, 3; respectively)

The Krantz quote is great, by the way.

Just think really simple. k has to be an integer. Which elements of A are related to 0? You are writing way too much.
 
  • #5
Dick said:
Just think really simple. k has to be an integer. Which elements of A are related to 0? You are writing way too much.

So 0R0, 0R3; 1R1, and 2R2?
 
  • #6
mliuzzolino said:
I guess I am lost in multiple concepts.

R is a relation on A given by xRy. Given that A only contains 0, 1, 2, and 3, does that mean that x and y can only take on these values? The relation x - y = 3k defines how x is related to y?

So say y = 0. Then the relation is saying that x is related to 0 when k = -x/3? or should it be written in terms of k: x = -3k?

I guess I'm really confused right here. x can only take on the value of 0, 1, 2, or 3. So that means k must be 0, -1/3, -2/3, or -1 to satisfy the relation? I don't really understand what this is saying. I understand that equivalence classes delineate different classes of the relation (maybe?), but I'm just not seeing this delineation. All I see is a giant mess of meaningless relationships.

If y = 0, then k = -x/3. So k = 0, -1/3, -2/3, -1. (for x = 0, 1, 2, 3; respectively)
If y = 1, then k = (1 - x)/3. So k = 1/3, 0, -1/3, -2/3. (for x = 0, 1, 2, 3; respectively)
If y = 2, then k = (2 - x)/3. So k = 2/3, 1/3, 0, -1/3. (for x = 0, 1, 2, 3; respectively)
If y = 3, then k = (3 - x)/3. So k = 1, 2/3, 1/3, 0. (for x = 0, 1, 2, 3; respectively)

The Krantz quote is great, by the way.

Yeah it's a quote that's stuck with me and I never forget it because of what it allowed me to realize.

Wiki has a really great definition of equivalence classes that's easy to understand i'll post it here if it helps you any :

In mathematics, given a set ##X## and an equivalence relation R on X, the equivalence class of an element ##n \in X## is the subset of all elements in X which are equivalent to n.

Equivalence classes among elements of a structure are often used to produce a smaller structure whose elements are the classes, distilling a relationship every element of the class shares with at least one other element of another class.

This is known as modding out by the class. The class may assume the identity of one of the original elements, as when fractions are put in reduced form.

I highlighted the important parts. This may or may not help you, but it's here for reference.

Also note that the equivalence class of any element of your set is defined as :

$$[a] = \{ x \in X \space | \space aRx \}$$

So let's do one for 0 from your set A.

##[0] = \{ a \in A \space | \space 0Ra \}##

Does this help?
 
  • #7
mliuzzolino said:
So 0R0, 0R3; 1R1, and 2R2?

Yes. So what are the equivalence classes? There are three of them.
 

Related to Equivalence Classes: Solve for x in A

1. What are equivalence classes?

Equivalence classes are a mathematical concept used to group elements that have the same properties or characteristics. It is used to analyze and compare objects or elements based on their equivalence relations.

2. How do you solve for x in A?

To solve for x in A, you need to first determine the equivalence relation that defines A. Then, you can use this relation to determine the elements that are equivalent to x. Finally, you can solve for x by finding the values of the equivalent elements.

3. Can you give an example of solving for x in A?

Sure, let's say A is the set of even numbers. The equivalence relation for A would be "having the same parity." So, to solve for x in A, you would find the elements that have the same parity as x (in this case, even numbers) and determine their values. For example, if x = 4, then the equivalent elements would be 2, 6, 10, etc. Therefore, x can be solved as any even number.

4. What are the applications of equivalence classes?

Equivalence classes have various applications in mathematics, computer science, and other fields. They are used in graph theory, group theory, and set theory to analyze and compare objects or elements based on their properties. In computer science, equivalence classes are used in data structures and algorithms, such as partitioning and clustering algorithms.

5. How are equivalence classes different from normal sets?

Equivalence classes are different from normal sets because they are based on equivalence relations, which determine the similarity or equality between elements. In contrast, normal sets are based on membership, where elements either belong or do not belong to the set. Additionally, equivalence classes can contain elements that are not the same, but have the same properties, while normal sets only contain identical elements.

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