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Interactive Audio Lesson
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Union of Equivalence Relations
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Today, we are going to explore the union of two equivalence relations. Who can remind me what properties we need a relation to retain in order to be classified as an equivalence relation?
It needs to be reflexive, symmetric, and transitive.
Exactly! Now let's examine if the union of two equivalence relations meets these criteria. Can someone explain why it remains reflexive?
Because if both relations are reflexive, then the pairs like (a, a), (b, b), and (c, c) will be in the union.
Well said! Now, how about symmetry? Does the union maintain that property too?
Yes, if (a, b) is in the union, then (b, a) must also be in one of the original relations.
Correct! However, what happens with transitivity? Can someone provide an example?
We could have (a, b) in R1 and (b, c) in R2, but (a, c) might not be in the union, so it's not necessarily transitive.
Great example! So, the union is reflexive and symmetric, but not guaranteed to be transitive. Let's summarize: the union of two equivalence relations retains reflexivity and symmetry but not always transitivity.
Intersection of Equivalence Relations
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Now, let's switch gears and discuss the intersection of two equivalence relations. Does anyone have an idea if the intersection also forms an equivalence relation?
It seems like it would because each relation is reflexive.
That's right! Let’s prove it maintains reflexivity first. If both R1 and R2 have (a, a), what can we say about (a, a) in the intersection?
It would be included in the intersection since it's present in both R1 and R2.
Exactly! And what about symmetry? How can we show that the intersection is symmetric?
If (a, b) is in the intersection, then it has to be in both R1 and R2, and since those are symmetric, (b, a) will also be in the intersection.
You all are getting the hang of it! Lastly, what can we say about transitivity?
If (a, b) and (b, c) are in the intersection, they are also in both relations, so (a, c) must be there too.
Exactly! Thus, we conclude that the intersection of two equivalence relations is always an equivalence relation. Great teamwork!
If and Only If Statement
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Let’s explore the statement regarding the union and composition of two equivalence relations - specifically when is the union an equivalence relation? What do you think?
It must be true if the composition equals the union.
Correct! If the composition equals the union, we can show transitivity. Can anyone help explain how we prove this?
We need to consider cases where the pairs belong to either relation, right?
Exactly! What happens when both pairs are from one relation?
In that case, we can apply the transitivity of that relation directly.
And if they come from both relations?
We can use the composition definition to show that if (a, b) is in one and (b, c) is in the other, then (a, c) must be in the union as well.
Precisely! So our key takeaway is: the union is an equivalence relation if and only if the composition equals the union. Let’s remember this important implication!
Number of Equivalence Relations
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Finally, let’s talk about how we can count equivalence relations. Have we discussed the function P(n) before?
Yes, it's related to the number of partitions of a set.
Exactly! So how do we derive P(n) recursively?
We consider the first element and decide how many others to include in its subset.
Correct! Can anyone express how choosing 'j' other elements leads us to the formula?
We select j from the remaining elements and then partition the rest, so it contributes to the total count.
Perfect! So how does that help us create the relation between the sets?
Combining all those ways of arranging the elements gives us P(n) = Σ C(n - 1, j) * P(n - j - 1).
Well said! So through this function, we can quantify equivalence relations effectively. Remember this formula for future reference.
Introduction & Overview
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Quick Overview
Standard
In this section, we examine the properties of equivalence relations, particularly how the union of two equivalence relations may fail to be an equivalence relation while the intersection always remains one. Several key concepts such as reflexivity, symmetry, and transitivity are highlighted with examples and counterexamples.
Detailed
Counterexamples and Properties of Equivalence Relations
This section delves into two fundamental properties of equivalence relations: the union and intersection of two equivalence relations. It begins by discussing the nature of equivalence relations, emphasizing that they must meet three criteria: reflexivity, symmetry, and transitivity.
Key Points Covered:
- Union of Equivalence Relations: The union of two equivalence relations is always reflexive and symmetric but may not be transitive. A counterexample is provided using a set X = {a, b, c} with two relations:
- R1 = {(a, a), (b, b), (c, c), (a, b), (b, a)}
- R2 = {(a, a), (b, b), (c, c), (b, c), (c, b)}
This shows that while R1 ∪ R2 retains reflexivity and symmetry, it fails to be transitive. - Intersection of Equivalence Relations: It is proven that the intersection of two equivalence relations is always an equivalence relation, maintaining all three properties.
- If and Only If Statement: A thorough proof is presented showing that the union of two equivalence relations is an equivalence relation if and only if the composition of the two relations equals their union.
- Number of Equivalence Relations: The number of equivalence relations can also be defined recursively with a function P(n) related to the partitions of sets.
This section provides a robust foundation for understanding the intricacies of equivalence relations in discrete mathematics.
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Union of Equivalence Relations
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This question you are given two equivalence relations on a non-empty set X. You are asked to prove or disprove whether R1 ∪ R2 is an equivalence relation or not. So, it turns out at a union of two equivalence relations need not be an equivalence relation. And this is demonstrated by this counter example. So, I consider my set X={a,b,c}. And let me have equivalence relations R1 and R2. So, R1 = {(a,a),(b,b),(c,c),(a,b),(b,a)} and R2 = {(a,a),(b,b),(c,c),(b,c),(c,b)}. It is easy to see that both of them are equivalence relations over the set X they are reflexive. Each of them is a reflexive relations symmetric relation and a transitive relation.
Detailed Explanation
When we consider the union of two equivalence relations R1 and R2, it might not be an equivalence relation itself. Let's take the set X={a,b,c} and define R1 = {(a,a),(b,b),(c,c),(a,b),(b,a)} and R2 = {(a,a),(b,b),(c,c),(b,c),(c,b)}. Both R1 and R2 satisfy the properties of equivalence relations: reflexive, symmetric, and transitive. The reflexive property is satisfied because each element is related to itself, and the symmetric property is satisfied because if (x,y) is in the relation, then (y,x) is also included. However, when we look at the union R1 ∪ R2, this union is reflexive and symmetric but fails to be transitive. For instance, (a,b) is in R1 and (b,c) is in R2, but (a,c) is not in R1 ∪ R2, hence violating transitivity.
Examples & Analogies
Think of two different clubs, each with its own set of members and a defined way to relate to each other. In Club A, every member is friends with each other, while Club B has some cross-friendships. If we consider all the friendships (the union of two clubs), people might know that they are friends with some but are not directly connected through the shared friendships. Thus, while you know some members are friends through one club or the other, not every friend relationship holds when combining both clubs.
Intersection of Equivalence Relations
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In part b of question 1, we are supposed to prove whether the intersection is an equivalence relation or not. And it turns out that intersection of two equivalence relations over the same set X is always an equivalence relation. So, let us prove the three required properties. If I take R1 and R2 to be equivalence relations over the set X, then since R1 and R2 are individually reflexive relations, you will have ordered pairs of the form (a,a) present in both R1 as well as in R2. And as a result, you will have ordered pairs of the form (a, a) present in the intersection of R1 and R2 as well. That shows that R1 ∩ R2 is a reflexive relation.
Detailed Explanation
When considering the intersection of two equivalence relations, R1 ∩ R2, it will always remain an equivalence relation. Let's verify this by examining the required properties: reflexive, symmetric, and transitive. First, for reflexivity: since both R1 and R2 have (a,a) for any element a in set X, their intersection must also have (a,a). Thus, the intersection is reflexive. For symmetry: if (a,b) exists in the intersection, then it exists in R1 and R2. Since both relations are symmetric, (b,a) must also be included in both R1 and R2, hence in the intersection. Lastly, for transitivity: if (a,b) and (b,c) are in the intersection, they are in both R1 and R2, and thus (a,c) would also be in both, ensuring transitivity is upheld. Therefore, R1 ∩ R2 is indeed an equivalence relation.
Examples & Analogies
Consider two overlapping groups of friends. If one group represents people who are friends through work and another through school, the intersection of these two groups consists of friends who are recognized through both environments. As a result, they share mutual friendships that retain properties like knowing each other (reflexivity), their friendships are mutual (symmetry), and friends of friends are still friends (transitivity), ensuring that the intersection forms a cohesive network.
Condition for Union of Relations
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Now on question number 2 again, you are given two equivalence relations over a non-empty set and we want to prove that R1 ∪ R2 will be an equivalence relation if and only if the composition R1 ∘ R2 is equal to the R1 ∪ R2. This is, an if and only if statement. So, we have to give two proofs. We have to prove the implication in both the directions.
Detailed Explanation
In this section, we are tasked with proving an important condition regarding the union of equivalence relations and their composition. Specifically, we need to show that the union, R1 ∪ R2, is an equivalence relation if and only if the composition R1 ∘ R2 equals R1 ∪ R2. First, we assume R1 ∘ R2 = R1 ∪ R2 and show R1 ∪ R2 is transitive. To establish this, consider pairs (a,b) and (b,c) from the union. There are several cases depending on which relation they belong to, and by verifying through these cases, we can show that (a,c) must also exist in the union. The reverse implication requires us to prove that if R1 ∪ R2 is an equivalence relation, then R1 ∘ R2 must equal R1 ∪ R2. This is done by showing that any relation in the composition also belongs to the union and vice versa.
Examples & Analogies
Imagine two local transit systems in a city that connect major points. If the routes of both systems allow for seamless transfers, we can say the union of their routes can serve every destination (R1 ∪ R2 is a complete relation). For simplicity’s sake, if you can go from point A to point C by transferring at point B, this interconnected route reflects the essence of transitive relationships in the context of our set relations.
Definitions & Key Concepts
Learn essential terms and foundational ideas that form the basis of the topic.
Key Concepts
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Union of Equivalence Relations: The union of two equivalence relations is always reflexive and symmetric but not necessarily transitive.
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Intersection of Equivalence Relations: The intersection of two equivalence relations over the same set is always an equivalence relation.
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If and Only If Statement: The union of two equivalence relations is an equivalence relation if and only if the composition of those relations equals their union.
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Counting Equivalence Relations: The recursive function P(n) quantifies the number of equivalence relations based on partitions of sets.
Examples & Real-Life Applications
See how the concepts apply in real-world scenarios to understand their practical implications.
Examples
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Given two equivalence relations on set X, R1 = {(a, a), (b, b), (c, c), (a, b), (b, a)} and R2 = {(a, a), (b, b), (c, c), (b, c), (c, b)}, the union R1 ∪ R2 is reflexive and symmetric but not transitive.
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The intersection of R1 and R2 yields only those pairs that are common to both relations, maintaining reflexivity, symmetry, and transitivity.
Memory Aids
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🎵 Rhymes Time
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If a union's not transitive, it won't be an equal friend, but an intersection's always a true blend.
📖 Fascinating Stories
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Imagine two neighborhoods (equivalence relations) that combine. While they share two-way streets (reflexivity and symmetry), they might not always connect through the alleys (transitivity) unless they specifically plan it. However, when they intersect at the park, they always work together (intersection is always reflexive, symmetric, and transitive).
🧠 Other Memory Gems
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R-S-T for Equivalence, where R = Reflexive, S = Symmetric, T = Transitive.
🎯 Super Acronyms
USE for Union
- Understand
- Symmetry
- Exception for transitivity.
Flash Cards
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Glossary of Terms
Review the Definitions for terms.
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Term: Equivalence relation
Definition:
A relation that is reflexive, symmetric, and transitive.
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Term: Reflexive relation
Definition:
A relation where every element is related to itself.
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Term: Symmetric relation
Definition:
A relation where if one element is related to another, then the second is related to the first.
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Term: Transitive relation
Definition:
A relation where if one element is related to a second, and the second is related to a third, then the first is related to the third.
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Term: Counterexample
Definition:
An example that disproves a statement or proposition.
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Term: Intersection
Definition:
The set of elements that are common to both sets.
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Term: Union
Definition:
The set that contains all elements from both sets.
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Term: Composition of relations
Definition:
The set of pairs (a, c) such that there is a b that relates a to b and b to c.