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Magazine Chapter 9: Problem 56
Suppose that \(R_{1}\) and \(R_{2}\) are equivalence relations on the set \(S .\) Determine whether each of these combinations of \(R_{1}\) and \(R_{2}\) must be an equivalence relation. $$ \begin{array}{llll}{\text { a) } R_{1} \cup R_{2}} & {\text { b) } R_{1} \cap R_{2}} & {\text { c) } R_{1} \oplus R_{2}}\end{array} $$
Short Answer
Expert verified
a) No, b) Yes, c) No
Step by step solution
01
- Define Equivalence Relation
An equivalence relation on a set S is a relation that is reflexive, symmetric, and transitive.
02
- Analyze Reflexivity
A relation is reflexive if every element is related to itself. Check if each of the combinations (a, b, c) satisfies this property.
03
Step 2a - Reflexivity of \(R_{1} \cup R_{2}\)
Since both \(R_{1}\) and \(R_{2}\) are reflexive, for any element \(x\) in \(S\), \((x, x) \in R_{1}\) and \((x, x) \in R_{2}\). Hence, \((x, x) \in R_{1} \cup R_{2}\). Thus, \(R_{1} \cup R_{2}\) is reflexive.
04
Step 2b - Reflexivity of \(R_{1} \cap R_{2}\)
Similarly, for \(R_{1} \cap R_{2}\), \((x, x) \in R_{1}\) and \((x, x) \in R_{2}\). Therefore, \((x, x) \in R_{1} \cap R_{2}\). So, \(R_{1} \cap R_{2}\) is reflexive.
05
Step 2c - Reflexivity of \(R_{1} \oplus R_{2}\)
For \(R_{1} \oplus R_{2}\), \((x, x) \in R_{1}\) or \((x, x) \in R_{2}\) means \((x, x) \in R_{1} \cap R_{2}\). Since reflexivity requires each element to relate to itself in both relations, \(R_{1} \oplus R_{2}\) is not reflexive.
06
- Symmetry
A relation is symmetric if \((a, b) \in R\) implies \((b, a) \in R\). Check symmetry for each combination.
07
Step 3a - Symmetry of \(R_{1} \cup R_{2}\)
Since both \(R_{1}\) and \(R_{2}\) are symmetric, if \((a, b) \in R_{1}\) or \((a, b) \in R_{2}\), then \((b, a) \in R_{1}\) or \((b, a) \in R_{2}\). Therefore, \(R_{1} \cup R_{2}\) is symmetric.
08
Step 3b - Symmetry of \(R_{1} \cap R_{2}\)
If \((a, b) \in R_{1} \cap R_{2}\), then \((a, b) \in R_{1}\) and \((a, b) \in R_{2}\). Hence, \((b, a) \in R_{1}\) and \((b, a) \in R_{2}\), meaning \((b, a) \in R_{1} \cap R_{2}\). Thus, \(R_{1} \cap R_{2}\) is symmetric.
09
Step 3c - Symmetry of \(R_{1} \oplus R_{2}\)
If \((a, b) \in R_{1} \oplus R_{2}\), then \((a, b) \in R_{1} \cap R_{2}\) does not necessarily ensure \((b, a) \in R_{1} \oplus R_{2}\). Therefore, \(R_{1} \oplus R_{2}\) is not symmetric.
10
- Transitivity
A relation is transitive if \((a, b) \in R\) and \((b, c) \in R\) imply \((a, c) \in R\). Check transitivity for each combination.
11
Step 4a - Transitivity of \(R_{1} \cup R_{2}\)
Even if \((a, b) \in R_{1} \cup R_{2}\) and \((b, c) \in R_{1} \cup R_{2}\), it does not guarantee \((a, c) \in R_{1} \cup R_{2}\). Hence, \(R_{1} \cup R_{2}\) is not transitive.
12
Step 4b - Transitivity of \(R_{1} \cap R_{2}\)
If \((a, b) \in R_{1} \cap R_{2}\) and \((b, c) \in R_{1} \cap R_{2}\), then \((a, b) \in R_{1}\) and \((b, c) \in R_{1}\) which implies \((a, c) \in R_{1}\); similarly for \(R_{2}\). Thus, \((a, c) \in R_{1} \cap R_{2}\) and \(R_{1} \cap R_{2}\) is transitive.
13
Step 4c - Transitivity of \(R_{1} \oplus R_{2}\)
No specific case ensures transitivity for \(R_{1} \oplus R_{2}\). Therefore, \(R_{1} \oplus R_{2}\) is not transitive.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Reflexivity
Reflexivity is one of the essential properties of an equivalence relation. A relation is said to be reflexive if every element in a set is related to itself. This means for any element x in a set S, the pair (x, x) must be in the relation. Let's break it down with some examples.
For example, considering the set of all integers:
For example, considering the set of all integers:
- The relation 'is equal to' is reflexive because any integer is always equal to itself.
- Another example is the relation 'is less than or equal to'. This is not reflexive because not every integer is less than or equal to itself.
- a) For the relation 饾憛鈧 鈭 饾憛鈧 to be reflexive, it means that any element in either 饾憛鈧 or 饾憛鈧 must relate to itself. Since both 饾憛鈧 and 饾憛鈧 are reflexive by themselves, the union of these relations will also be reflexive.
- b) For the relation 饾憛鈧 鈭 饾憛鈧 to be reflexive, each element in 饾憛鈧 鈭 饾憛鈧 will relate to itself if it is present in both 饾憛鈧 and 饾憛鈧, thus maintaining reflexivity.
- c) 饾憛鈧 鈯 饾憛鈧, however, is different. Since it requires the elements to be in either one of the relations and not both, it doesn't necessarily ensure every element will relate to itself. Thus, 饾憛鈧 鈯 饾憛鈧 is not reflexive.
Symmetry
Symmetry is another important aspect of equivalence relations. A relation is symmetric if, for any pair of elements (a, b) in the relation, the pair (b, a) must also be in the relation. In simpler terms, if a is related to b, then b must be related to a.
For example:
For example:
- The relationship 'is a sibling of' is symmetric because if Alex is a sibling of Jamie, then Jamie is also a sibling of Alex.
- However, the relationship 'is a parent of' is not symmetric because if Alex is a parent of Jamie, Jamie is not a parent of Alex.
- a) If either 饾憛鈧 or 饾憛鈧 is symmetric, meaning any pair (a, b) in 饾憛鈧 鈭 饾憛鈧 will include (b, a). Hence, the union of 饾憛鈧 and 饾憛鈧 will be symmetric.
- b) In the case of intersecting 饾憛鈧 鈭 饾憛鈧, both 饾憛鈧 and 饾憛鈧 need to contain the pairs (a, b) and (b, a) for the same elements. Thus, their intersection will also be symmetric.
- c) For symmetric difference 饾憛鈧 鈯 饾憛鈧, if (a, b) is in one relation but not the other, there is no guarantee that (b, a) will exist under the same condition. Therefore, 饾憛鈧 鈯 饾憛鈧 is not symmetric.
Transitivity
Transitivity is the third key property defining an equivalence relation. A relation is transitive if, whenever an element a is related to b and b is related to c, then a must also be related to c. This property chains relationships together.
For example:
For example:
- The 'is an ancestor of' relation is transitive because if Alice is an ancestor of Bob and Bob is an ancestor of Charlie, then Alice is also an ancestor of Charlie.
- Conversely, 'is friends with' is not necessarily transitive. If Alice is friends with Bob and Bob is friends with Charlie, it doesn't guarantee Alice is friends with Charlie.
- a) In the case of 饾憛鈧 鈭 饾憛鈧, if (a, b) exists in 饾憛鈧 or 饾憛鈧 and (b, c) exists in 饾憛鈧 or 饾憛鈧, that doesn't ensure (a, c) exists in any one of 饾憛鈧 or 饾憛鈧. Thus, the union is not necessarily transitive.
- b) For 饾憛鈧 鈭 饾憛鈧, if (a, b) exists in both 饾憛鈧 and 饾憛鈧 and (b, c) does as well, then both relations will ensure (a, c) exists in 饾憛鈧 鈭 饾憛鈧. Therefore, the intersection is transitive.
- c) For 饾憛鈧 鈯 饾憛鈧, there is no consistent rule ensuring that if (a, b) and (b, c) are in the relation, (a, c) will be too. Hence, 饾憛鈧 鈯 饾憛鈧 is not transitive.
