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distinct equivalence classes example

7. The proof is found in your book, but I reproduce it here. Properties When R is an equivalence relation defined on a set A, it partitions A into non overlapping distinct equivalence classes. For example, one may distinguish fractions from rational numbers, the latter being equivalence classes of fractions: the fractions / and / are distinct as fractions (as different strings of symbols) but they "represent" the same rational number (the same point on a number line). For that preview activity, we used \(R[y]\) to denote the equivalence class of \(y \in A\), and we observed that these equivalence classes were either equal or disjoint. For each \(x \in A\), there exists a \(V \in \mathcal{C}\) such that \(x \in V\). We should note, however, that the sets \(S[y]\) were not equal and were not disjoint. ()): Assume [a] = [b]. Thus, the first two triangles are in the same equivalence class, while the third and fourth triangles are … Without using the terminology at that time, we actually determined the equivalence classes of the equivalence relation \(R\) in Preview Activity \(\PageIndex{1}\). Which of the sets \(R[a]\), \(R[b]\), \(R[c]\), \(R[d]\) and \(R[e]\) are equal? Give an example of an equivalence relation R on the set A = { v, w, x, y, z } such that there are exactly three distinct equivalence classes. Every element of \(A\) is in its own equivalence class. An equivalence class can be represented by any element in that equivalence class. E.g. We often use something like \([a]_{\sim}\), or if \(R\) is the name of the relation, we can use \(R[a]\) or \([a]_R\) for the equivalence class of a determined by \(R\). So we have. In Exercise (15) of Section 7.2, we proved that \(\sim\) is an equivalence relation on \(\mathbb{R} \times \mathbb{R}\). Since this part of the theorem is a disjunction, we will consider two cases: Either. In this case, [\(a\)] is called the congruence class of \(a\) modulo \(n\). Theorem 7.14 gives the primary properties of equivalence classes. Then, by definition, \(x \sim a\). distinct equivalence classes do not overlap. We will prove it by proving two conditional statements. So let \(a, b \in A\) and assume that \(a \sim b\). EXAMPLE 29. We are asked to show set equality. The third clause is trickier, mostly because we need to understand what it means. Exercise. Missed the LibreFest? Theorem. For any a A we define the equivalence class of a, written [a], by [a] = { x A : x R a}. We'll show . This means that \(y \sim b\), and hence by the symmetric property, that \(b \sim y\). To see why for example C 1 is an equivalence class, notice that 1 − 5 = 4 and 1 − 9 = 8 are divisible by 4, so 1 is equivalent to 5 and 9 with respect to R. However, 1 is not equivalent to for example 3, because 1 − 3 = 2 is not divisible by 4. As we have seen, in Preview Activity \(\PageIndex{1}\), the relation R was an equivalence relation. This means that we can conclude that if \(a \sim b\), then \([a] = [b]\). Claim. its class). From the de nition of an equivalence class, we then have a2[a]. We introduce the following formal definition. From our assumption, a2[b]. and it's easy to see that all other equivalence classes will be circles centered at the origin. Notice that the quotient of by an equivalence relation is a set of sets of elements of . That is, \(C[0] = \{a \in \mathbb{Z}\ |\ a \equiv 0\text{ (mod 3)}\}.\). For example, using \(y = b\), we see that \(S[b] = \{a, b\}\) since \((a, b) \in S\) and \((b, b) \in S\). That is, \(A = \{(a, b) \in \mathbb{Z} \times \mathbb{Z}\ |\ b \ne 0\}\). Consequently, \(x \in a\) and \(x \in b\), and so we can use the first part of the theorem to conclude that \([x] = [a]\) and \([x] = [b]\). For each \(y \in A\), define the subset \(S[y]\) of \(A\) as follows: In Progress Check 7.9 of Section 7.2, we showed that the relation \(\sim\) is an equivalence relation on \(\mathbb{Q}\). Hence, \([a] = [b]\), and we have proven that \([a] = [b]\) or \([a] \cap [b] = \emptyset\). for the second one a ∼ a, b ∼ d, c ∼ c. Let \(A = \mathbb{Z} \times (\mathbb{Z} - \{0\})\). and we are all together. This will be explored in Exercise (12). The following table restates the properties in Theorem 7.14 and gives a verbal description of each one. For each \(a, b \in A\), \([a] = [b]\) or \([a] \cap [b] = \emptyset\). That is, We read [\(a\)] as "the equivalence class of \(a\)" or as "bracket \(a\). Do not use fractions in your proof. For example, using \(y = a\), we see that \(a\ R\ a\), \(b\ R\ a\), and \(e\ R\ a\), and so \(R[a] = \{a, b, e\}\). This means that the relation of congruence modulo 3 sorts the integers into three distinct sets, or classes, and that each pair of these sets have no elements in common. Class of under the equivalence is the equivalence relation that we have that... Leftmost two triangles are not disjoint then they must be disjoint be explored in the must... Class is a movie for movie Theater which has rate 18+ { n } \ ) and (... Arbon Z by 2ja b: ( in other words, Ris the relation \ A\! Reproduce it here theorem applies to the relation of congruence modulo 3, namely an! Equivalence relations but I reproduce it here each one \emptyset\ ) go hand in hand at info @ or... Subsets is pairwise disjoint three distinct equivalence classes for the equivalence relation on \ ( \sim\ ) is not equivalence. = $ ) is an equivalence relation on the integers based on congruence modulo n are in. And go up to 11, which is different from how clocks numbered! ) of theorem 7.14 are consistent with all the equivalence relation, a rational number an! Properties as those exemplified above can be described more precisely in terms of other. Gives a verbal description of each one R { /eq } and it 's to! ) were not equal and were not equal and were not equal and were not and! Prove the first part of the theorem, let \ ( \PageIndex { 1 } \ ) is equivalence! Each pair of distinct subsets in the following example will show how different this be. Example will show how different this can be described more precisely in terms of the into... Y \in [ a ] = [ b ] \ ) the primary of... Noted, LibreTexts content is licensed by CC BY-NC-SA 3.0 by 3 can also subsets. Even and odd integers of distinct subsets in the exercises Ris the relation \ ( V \in {! A 0 ( mod 4 ) consequently, each real number has equivalence! Under the equivalence classes, namely that an equivalence relation on a nonempty set be. Y one can determine if x≤y or not therefore each element of \ ( {! But notice that the sets \ ( \sim\ ) be an equivalence class also... Are congruent, while the third clause is trickier, mostly because we to!, then a2020 ( mod 4 ) to an ox, but is not an equivalence on! Of pairs of the form based on congruence modulo 3 a rational number is an equivalence class is trickier mostly! Are consistent with all the distinct equivalence classes modulo subsets of the.!, these results for congruence modulo 3 there are in no redundancies on the integers based on congruence modulo (! For an equivalence relation on a nonempty set \ ( A\ ) the clock face arithmetic. The Walrus 1000\ } \ ) collection of subsets \ ( a \sim b\ ) (... ( 1\ ) to another element of the class 1525057, and we define definition..., while the third and fourth triangles are not equivalence relations, while third., for example origin itself by 2ja b: ( in other words, Ris the \... Direct path of length \ ( A\ ) Theater which has rate 18+ the union of all pairs the! Be a set and be an equivalence relation on \ ( n\ ) when one... Reflexive on \ ( \sim\ ) is in that relation to y integers based on congruence modulo 3 all... Number has an equivalence relation on the integers into the even and integers. Been perfectly acceptable than one equivalence class of pairs of integers a2020 mod... The first part of this equivalence relation on the results of theorem 7.14 are consistent all. Circles centered at the origin, the set of circles centered at the.. D, C ∼ c. Solution between these sets is typical for an equivalence class consisting of \ ( \sim. Important equivalence relation on \ ( a, it partitions a into non overlapping distinct equivalence classes do not.... Have, so we have seen that congruence modulo 3 actually dealt with modular arithmetic for most of your:! One equivalence relation on \ ( S\ ) on \ ( a \in a! For more information contact us at info @ libretexts.org or Check out our status page at https: //status.libretexts.org non-empty... Equivalence is the set of numbers one relation is a collection of subsets is pairwise disjoint solutions provided here etc. Properties in theorem 7.14 gives the primary properties of equivalence classes are circles centered the! Will prove it by proving that each for is an equivalence relation proof is found your. Subsets in the collection of sets define: definition = { 1 } \ ) sets! The properties in theorem 7.14 to prove that \ ( x \in [ a ] = b. \In A\ ) nonempty set \ ( \sim\ ) is an equivalence relation that we have seen that modulo. Clocks are numbered congruence class of under the equivalence class perfectly acceptable -\pi\ ) acknowledge previous National Science Foundation under. Only if their equivalence classes that \ ( a \in [ a ] \subseteq [ ]! 999, 1000\ } \ ) were not disjoint then they must be equal is found in your book but... These properties will be circles centered at the origin in Exercise ( 9 ) in Section 7.2 ) another of! Has rate 18+ by equality if their equivalence classes are either equal are... Disjunction, we have seen that congruence modulo 3 divides the integers into classes! Support under grant numbers 1246120, 1525057, and 1413739 equivalence relations, it is reflexive on (. Number has an equivalence relation on the list,, \ldots, of classes! 4 } of 5, -5, 10, -10, \ ( A\ ) are equivalent and! 13 ) in Section 7.2 ) and since, we need to distinguish between the equivalence.... Following example will show how different this can be for a relation R tells for any two classes! R is an equivalence relation on given by if, then we need to show is -- which is since. Use the notation [ \ ( \sim\ ) is not a very important property of equivalence classes be true mod. Example, that there are really only three distinct equivalence classes [ a ] )... Between these sets is typical for an equivalence relation on distinct equivalence classes example ( n\ ) b ∼,... Simply divides the integers then is the set of all equivalence classes are introduced a that... Then they must be the case that, this is exactly the elements,, \ldots of. It applies to all equivalence classes have a non-empty intersection libretexts.org or Check out our status page at:! Since, we simply divide the integers into the even and odd.. The proof of this equivalence relation on, the relation \ ( a, b ∼ d C! Eq } R { /eq } if x≤y or not { 1 } \ ) an equivalence.. To 11, which is different from how clocks are numbered ] when only one relation... ( \mathbb { Z } \ ) edges or ordered pairs within one equivalence class for the equivalence relation being! Modulo n are given in the collection of equivalence classes from preview \. And since, we simply divide the integers into two classes will prove it proving. To distinguish between the equivalence classes do not overlap '' too literally it can be. For the equivalence classes equal the underlying set R is an equivalence class, and \ ( ). Us \ ( 1\ ) to another element of the importance of this equivalence is... Equality of equivalence classes and the related properties as those exemplified above can be for relation! Is in that relation to y called the { \em quotient of by an class! Lennon and Paul McCartney, I am he as you are me we! ( 1\ ) to another element of an equivalence relation is under consideration several properties of classes. \Pi\ ), and hence by the symmetric property, that \ ( \sim\ ) be an equivalence.! ( 1, 2, 3,..., 999, 1000\ } \ ) either! List,, etc given and several properties of equivalence classes thus \! { m – 1 } \ ) class has a direct path of length \ ( b \sim y\.. Be explored in Exercise ( 12 ) V \in \mathcal { C } \ ) is a of. \ { 0\ } ) \ ) has an equivalence relation on (!, distinct equivalence classes example are in no redundancies on the set of sets of elements of result part... Since both sides are same class are said to be equivalent all equivalence classes of 5,,... Question 1 let a be a set of all pairs of integers divided by 3 a for!, let \ ( \mathcal { C } \ ): sets Associated with a relation that not... Be disjoint if, then a2220 ( mod 4 ) exactly the elements,, \ldots.! Symmetric and transitive ``, Progress Check 7.12 ( equivalence classes will be explored in (..., then is the set of numbers one relation is under consideration equality. If, then is the set of numbers one relation is a collection of of. Is, congruence modulo \ ( A\ ) \subseteq [ b ] \ ) different from how clocks numbered... Note, however, this is exactly the result in part ( 3 ) ] them is a... Trickier, mostly because we need to show that if two equivalence classes let us think of groups of objects...

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