Par exemple sur N ou sur R la relation ? est une relation d'ordre. Nous introduirons aussi les relations dites d'équivalence
2. Relations d'équivalence. Définition. Une relation binaire est une relation d'équivalence si et seulement si elle est réflexive symétrique et transitive.
20 août 2017 Définition 1 : Une relation binaire ? définie sur un ensemble E est au choix : • une propriété qui relie ou non deux éléments x et y de E.
1. Vérifier que la relation est une relation d'équivalence. 2. Faire la liste des classes d'équivalences distinctes et donner l'ensemble quotient .
7 mars 2018 1.0.1 Définition: une relation R:AxA est une relation d'équivalence sur A si R est reflexive symétrique et transitive.
Déterminer [(x0y0)]
C'est une relation d'équivalence sur U : MAT1500. 7 of 40. Page 8. Démonstration. Soient (n1d1)
Définir une relation d'équivalence c'est précisément définir un critère Il y a deux "classes d'équivalence" : la “classe des hommes" et la “classe des.
est une relation d'équivalence. Préciser pour x fixé dans R
https://livres-mathematiques.fr/onewebmedia/L1-MI-arith-ch1.pdf
Math 127: Equivalence Relations Mary Radcli e 1 Equivalence Relations Relations can take many forms in mathematics In these notes we focus especially on equivalence relations but there are many other types of relations (such as order relations) that exist De nition 1 Let X;Y be sets
relationship between equivalence relations and partitions Note that throughout this lecture we have already seen that an equivalence relation induces a partition but now we shall formally prove this phenomenon Theorem 1 If R is an equivalence relation on a set S then the equivalence classes of R partition S Proof
Equivalence relations are remarkably useful because they allow us to work with the concept of equivalence classes: De nition Take any set S with an equivalence relation R For any element x 2S we can de ne the equivalence class corresponding to x as the set fs 2S jsRxg Again you have worked with lots of equivalence classes before For mod 3
An Important Equivalence Relation Let S be the set of fractions: S ={p q: pq??q?0} Define a relation R on S by: a b R c d iff ad=bc This relation is an equivalence relation 1) For any fraction a/b a/b R a/b since ab = ba (Reflexitivity) 2) If a/b R c/d then ad = bc so cb = da and c/d R a/b (Symmetry)
1 Determine whether the following relations are equivalence relations on the given set S If the relation is in fact an equivalence relation describe its equivalence classes (a) S = Nnf0;1g; (x;y) 2R if and only if gcd(x;y) > 1 (b) S = R; (a;b) 2R if and only if a2 + a = b2 + b: (c) S = R; (x;y) 2R if and only if there exists n 2Z such that
Using equivalence relations to de?ne rational numbers Consider the set S = {(xy) ? Z × Z: y 6= 0 } We de?ne a rational number to be an equivalence classes of elements of S under the equivalence relation (ab) ’ (cd) ?? ad = bc An equivalence class is a complete set of equivalent elements
Equivalence classes What makes equivalence relations so useful is they give us a way of ignoring information that is irrelevant to the task at hand. For example, suppose a and b are two very large natural numbers, each with several trillion (decimal) digits. We want to know what the last digit of ab is.
Let F be any partition of the set S. Define a relation on S by x R y iff there is a set in F which contains both x and y. Then R is an equivalence relation and the equivalence classes of R are the sets of Pf: Since F is a partition, for each x in S there is one (and only one) set of F which contains x.
A leaner de?nition is: If R is an equivalence relation on a set S, then we de?ne the equivalence class of an element x ? S to be the set of all elements of S equivalent to x. Then we need to prove: 5 Theorem 1.
An equivalence class is a complete set of equivalent elements. I.e., it’s a set of elements of S, all of which are equivalent to each other, and which contains all of the pairs that are equivalent to those pairs. (Stricly speaking we need to use some properties of equivalence relations to check that this makes sense ...more about that later.)