Equivalence Classes and Partitions

1. First confirm or use the given fact that R is an equivalence relation. 2. Choose an element a and write the condition x R a. 3. Solve the condition to list or describe all x in the class. 4. Avoid repeating a class already obtained. 5. Verify partition by checking union and no overlap between distinct classes.

A partition diagram with disjoint boxes can support the idea, but exam answers mainly require set notation and a clear representative-based description.

Writing [1] = {1} for a congruence-type relation, ignoring other elements that satisfy the same remainder or property.

If R is an equivalence relation on a set A, the equivalence class of an element a in A is [a] = {x in A : x R a}. Equivalence classes are non-empty, pairwise disjoint or identical, and together they partition A.

If R is an equivalence relation on a set A, the equivalence class of an element a in A is [a] = {x in A : x R a}. Equivalence classes are non-empty, pairwise disjoint or identical, and together they partition A. Equivalence class: [a] = {x in A : x R a}. Partition property: union of all distinct equivalence classes is A, and any two distinct classes have empty intersection. If [a] and [b] have one common element, then [a] = [b]. For integers under the relation a R b if a - b is divisible by 3, the equivalence classes are [0], [1], and [2], representing integers that leave remainders 0, 1, and 2 on division by 3.

An equivalence class collects all elements that are related to a chosen representative. Different representatives can name the same class. The important idea is that an equivalence relation separates the set into blocks with no overlap between distinct blocks. Equivalence class: [a] = {x in A : x R a}. Partition property: union of all distinct equivalence classes is A, and any two distinct classes have empty intersection. If [a] and [b] have one common element, then [a] = [b]. Let A = {1,2,3,4,5,6} and define a R b if a and b have the same parity. Class [1] = {1,3,5} because these elements are odd like 1. Class [2] = {2,4,6} because these elements are even like 2. Now [3] = {1,3,5}, same as [1], so it is not a new class. Distinct classes are {1,3,5} and {2,4,6}; their union is A and their intersection is empty. Questions ask students to form equivalence classes for a given relation, show that the classes partition the set, or identify when two representatives give the same class. Writing [1] = {1} for a congruence-type relation, ignoring other elements that satisfy the same remainder or property.