Math Problem Statement

Let R be the relation on the set Z of integers defined by the rule: aRb if a–b is divisible by 2 (that is, a – b =2n for some integer n). Describe the equivalence classes of an equivalence relation.

Solution

The relation RR on the set of integers Z\mathbb{Z}, defined by aRbaRb if and only if aba - b is divisible by 2, forms an equivalence relation. To verify that this relation is an equivalence relation, we need to check the three necessary properties: reflexivity, symmetry, and transitivity.

1. Reflexivity:

For all aZa \in \mathbb{Z}, aa=0a - a = 0, and 0 is divisible by 2. Therefore, aRaaRa, so the relation is reflexive.

2. Symmetry:

If aRbaRb, then ab=2na - b = 2n for some integer nn. This implies ba=2nb - a = -2n, which is also divisible by 2, so bRabRa. Thus, the relation is symmetric.

3. Transitivity:

If aRbaRb (i.e., ab=2na - b = 2n for some nn) and bRcbRc (i.e., bc=2mb - c = 2m for some mm), then: ac=(ab)+(bc)=2n+2m=2(n+m).a - c = (a - b) + (b - c) = 2n + 2m = 2(n + m). Since aca - c is divisible by 2, we have aRcaRc. Therefore, the relation is transitive.

Since the relation satisfies reflexivity, symmetry, and transitivity, it is an equivalence relation.

Equivalence Classes:

The equivalence classes of this relation group integers based on whether their difference is divisible by 2. Specifically, the integers are classified based on their parity (whether they are even or odd).

  • Class of even integers: All integers that are congruent to 0 modulo 2 (i.e., the even integers) form one equivalence class. This class can be represented by [0][0], and it includes all even integers: {,4,2,0,2,4,6,}\{ \dots, -4, -2, 0, 2, 4, 6, \dots \}.

  • Class of odd integers: All integers that are congruent to 1 modulo 2 (i.e., the odd integers) form another equivalence class. This class can be represented by [1][1], and it includes all odd integers: {,3,1,1,3,5,}\{ \dots, -3, -1, 1, 3, 5, \dots \}.

Thus, the two equivalence classes under this relation are:

  1. The set of all even integers: [0]={,4,2,0,2,4,}[0] = \{ \dots, -4, -2, 0, 2, 4, \dots \}
  2. The set of all odd integers: [1]={,3,1,1,3,5,}[1] = \{ \dots, -3, -1, 1, 3, 5, \dots \}

Conclusion:

The equivalence classes of the relation RR partition the set of integers Z\mathbb{Z} into two distinct classes:

  • One class for the even integers.
  • One class for the odd integers.

Would you like more details or have any questions?


Here are 5 related questions to expand your understanding:

  1. How would the equivalence classes change if the condition was divisibility by 3 instead of 2?
  2. Can you describe an equivalence relation on R\mathbb{R} (real numbers) similar to this one?
  3. How can you define an equivalence relation based on divisibility by any integer nn?
  4. How do equivalence relations relate to partitions of a set?
  5. Can you provide an example of a relation that is not an equivalence relation?

Tip: Equivalence relations always partition a set into disjoint equivalence classes, and each element belongs to exactly one equivalence class.

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Math Problem Analysis

Mathematical Concepts

Equivalence Relation
Set Theory
Modular Arithmetic

Formulas

a - b = 2n, where n is an integer

Theorems

Properties of Equivalence Relations (Reflexivity, Symmetry, Transitivity)

Suitable Grade Level

Grades 9-12