Math Problem Statement

Let A = {1, 2, 3, 4, 5, 6} with the relation given by R = {(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5), (2, 2), (2, 6), (6, 2), (6, 6), (4, 4)} ⊂ A × A. find E(1),E(2),E(3),E(4),E(5),E(5)

Solution

To determine the equivalence classes E(x)E(x) for each element xx in AA with respect to the equivalence relation RR, we need to analyze the pairs in RR and group the elements that are equivalent.

Given Relation RR:

R={(1,1),(1,3),(1,5),(3,1),(3,3),(3,5),(5,1),(5,3),(5,5),(2,2),(2,6),(6,2),(6,6),(4,4)}R = \{(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5), (2, 2), (2, 6), (6, 2), (6, 6), (4, 4)\}

An equivalence relation satisfies reflexivity, symmetry, and transitivity. Using this relation, let's find the equivalence classes E(x)E(x).


Steps:

  1. Find E(1)E(1):

    • Look for all yy such that (1,y)R(1, y) \in R or (y,1)R(y, 1) \in R.
    • From RR: (1,1),(1,3),(1,5),(3,1),(5,1)(1, 1), (1, 3), (1, 5), (3, 1), (5, 1).
    • Thus, 1,3,1, 3, and 55 are related.
    • Equivalence class E(1)={1,3,5}E(1) = \{1, 3, 5\}.
  2. Find E(2)E(2):

    • Look for all yy such that (2,y)R(2, y) \in R or (y,2)R(y, 2) \in R.
    • From RR: (2,2),(2,6),(6,2)(2, 2), (2, 6), (6, 2).
    • Thus, 22 and 66 are related.
    • Equivalence class ( E(2) = {2, 6}.
  3. Find E(3)E(3):

    • Look for all yy such that (3,y)R(3, y) \in R or (y,3)R(y, 3) \in R.
    • From RR: (3,1),(3,3),(3,5),(1,3),(5,3)(3, 1), (3, 3), (3, 5), (1, 3), (5, 3).
    • Thus, 1,3,1, 3, and 55 are related.
    • Equivalence class E(3)={1,3,5}E(3) = \{1, 3, 5\}.
  4. Find E(4)E(4):

    • Look for all yy such that (4,y)R(4, y) \in R or (y,4)R(y, 4) \in R.
    • From RR: (4,4)(4, 4).
    • Thus, 44 is only related to itself.
    • Equivalence class ( E(4) = {4}.
  5. Find E(5)E(5):

    • Look for all yy such that (5,y)R(5, y) \in R or (y,5)R(y, 5) \in R.
    • From RR: (5,1),(5,3),(5,5),(1,5),(3,5)(5, 1), (5, 3), (5, 5), (1, 5), (3, 5).
    • Thus, 1,3,1, 3, and 55 are related.
    • Equivalence class E(5)={1,3,5}E(5) = \{1, 3, 5\}.
  6. Find E(6)E(6):

    • Look for all yy such that (6,y)R(6, y) \in R or (y,6)R(y, 6) \in R.
    • From RR: (6,2),(6,6),(2,6)(6, 2), (6, 6), (2, 6).
    • Thus, 22 and 66 are related.
    • Equivalence class ( E(6) = {2, 6}.

Final Results:

  • E(1)={1,3,5}E(1) = \{1, 3, 5\}
  • E(2)={2,6}E(2) = \{2, 6\}
  • E(3)={1,3,5}E(3) = \{1, 3, 5\}
  • E(4)={4}E(4) = \{4\}
  • E(5)={1,3,5}E(5) = \{1, 3, 5\}
  • E(6)={2,6}E(6) = \{2, 6\}

Would you like more details on how these equivalence classes were derived or any additional examples?


5 Related Questions:

  1. How do equivalence relations differ from general binary relations?
  2. What are the key properties of equivalence relations?
  3. How would RR change if we added new elements to AA?
  4. What would happen if reflexivity, symmetry, or transitivity failed in RR?
  5. Can equivalence classes overlap in any relation?

Tip:

When analyzing equivalence relations, always verify the reflexivity, symmetry, and transitivity properties before determining equivalence classes.

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Math Problem Analysis

Mathematical Concepts

Set Theory
Equivalence Relations
Equivalence Classes

Formulas

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Theorems

Definition of an equivalence relation (reflexivity, symmetry, transitivity)

Suitable Grade Level

Grades 11-12