Math Problem Statement

Determine whether the relation {(1,1),(2,2),(3,3),(4,4),(5,5),(1,5),(5,1),(3,5),(5,3),(1,3),(3,1)} is an equivalence relation on {1,2,3,4,5}. If the relation is an equivalence relation, list the equivalence classes.

Solution

To determine whether the given relation R={(1,1),(2,2),(3,3),(4,4),(5,5),(1,5),(5,1),(3,5),(5,3),(1,3),(3,1)}R = \{(1,1), (2,2), (3,3), (4,4), (5,5), (1,5), (5,1), (3,5), (5,3), (1,3), (3,1)\} on the set {1,2,3,4,5}\{1, 2, 3, 4, 5\} is an equivalence relation, we need to check if it satisfies the three necessary properties of an equivalence relation:

  1. Reflexivity: A relation RR is reflexive if for every element aAa \in A, the pair (a,a)(a, a) is in RR.
  2. Symmetry: A relation RR is symmetric if for every pair (a,b)R(a, b) \in R, the pair (b,a)(b, a) is also in RR.
  3. Transitivity: A relation RR is transitive if whenever (a,b)R(a, b) \in R and (b,c)R(b, c) \in R, then (a,c)R(a, c) \in R.

1. Reflexivity:

For the set {1,2,3,4,5}\{1, 2, 3, 4, 5\}, we need to ensure that each element is related to itself. From the given relation, we have the pairs (1,1),(2,2),(3,3),(4,4),(5,5)(1,1), (2,2), (3,3), (4,4), (5,5), which satisfy reflexivity.

2. Symmetry:

The given relation contains the pairs:

  • (1,5)(1,5) and (5,1)(5,1)
  • (3,5)(3,5) and (5,3)(5,3)
  • (1,3)(1,3) and (3,1)(3,1)

Since for every pair (a,b)R(a, b) \in R, we also have (b,a)R(b, a) \in R, the relation is symmetric.

3. Transitivity:

We need to verify whether the relation satisfies transitivity. That is, if (a,b)R(a, b) \in R and (b,c)R(b, c) \in R, then (a,c)(a, c) should also be in RR.

  • (1,5)(1, 5) and (5,3)(5, 3) imply (1,3)(1, 3), which is in RR.
  • (3,5)(3, 5) and (5,1)(5, 1) imply (3,1)(3, 1), which is in RR.
  • (1,3)(1, 3) and (3,5)(3, 5) imply (1,5)(1, 5), which is in RR.
  • All other possible combinations of pairs from the relation are also satisfied.

Thus, the relation is transitive.

Conclusion:

Since the relation is reflexive, symmetric, and transitive, it is an equivalence relation.

Equivalence Classes:

The equivalence classes are formed by grouping all elements that are related to each other. From the given pairs in the relation, we have the following equivalence classes:

  • {1,3,5}\{1, 3, 5\} (since 11, 33, and 55 are related to each other)
  • {2}\{2\} (since 22 is only related to itself)
  • {4}\{4\} (since 44 is only related to itself)

Thus, the equivalence classes are:

  • {1,3,5}\{1, 3, 5\}
  • {2}\{2\}
  • {4}\{4\}

Let me know if you'd like further details or have any questions!

Further Related Questions:

  1. How do you prove that a relation is symmetric?
  2. What is the significance of an equivalence relation in set theory?
  3. Can a relation be reflexive and symmetric but not transitive?
  4. How do you compute the equivalence classes for a larger set or relation?
  5. What are the applications of equivalence relations in mathematics?

Tip: Always check the reflexive property first when determining if a relation is an equivalence relation, as it’s the easiest to verify for all elements in the set.

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Math Problem Analysis

Mathematical Concepts

Set Theory
Relations
Equivalence Relations

Formulas

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Theorems

Reflexivity
Symmetry
Transitivity

Suitable Grade Level

Grades 10-12