Math Problem Statement
Solution
Let's go through each part of this problem, step-by-step.
Given Information:
- Set .
- Relation on is defined such that if and only if , which means that is even.
Solution:
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(a) Directed Graph of : To construct the directed graph, we need to determine which pairs satisfy the relation .
- For two elements , the sum is even if both and are either both even or both odd.
- Even numbers in :
- Odd numbers in :
Therefore:
- holds for pairs where both and are either even or both are odd.
Pairs that satisfy :
- For odd elements : pairs are .
- For even elements : pairs are .
Using these pairs, you can create a directed graph with nodes labeled to and arrows connecting nodes according to the pairs above.
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(b) Equivalence Class : The equivalence class consists of all elements such that .
Since is odd, we need to find all odd elements in . These are and .
Thus, .
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(c) Number of Distinct Equivalence Classes: We observe that groups elements into two classes based on parity:
- Odd elements:
- Even elements:
These two groups represent two distinct equivalence classes under .
Therefore, the relation has 2 distinct equivalence classes.
Would you like further details on any part of the solution?
Related Questions
- What makes a relation an equivalence relation, and how can you check it?
- How do you interpret equivalence classes in terms of partitions of a set?
- How would the solution change if had a different number of elements?
- Can you find equivalence classes in non-numeric sets, such as sets of words or other objects?
- How would you generalize this relation for any finite set with an arbitrary rule for even sums?
Tip
To verify that a relation is an equivalence relation, always check that it is reflexive, symmetric, and transitive. These properties are essential for defining equivalence classes.
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Math Problem Analysis
Mathematical Concepts
Relations
Equivalence Relations
Directed Graphs
Equivalence Classes
Formulas
m R n ⟺ 2 | (m + n)
Theorems
Properties of Equivalence Relations (Reflexive, Symmetric, Transitive)
Suitable Grade Level
Grades 10-12
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