Math Problem Statement
Solution
To prove that the divisibility relation |
is a partial ordering on the set (the set of positive integers), we need to show that it satisfies the three properties of a partial order: reflexivity, antisymmetry, and transitivity.
1. Reflexivity
For all , . This is true because any integer divides itself.
2. Antisymmetry
If and and , then . This holds because if , then for some , and if , then for some . Substituting into , we get , which implies . Since , the only possibility is , so .
3. Transitivity
If , and , then . If , then for some , and if , then for some . Substituting into , we get , so .
Thus, the divisibility relation |
is a partial order on .
Would you like further clarification or examples?
Related Questions:
- How does divisibility differ from a total ordering?
- Can we extend this partial order to a total order on ?
- What is an example of a set where divisibility is not antisymmetric?
- How does the divisibility relation compare to other types of partial orders like subset inclusion?
- How can we visualize the divisibility relation using a Hasse diagram?
Tip:
To verify partial ordering, always focus on the three core properties: reflexivity, antisymmetry, and transitivity.
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Math Problem Analysis
Mathematical Concepts
Divisibility
Partial Ordering
Properties of Relations
Formulas
a | b implies b = ak for some k ∈ \( \mathbb{Z}^+ \)
Theorems
Reflexivity, Antisymmetry, and Transitivity of Relations
Suitable Grade Level
Undergraduate (Math majors)
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