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Let's break down the problem step by step.

The set \( A_1 = \{ 1, 2, 3, 4, 5, 6 \} \) is ordered by the relation "x divides y." This is a **divisibility** relation where \( x \leq y \) if and only if \( x \) divides \( y \) (denoted as \( x \mid y \)).

### 1. Proving that \( A_1 \) is a poset:
A **poset** (partially ordered set) must satisfy three properties:
- **Reflexivity**: Every element divides itself, i.e., \( x \mid x \) for all \( x \in A_1 \).
- **Antisymmetry**: If \( x \mid y \) and \( y \mid x \), then \( x = y \).
- **Transitivity**: If \( x \mid y \) and \( y \mid z \), then \( x \mid z \).

Let's check each property for \( A_1 \):
1. **Reflexivity**: Clearly, for every \( x \in A_1 \), \( x \mid x \). Hence, reflexivity holds.
2. **Antisymmetry**: If \( x \mid y \) and \( y \mid x \), the only way this can happen in terms of divisibility is if \( x = y \). Hence, antisymmetry holds.
3. **Transitivity**: If \( x \mid y \) and \( y \mid z \), then \( x \mid z \) because divisibility is transitive. Hence, transitivity holds.

Since all three properties are satisfied, \( A_1 \) is a poset.

### 2. Drawing the Hasse Diagram:
The Hasse diagram is a graphical representation of the poset, where each element is represented as a node, and there is an edge between two elements \( x \) and \( y \) if \( x \mid y \) and there is no intermediate element \( z \) such that \( x \mid z \mid y \). This eliminates redundant edges, capturing only the direct divisibility relationships.

- Elements of \( A_1 \): \( 1, 2, 3, 4, 5, 6 \)
- Divisibility relationships:
  - \( 1 \) divides every number: \( 1 \mid 2, 1 \mid 3, 1 \mid 4, 1 \mid 5, 1 \mid 6 \)
  - \( 2 \mid 4 \) and \( 2 \mid 6 \)
  - \( 3 \mid 6 \)

Now, let's create the Hasse diagram.

We will use the following relations for the diagram:
- \( 1 \) divides all elements.
- \( 2 \mid 4 \), \( 2 \mid 6 \).
- \( 3 \mid 6 \).
- There are no other direct divisibility relations.

The Hasse diagram will look something like this:

```
    6
   / \
  4   3
   \ /
    2
    |
    1
```

This is the Hasse diagram for the poset defined by the divisibility relation on the set \( A_1 = \{ 1, 2, 3, 4, 5, 6 \} \).

Would you like more details or have any questions?

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Here are 5 questions to consider:
1. What is the difference between a poset and a totally ordered set?
2. How would the Hasse diagram change if we included more numbers in the set, such as 12 or 15?
3. How do we represent reflexivity and transitivity in a Hasse diagram?
4. Can a poset have cycles in its Hasse diagram? Why or why not?
5. How do we generalize this divisibility poset for larger sets of numbers?

**Tip**: To create a Hasse diagram, remember to only draw edges for direct relations and omit reflexive and transitive connections.

Let A1 = {1, 2, 3, 4, 5, 6} be ordered by the relation 'x divides y.' Prove that A1 is a poset and draw the Hasse diagram.

Explore the process of proving that the set A1 = {1, 2, 3, 4, 5, 6} forms a poset under divisibility. The solution covers key concepts like reflexivity, antisymmetry, and transitivity, and includes a Hasse diagram for visual representation. Suitable for undergraduate students.

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