Let x, y ∈ Z. Prove or disprove that the following relations are equivalence relations. (a) (0.5 points) x ∼ y if and only if x − y is greater than −1.

To determine whether the given relation \( x \sim y \) (where \( x, y \in \mathbb{Z} \)) is an equivalence relation, we must check whether it satisfies the three defining properties of equivalence relations:

1. **Reflexivity**: \( x \sim x \) for all \( x \in \mathbb{Z} \).
2. **Symmetry**: If \( x \sim y \), then \( y \sim x \).
3. **Transitivity**: If \( x \sim y \) and \( y \sim z \), then \( x \sim z \).

The given relation is:
\[
x \sim y \text{ if and only if } x - y > -1.
\]
This is equivalent to \( x - y \geq 0 \), or simply \( x \geq y \).

### 1. Reflexivity:
For reflexivity, we need to check whether \( x \sim x \). That is, does \( x - x \geq 0 \)?

- \( x - x = 0 \), which is greater than or equal to 0.
- Thus, the relation is **reflexive**.

### 2. Symmetry:
For symmetry, we need to check whether \( x \sim y \) implies \( y \sim x \). In other words, if \( x - y \geq 0 \), does \( y - x \geq 0 \)?

- Suppose \( x \sim y \), meaning \( x - y \geq 0 \), or equivalently \( x \geq y \).
- If \( x \geq y \), then \( y - x \leq 0 \). However, the relation \( y \sim x \) requires \( y - x \geq 0 \), which is not guaranteed.
- Thus, the relation is **not symmetric**.

### 3. Transitivity:
For transitivity, we need to check whether \( x \sim y \) and \( y \sim z \) imply \( x \sim z \). In other words, if \( x \geq y \) and \( y \geq z \), does \( x \geq z \)?

- If \( x \geq y \) and \( y \geq z \), then by the transitive property of inequality, \( x \geq z \).
- Thus, the relation is **transitive**.

### Conclusion:
The relation \( x \sim y \) (defined as \( x - y \geq 0 \) or \( x \geq y \)) is **reflexive** and **transitive** but **not symmetric**. Therefore, the relation is **not an equivalence relation**.

Would you like more clarification on any part of the proof or the definitions?

### Related Questions:
1. What are the common examples of equivalence relations?
2. How do you determine whether a relation is symmetric but not transitive?
3. Can a relation be reflexive and symmetric but not transitive?
4. What is the geometric interpretation of equivalence relations?
5. How does partial ordering differ from equivalence relations?

### Tip:
When proving that a relation is an equivalence relation, always check for reflexivity, symmetry, and transitivity individually. Failure of any one of these properties disqualifies the relation as an equivalence relation.

Learn how to determine if the relation x ∼ y, defined as x - y > -1, is an equivalence relation. This problem covers reflexivity, symmetry, and transitivity in equivalence relations, with a detailed explanation and proof. Suitable for college-level students studying discrete mathematics or algebra.

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