Exercise 6.18. Let Z/nZ := Z/ ∼n be the set of ∼n-equivalence classes on Z, where a ∼n b if a − b is divisible by n. Show that “If x = [a] and y = [b], then x + y := [a + b]. ′′

To prove that if \( x = [a] \) and \( y = [b] \), then the operation \( x + y := [a + b] \) is well-defined on \( \mathbb{Z}/n\mathbb{Z} \), let’s carefully go through each component of this statement.

### Step-by-Step Proof

1. **Understand the Definition of Equivalence Classes in \( \mathbb{Z}/n\mathbb{Z} \)**:
   - The notation \( \mathbb{Z}/n\mathbb{Z} \) represents the set of equivalence classes of integers under the equivalence relation \( \sim_n \).
   - Two integers \( a \) and \( b \) are equivalent (written as \( a \sim_n b \)) if and only if their difference \( a - b \) is divisible by \( n \). This means there exists an integer \( k \) such that:
     \[
     a - b = kn.
     \]

2. **Define Addition in \( \mathbb{Z}/n\mathbb{Z} \)**:
   - Suppose \( x = [a] \) and \( y = [b] \) are equivalence classes in \( \mathbb{Z}/n\mathbb{Z} \).
   - We define the sum of \( x \) and \( y \) as:
     \[
     x + y := [a + b].
     \]
   - Our goal is to show that this definition does not depend on the representatives \( a \) and \( b \) chosen for \( x \) and \( y \), meaning that if we chose different representatives, the resulting equivalence class would still be the same.

3. **Verify Well-Definedness**:
   - Suppose \( a' \) and \( b' \) are other representatives of the classes \( x \) and \( y \), respectively, such that \( a' \sim_n a \) and \( b' \sim_n b \).
   - By the definition of \( \sim_n \), we have:
     \[
     a' = a + k_1 n \quad \text{and} \quad b' = b + k_2 n
     \]
     for some integers \( k_1 \) and \( k_2 \).
   - We want to show that:
     \[
     [a + b] = [a' + b'].
     \]
   - Substituting \( a' \) and \( b' \) in terms of \( a \) and \( b \), we get:
     \[
     a' + b' = (a + k_1 n) + (b + k_2 n) = (a + b) + (k_1 + k_2) n.
     \]
   - Therefore, \( a' + b' \equiv a + b \pmod{n} \), which implies that \( [a' + b'] = [a + b] \).

4. **Conclusion**:
   - This calculation shows that the equivalence class \( [a + b] \) does not depend on the choice of representatives \( a \) and \( b \).
   - Hence, the operation \( x + y := [a + b] \) is well-defined in \( \mathbb{Z}/n\mathbb{Z} \).

### Summary
We've shown that addition in \( \mathbb{Z}/n\mathbb{Z} \) using the operation \( x + y := [a + b] \) is well-defined, meaning that the sum of two equivalence classes does not depend on the choice of representatives. This allows us to treat \( \mathbb{Z}/n\mathbb{Z} \) as a set with a consistent addition operation.

Would you like further details, or do you have any questions?

---

1. How can we define multiplication in \( \mathbb{Z}/n\mathbb{Z} \)?
2. What are some examples of equivalence classes in \( \mathbb{Z}/n\mathbb{Z} \)?
3. How does \( \mathbb{Z}/n\mathbb{Z} \) relate to modular arithmetic?
4. Can you prove that \( \mathbb{Z}/n\mathbb{Z} \) forms a group under addition?
5. What additional structure makes \( \mathbb{Z}/n\mathbb{Z} \) a ring?

**Tip**: When proving that an operation is well-defined, always check if the operation yields consistent results independent of chosen representatives.

Exercise 6.18. Let Z/nZ := Z/ ∼n be the set of ∼n-equivalence classes on Z, where a ∼n b if a − b is divisible by n. Show that 'If x = [a] and y = [b], then x + y := [a + b].'

This solution shows how to prove that the addition operation on equivalence classes in Z/nZ is well-defined, where a ∼n b means a - b is divisible by n. The proof covers key concepts such as equivalence relations and modular arithmetic. Suitable for introductory-level undergraduate students studying group theory and number theory.

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