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The problem asks us to show that \( \mathbb{R} \), the set of real numbers, equipped with the binary operation \( x * y = x + y + k \) (for some fixed value \( k \)) is an Abelian group. Additionally, we are told that \( * \) is associative and that an Abelian group is a group where \( x * y = y * x \).

### Step-by-Step Solution:

#### 1. **Binary Operation Definition**:
   The binary operation is given by:
   \[
   x * y = x + y + k
   \]
   where \( k \) is a fixed real number.

#### 2. **Abelian Group Properties**:
   To prove that this structure is an Abelian group, we need to verify the following group properties:

   **Closure**: For any \( x, y \in \mathbb{R} \), \( x * y \in \mathbb{R} \).

   **Associativity**: For any \( x, y, z \in \mathbb{R} \), \( (x * y) * z = x * (y * z) \).

   **Identity Element**: There must exist an element \( e \in \mathbb{R} \) such that for all \( x \in \mathbb{R} \), \( x * e = e * x = x \).

   **Inverse Element**: For each \( x \in \mathbb{R} \), there exists \( x' \in \mathbb{R} \) such that \( x * x' = x' * x = e \), where \( e \) is the identity element.

   **Commutativity (Abelian Property)**: For any \( x, y \in \mathbb{R} \), \( x * y = y * x \).

#### 3. **Closure**:
   The operation \( x * y = x + y + k \) is closed in \( \mathbb{R} \) because the sum of two real numbers plus a constant \( k \) is still a real number.

#### 4. **Associativity**:
   We are told to assume that the operation is associative, meaning:
   \[
   (x * y) * z = x * (y * z)
   \]
   for all \( x, y, z \in \mathbb{R} \). Hence, associativity holds.

#### 5. **Identity Element**:
   We need to find an element \( e \in \mathbb{R} \) such that for any \( x \in \mathbb{R} \),
   \[
   x * e = e * x = x
   \]
   Using the binary operation definition:
   \[
   x * e = x + e + k = x
   \]
   This simplifies to:
   \[
   e + k = 0 \quad \Rightarrow \quad e = -k
   \]
   So, the identity element is \( e = -k \).

#### 6. **Inverse Element**:
   For each \( x \in \mathbb{R} \), we need to find an element \( x' \in \mathbb{R} \) such that:
   \[
   x * x' = e
   \]
   Using the binary operation definition:
   \[
   x * x' = x + x' + k = -k
   \]
   This simplifies to:
   \[
   x' = -x - 2k
   \]
   Hence, the inverse of \( x \) is \( x' = -x - 2k \).

#### 7. **Commutativity (Abelian Property)**:
   To show that the group is Abelian, we need to verify that \( x * y = y * x \). Using the binary operation:
   \[
   x * y = x + y + k
   \]
   and
   \[
   y * x = y + x + k
   \]
   Since \( x + y + k = y + x + k \), the operation is commutative.

### Conclusion:
All the group axioms (closure, associativity, identity, inverse, and commutativity) are satisfied. Therefore, \( \mathbb{R} \) equipped with the operation \( x * y = x + y + k \) is an Abelian group.

Do you have any questions or need more details on any step?

### Related Questions:
1. What would happen if \( k \) were a variable instead of a constant?
2. How does the structure of the group change if we alter the operation slightly?
3. What is the effect of choosing different values for \( k \)?
4. How can we generalize this result to other sets besides \( \mathbb{R} \)?
5. Can we define similar operations that form non-Abelian groups?

### Tip:
Always verify each group property when working with an unfamiliar binary operation to ensure the structure you're studying forms a group.

Show that ℝ, when equipped with the binary operation x*y = x + y + k for some fixed value k is an abelian group. You may assume that * is associative.

This problem demonstrates how to prove that the set of real numbers equipped with the binary operation x * y = x + y + k forms an Abelian group. It covers group theory concepts such as identity and inverse elements, closure, and commutativity.

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