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.