Let G = {q ∈Q / q ≠ -1}. Define the binary operation o on G by
x o y = x + y + xy. Prove that (G, o) is an abelian group.

To prove that \((G, \circ)\), where \(G = \{q \in \mathbb{Q} \mid q \neq -1\}\) and the binary operation is defined as \(x \circ y = x + y + xy\), forms an abelian group, we need to show that the following group properties hold:

1. **Closure**: \(x \circ y \in G\) for all \(x, y \in G\),
2. **Associativity**: \(x \circ (y \circ z) = (x \circ y) \circ z\) for all \(x, y, z \in G\),
3. **Identity element**: There exists an element \(e \in G\) such that \(x \circ e = e \circ x = x\) for all \(x \in G\),
4. **Inverse element**: For each \(x \in G\), there exists an element \(x^{-1} \in G\) such that \(x \circ x^{-1} = x^{-1} \circ x = e\),
5. **Commutativity**: \(x \circ y = y \circ x\) for all \(x, y \in G\).

### Step 1: Closure
We need to verify that \(x \circ y \in G\) for all \(x, y \in G\), i.e., that \(x \circ y \neq -1\) for all \(x, y \in G\).

The operation is defined as:
\[
x \circ y = x + y + xy.
\]
We want to ensure that \(x \circ y \neq -1\):
\[
x + y + xy \neq -1.
\]
Rewriting this expression, we get:
\[
x + y + xy = -1 \implies (x + 1)(y + 1) = 0.
\]
This implies that either \(x = -1\) or \(y = -1\), but since \(x, y \in G\), we know \(x \neq -1\) and \(y \neq -1\). Therefore, \(x \circ y \neq -1\), so closure holds.

### Step 2: Associativity
We need to verify that \(x \circ (y \circ z) = (x \circ y) \circ z\) for all \(x, y, z \in G\).

First, calculate \(x \circ (y \circ z)\):
\[
y \circ z = y + z + yz,
\]
so
\[
x \circ (y \circ z) = x + (y + z + yz) + x(y + z + yz).
\]
Now, calculate \((x \circ y) \circ z\):
\[
x \circ y = x + y + xy,
\]
so
\[
(x \circ y) \circ z = (x + y + xy) + z + (x + y + xy)z.
\]
After simplifying both expressions, we observe that \(x \circ (y \circ z) = (x \circ y) \circ z\), so the operation is associative.

### Step 3: Identity Element
We need to find an element \(e \in G\) such that \(x \circ e = e \circ x = x\) for all \(x \in G\).

We solve \(x \circ e = x + e + xe = x\), which simplifies to:
\[
x + e + xe = x \implies e(1 + x) = 0.
\]
Since \(x \neq -1\), we have \(1 + x \neq 0\), so \(e = 0\). Therefore, 0 is the identity element.

### Step 4: Inverse Element
We need to find an element \(x^{-1} \in G\) such that \(x \circ x^{-1} = 0\), where 0 is the identity element.

We solve \(x \circ x^{-1} = x + x^{-1} + xx^{-1} = 0\), which simplifies to:
\[
x + x^{-1} + xx^{-1} = 0 \implies (x + 1)(x^{-1} + 1) = 1.
\]
Thus, \(x^{-1} = \frac{-x}{x+1}\). Since \(x \neq -1\), \(x^{-1}\) is well-defined and belongs to \(G\), so every element has an inverse.

### Step 5: Commutativity
We need to verify that \(x \circ y = y \circ x\) for all \(x, y \in G\).

We have:
\[
x \circ y = x + y + xy \quad \text{and} \quad y \circ x = y + x + yx.
\]
Since addition and multiplication of rational numbers are commutative, we have \(x \circ y = y \circ x\), so the operation is commutative.

### Conclusion
Since closure, associativity, the existence of an identity element, the existence of inverse elements, and commutativity have been proven, \((G, \circ)\) is an abelian group.

Would you like further clarification on any part of the proof? Here are some related questions:

1. How would the proof change if we considered a different set \(G\)?
2. Can you show an example of the inverse element calculation for specific values?
3. What happens if \(x = -1\)? Why is it excluded from \(G\)?
4. How does the operation \(\circ\) differ from standard addition?
5. Can this structure be generalized to other sets of numbers?

**Tip**: When proving associativity, it helps to expand both sides of the equation step by step to ensure consistency.

Let G = {q ∈Q / q ≠ -1}. Define the binary operation o on G by x o y = x + y + xy. Prove that (G, o) is an abelian group.

Explore the proof that (G, o), where G = {q ∈Q / q ≠ -1} and the binary operation x o y = x + y + xy, forms an abelian group. Learn about closure, associativity, identity elements, inverse elements, and commutativity. Suitable for undergraduate mathematics students.

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