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.