Integers under Addition form Abelian Group

Theorem
The set of integers under addition $\left({\Z, +}\right)$ forms an abelian group.

Proof
From the definition of the integers, the algebraic structure $\left({\Z, +}\right)$ is an isomorphic copy of the inverse completion of $\left ({\N, +}\right)$.

As the Natural Numbers are a Naturally Ordered Semigroup, it follows that:
 * $\left ({\N, +}\right)$ is a commutative semigroup;
 * all elements of $\left ({\N, +}\right)$ are cancellable.

The result follows from Inverse Completion of Commutative Semigroup is Abelian Group.

Thus addition on $\Z$ is well-defined, closed, associative and commutative on $\Z$.

Let us define $\left[\!\left[{\left({a, b}\right)}\right]\!\right]_\boxminus$ as in the formal definition of integers.

That is, $\left[\!\left[{\left({a, b}\right)}\right]\!\right]_\boxminus$ is an equivalence class of ordered pairs of natural numbers under the congruence relation $\boxminus$.

$\boxminus$ is the congruence relation defined on $\N \times \N$ by:
 * $\left({x_1, y_1}\right) \boxminus \left({x_2, y_2}\right) \iff x_1 + y_2 = x_2 + y_1$

In order to streamline the notation, we will use $\left[\!\left[{a, b}\right]\!\right]$ to mean $\left[\!\left[{\left({a, b}\right)}\right]\!\right]_\boxminus$, as suggested.

Identity is Zero
From Construction of Inverse Completion: Identity of Quotient Structure, the identity of $\left({\Z, +}\right)$ is $\left[\!\left[{c, c}\right]\!\right]$ for any $c \in \N$:

$\left[\!\left[{c, c}\right]\!\right]$ is the equivalence class of pairs of elements $\N \times \N$ whose difference is zero.

Thus the identity of $\left({\Z, +}\right)$ is seen to be $0$.

Note that a perfectly good representative of $\left[\!\left[{c, c}\right]\!\right]$ is $\left[\!\left[{0, 0}\right]\!\right]$. This usually keeps to a minimum the complexity of any arithmetic that is needed.

Construction of Inverses
From Construction of Inverse Completion: Invertible Elements in Quotient Structure, we see that every element of $\left({\Z, +}\right)$ has an inverse.

We can see that:

The above construction is valid because $a$ and $b$ are both in $\N$ and hence cancellable.

From Construction of Inverse Completion: Identity of Quotient Structure, $\left[\!\left[{a + b, a + b}\right]\!\right]$ is a member of the equivalence class which is the identity of $\left({\Z, +}\right)$.

Thus the inverse of $\left[\!\left[{a, b}\right]\!\right]$ is $\left[\!\left[{b, a}\right]\!\right]$.