(related to Proposition: Algebraic Structure of Integers Together with Addition)
A proof of this proposition is a direct application of the theorem construction of groups from commutative cancellative semigroups \(( * )\), in which we will construct the group \((\mathbb Z, +)\) from the semigroup \((\mathbb N, +)\).
We first have to check, if all preconditions necessary to apply the above-mentioned theorem are fulfilled. We do this by showing that \((\mathbb N, +)\) is a commutative and a cancellative semigroup. Please note that this follows immediately from the corresponding proposition stating that \((\mathbb N, +)\) is, in fact, a special case of a commutative and cancellative monoid. Since by definition every monoid is also a semigroup, all preconditions to apply theorem \(( * )\) are fulfilled.
In the proof of the theorem \(( * )\), a group \((G,\ast)\) is constructed from a semigroup \((H,\circ)\) as the set of all equivalence classes \([a,b]\), which are represented by ordered pairs \((a,b)\in H\times H\). In this proof, the operation \(\ast\) is defined as
\[[a,b]\ast[c,d]:=[a\circ c,b\circ d].\]
In our special case, \((H,\circ)=(\mathbb N, + )\), and the operation \(\ast\) is given by
\[[a,b]\ast[c,d]=[a,b] + [c,d]:=[a + c,b + d],\]
in which, for simplicity reasons and in accordance with usual notation of addition of integers, we replace the operation \(\ast\) (general case) by \( +\) (special case).
Furthermore, according to the proof of \(( * )\), the semigroup \((\mathbb N, + )\) is isomorphic to a proper subset \(S\subset G\) with (using our new notation) \[S:=\{[a+h,h]~|~a\in\mathbb N\}.\] Please note, that because \((\mathbb N, + )\) is cancellative and a monoid, we can even write \[S=\{[a+h,h]~|~a\in\mathbb N\}=\{[a+\cancel h,\cancel h]~|~a\in\mathbb N\}=\{[a+0,0]~|~a\in\mathbb N\}=\{[a,0]~|~a\in\mathbb N\}.\] Thus, we have just identified all natural numbers \(a\in\mathbb N\) with the equivalence classes \([a,0]\in G\) , represented by the ordered pair \((a,0)\in\mathbb N\times \mathbb N\).
We also learn from the proof of theorem \(( * )\) that for each \([a,0]\in G\) there is a unique inverse element \([0,a]\in G\), i.e. an element fulfilling the equation \[[a,0]+[0,a]=[a+0,0+a]=[a,a]=[\cancel a,\cancel a]=[0,0],\] applying the cancellation and monoid properties of \(\mathbb N\) once again and showing us that \([0,0]\) is the identity element of the group \(G\).
Altogether, we can now introduce the set of integers \((\mathbb Z, + )\) as follows:
\[\begin{array}{ccc} a\in \mathbb Z&\Longleftrightarrow&[a,0]\in G~\forall a\in\mathbb N, a\neq 0\\ - a\in \mathbb Z&\Longleftrightarrow&[0,a]\in G~\forall a\in\mathbb N, a\neq 0\\ 0\in \mathbb Z&\Longleftrightarrow&[0,0]\in G,~0\in\mathbb N. \end{array}~~~~~~~(**)\] For simplicity reasons, we write \(a-b\) instead of \(a + (-b)\).
Finally, we have to prove that there are no other elements in \(\mathbb Z\), instead of those listed in the three cases \(( * *)\). We can do so by demonstrating that any equivalence class \([a,b]\in G\) with both \(a\neq 0\) and \(b\neq 0\) can be identified with one of the integers already defined in \(( * *)\). Assume \(a -b\in\mathbb N\). Then we can identify \([a,b]\in G\) with \[[a,b]=[a-b,b-b]=[a-b,0],\] which shows us that the equivalence class \([a,b]\) is independent of the representative, since we have replaced the representative \((a,b)\in\mathbb N\times\mathbb N\) by another representative \((a-b,0)\in\mathbb N\times\mathbb N\). In a similar way, we can identify \([a,b]\) with \([0,b-a]\), if we assume \(b -a\in\mathbb N\). This completes the proof.
