# Proof

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.

### Application of the theorem $$( * )$$ to construct the group $$(\mathbb Z, +)$$ from the semigroup $$(\mathbb N, +)$$

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.

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### References

#### Bibliography

1. Kramer Jürg, von Pippich, Anna-Maria: "Von den natürlichen Zahlen zu den Quaternionen", Springer-Spektrum, 2013