# Proof

We have to show that $$(\mathbb Z, +, \cdot)$$, i.e. the set of integers $$\mathbb Z$$, together with the operations addition "$$+$$" and multiplication "$$\cdot$$" forms an integral domain. We will do so by demonstrating, that $$(\mathbb Z, +, \cdot)$$ is a commutative ring, which is not the zero ring and in which the integer $$0$$ is the only zero divisor.

In order to show that $$(\mathbb Z, +,\cdot)$$ is a commutative ring, we check the following properties:

• $$(\mathbb Z, + )$$ is a commutative group. This has been shown in a corresponding proposition.
• $$(\mathbb Z, \cdot )$$ is a commutative monoid. This requires the following (sub)properties:
• $$(\mathbb Z, +, \cdot)$$ is distributive, which has been shown in the corresponding proposition. We have just shown that $$(\mathbb Z, + ,\cdot)$$ is a commutative ring with $$1:=[1,0]$$ as the identity of the commutative monoid $$(\mathbb Z,\cdot)$$. Obviously, $$(\mathbb Z, + ,\cdot)$$ is not the zero ring, because $$1\in\mathbb Z$$. It remains to be shown that $$0_{\in\mathbb Z}:=[0_{\in\mathbb N},0_{\in\mathbb N}]$$ is the only zero divisor.

Suppose, there would be other (but $$0$$) zero divisors, say $$x,y\in\mathbb Z$$ with $$x\neq 0$$, $$y\neq 0$$, and still $$x\cdot y=0$$. By definition of integers, suppose $$x=[a,b],y=[c,d]$$ for some natural numbers $$a,b,c,d\in\mathbb N$$. By definition of multiplying integers, we also have that $\begin{array}{ccl} x\cdot y=[a,b] \cdot [c,d] := [ac + bd, ad + bc]. \end{array}$

Because both $$x$$ and $$y$$ are not equal $$0$$, we have that $$[a,b]\neq[0,0]$$ and $$[c,d]\neq[0,0]$$, which means that $$a$$ and $$b$$ are not simultaneously equal $$0$$, and that $$c$$ and $$d$$ are not simultaneously equal $$0$$. There are $$7$$ possibilities for this and we have to check, in which cases the natural numbers $$ac + bd$$ and $$ad + bc$$ both equal zero:

$$a$$ $$b$$ $$c$$ $$d$$ $$ac$$ $$bd$$ $$ad$$ $$bc$$ $$ac+bd$$ $$ad+bc$$
$$\neq 0$$ $$\neq 0$$ $$\neq 0$$ $$\neq 0$$ $$\neq 0$$ $$\neq 0$$ $$\neq 0$$ $$\neq 0$$ $$\neq 0$$ $$\neq 0$$
$$= 0$$ $$\neq 0$$ $$\neq 0$$ $$\neq 0$$ $$=0$$ $$\neq 0$$ $$= 0$$ $$\neq 0$$ $$\neq 0$$ $$\neq 0$$
$$\neq 0$$ $$=0$$ $$\neq 0$$ $$\neq 0$$ $$\neq 0$$ $$=0$$ $$\neq 0$$ $$= 0$$ $$\neq 0$$ $$\neq 0$$
$$= 0$$ $$\neq 0$$ $$= 0$$ $$\neq 0$$ $$=0$$ $$\neq 0$$ $$= 0$$ $$=0$$ $$\neq 0$$ $$= 0$$
$$= 0$$ $$\neq 0$$ $$\neq 0$$ $$= 0$$ $$=0$$ $$= 0$$ $$= 0$$ $$\neq 0$$ $$= 0$$ $$\neq 0$$
$$\neq 0$$ $$= 0$$ $$= 0$$ $$\neq 0$$ $$=0$$ $$=0$$ $$\neq 0$$ $$=0$$ $$= 0$$ $$\neq 0$$
$$\neq 0$$ $$= 0$$ $$\neq 0$$ $$= 0$$ $$\neq 0$$ $$=0$$ $$=0$$ $$=0$$ $$\neq 0$$ $$= 0$$

We have seen that in all $$7$$ cases $\begin{array}{ccl} [ac + bd, ad + bc]\neq[0,0], \end{array}$ which demonstrates that the product of integers $$x\cdot y$$ is never equal (integer) zero. This is a contradiction to our hypothesis $$x\cdot y=0$$. 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