# Proof

By the definition of rational numbers, a rational number $$x\in\mathbb Q$$ can be represented by a pair of integers $$a,b\in\mathbb Z$$: $x:=\frac ab,\quad a,b\in\mathbb Z,~b\neq 0.$

Let $$x\neq 0$$, i.e. let $$x$$ not be the rational zero. Therefore, the integer $$a$$ does not equal the integer number zero $$0_{\in\mathbb Z}$$. It is now obvious that we can use the integers $$a,b$$ to define a new rational number

$y:=\frac ba,\quad a,b\in\mathbb Z,~a\neq 0.$

We will show that $$x^{-1}=y$$, i.e. that $$y$$ is an inverse with respect to the multiplication of rational numbers of the rational number $$x$$. We will do so applying the commutativity of multiplying real numbers, and the existence of the rational number one:

$\begin{array}{rcll} x \cdot y&=&\frac ab\cdot\frac ba&\text{by definition of rational numbers}\\ &=&\frac {ab}{ba}&\text{by definition of multiplying rational numbers}\\ &=&\frac {ab}{ab}&\text{by commutativity of multiplying integers numbers}\\ &=&1&\text{by the definition of rational one and because from }(a\neq 0\text{ and }b\neq 0)\text{ it follows } ab\neq 0. \end{array}$

The last conclusion $\text{[From }(a\neq 0\text{ and }b\neq 0)\text{ it follows } ab\neq 0\text{"}$ is a consequence of the fact that integers form an integral domain. This completes the proof.

Github: ### References

#### Bibliography

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