(related to Proposition: Existence of Inverse Rational Numbers With Respect to Multiplication)
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.
