(related to Proposition: Existence of Inverse Real Numbers With Respect to Multiplication)
We have to show that each real number, which does not equal zero, formally \(x\neq 0\), has an inverse element \(x^{-1}:=y\) such that the multiplication of both numbers equals one, formally \(x\cdot y=1\).
Let \(x\) be represented by the rational Cauchy sequence \((x_n)_{n\in\mathbb N}\) and let \(I\) be the set of all rational Cauchy sequences convergent to \(0\in\mathbb Q\), in other words, \(x=(x_n)_{n\in\mathbb N} + I\).
Then, by hypothesis, \((x_n)_{n\in\mathbb N}\not\in I\). Following the definition of convergence of rational sequences, this means that there are numbers \(\epsilon_0 > 0, \epsilon_0\in\mathbb Q\) and \(N(\epsilon_0)\in\mathbb N\) such that \(|x_n|>\epsilon_0\) for all \(n > N(\epsilon_0)\). Therefore, \(|x_n|\neq 0\) for sufficiently large indices \(n > N(\epsilon_0)\). With this result, we define a rational sequence \((y_n)_{n\in\mathbb N}\) with
\[y_n:=\begin{cases} 0,&\quad 0\le n\le N(\epsilon_0)\\ \frac 1{x_n},&\quad n > N(\epsilon_0). \end{cases}\quad\quad\quad( * )\] We will now show:
Equivalently, we have to show that \((y_n)_{n\in\mathbb N}\) is a rational Cauchy sequence. For \(n,m > N(\epsilon_0)\) we have \[|y_n - y_m|=\left|\frac 1{x_n}-\frac 1{x_m}\right|=\frac {|x_n - x_m|}{|x_n\cdot x_m|} < \frac {|x_n - x_m|}{\epsilon_0^2}.\] Because \((x_n)_{n\in\mathbb N}\) is a rational Cauchy sequence by hypothesis, for any arbitrarily small \(\epsilon > 0, \epsilon \in\mathbb Q\) there exists an \(N(\epsilon_0^2\cdot \epsilon)\) such that for all \(n,m > N(\epsilon_0^2\cdot \epsilon)\) we have \[|x_n - x_m| < \epsilon_0^2\cdot \epsilon.\] Thus, for all \(n,m > \max(N(\epsilon_0),N(\epsilon_0^2\cdot \epsilon))\) we have \[|y_n - y_m| < \frac {|x_n - x_m|}{\epsilon_0^2} < \frac {\epsilon_0^2\cdot \epsilon}{\epsilon_0^2}=\epsilon,\] which proves that \((y_n)_{n\in\mathbb N}\) is a rational Cauchy sequence, and \(y\) is a real number.
By the definition \( ( * ) \) above, we have \[x_n\cdot y_n=\begin{cases} 0,&\quad 0\le n\le N(\epsilon_0)\\ 1,&\quad n > N(\epsilon_0). \end{cases}\] Therefore, \[((x_n)_{n\in\mathbb N} + I)\cdot((y_n)_{n\in\mathbb N} + I)=(x_n\cdot y_n)_{n\in\mathbb N} + I=(1)_{n\in\mathbb N} + I\quad\quad\text{for all }n > N(\epsilon_0).\] This means that all except the first finitely many \(N(\epsilon_0)\)) rational Cauchy sequence members \(x_n\cdot y_n\) equal \(1\in\mathbb Q\), and therefore \(x\cdot y=1\in\mathbb R\).
