# Proof

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:

### $$(i)$$ $$y=(y_n)_{n\in\mathbb N} + I$$ is a real number.

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.

### $$(ii)$$ $$y$$ is the multiplicative inverse of $$x$$.

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 $$x\cdot y=1\in\mathbb R$$.

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### References

#### Bibliography

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