A structure that fulfils all the previous axioms is called an ordered field.
\end{rem}
\begin{defi}
Let $A \subset\mathbb{R}$, $x \in\realn$.
\begin{enumerate}[(i)]
\item$x$ is called an upper bound of $A$ if $\forall y \in A: ~y \le x$
\item$x$ is called a maximum of $A$ if $x$ is an upper bound of $A$ and $x \in A$
\item$x$ is called supremum of $A$ is $x$ is an upper bound of $A$ and if for every other upper bound $y \in\realn$ the statement $x \le y$ holds. In other words, $x$ is the smallest upper bound of $A$.
\end{enumerate}
$A$ is called bounded above if it has an upper bound. Analogously, there exists a lower bound, a minimum and an infimum. We introduce the notation $\sup A$ for the supremum and $\inf A$ for the infimum.
\end{defi}
\begin{defi}
$a, b \in\realn$, $a < b$. We define
\begin{itemize}
\item$(a, b) :=\{x \in\realn\setvert a < x \wedge x < b\}$
\item$[a, b] :=\{x \in\realn\setvert a \le x \wedge x \le b\}$
\item$(a, \infty) :=\{x \in\realn\setvert a < x\}$
\end{itemize}
\end{defi}
\begin{eg}
$(-\infty, 1)$ is bounded above ($1$, $2$, $1000$, $\cdots$ are upper bounds), but has no maximum. $1$ is the supremum.
\end{eg}
\begin{defi}[Completeness of the real numbers]
Every non-empty subset of $\realn$ with an upper bound has a supremum.
\end{defi}
\begin{defi}
A set $A \subset\realn$ is called inductive if $1\in A$ and
\[
x \in A \implies x + 1 \in A
\]
\end{defi}
\begin{lem}
Let $I$ be an index set, and let $A_i$ be inductive sets for every $i \in I$. Then $\bigcap_{i \in I} A_i$ is also inductive.
\end{lem}
\begin{proof}
Since $A_i$ is inductive $\forall i \in I$, we know that $1\in A_i$. Therefore
\begin{equation}
1 \in\bigcap_{i \in I} A_i
\end{equation}
Now let $x \in\bigcap_{i \in I} A_i$, this means that $x \in A_i ~~\forall i \in I$.
\begin{equation}
\implies x + 1 \in A_i ~~\forall i \in I \implies x + 1 \in\bigcap_{i \in I} A_i
\end{equation}
\end{proof}
\begin{defi}
The natural numbers are the smallest inductive subset of $\realn$. I.e.
\[
\bigcap_{A \text{ inductive}} A =: \natn
\]
\end{defi}
\begin{thm}[The principle of induction]
Let $\Phi(x)$ be a statement with a free variable $x$. If $\Phi(1)$ is true, and if $\Phi(x)\implies\Phi(x +1)$, then $\Phi(x)$ holds for all $x \in\natn$.
\end{thm}
\begin{proof}
Define $A =\{x \in\realn\setvert\Phi(x)\}$. According to the assumptions, $A$ is inductive and therefore $\natn\subset A$. This means that $\forall n \in\natn: ~~\Phi(n)$.
\end{proof}
\begin{cor}
$m, n \in\natn$
\begin{enumerate}[(i)]
\item$m + n \in\natn$
\item$mn \in\natn$
\item$1\le n ~~\forall n \in\natn$
\end{enumerate}
\end{cor}
\begin{proof}
We will only proof (i). (ii) and (iii) are left as an exercise for the reader. Let $n \in\natn$. Define $A =\{m \in\natn\setvert m + n \in\natn\}$. Then $1\in A$, since $\natn$ is inductive. Now let $m \in A$, therefore $n + m \in\natn$.
\begin{align}
&\implies n + m + 1 \in\natn\\
&\iff m + 1 \in A
\end{align}
Hence $A$ is inductive, so $\natn\subset A$. From $A \subset N$ follows that $\natn= A$.
\end{proof}
\begin{thm}
$n \in\natn$. There are no natural numbers between $n$ and $n +1$.
\end{thm}
\begin{hproof}
Show that $x \in\natn\cap(1, 2)$ implies that $\natn\setminus\{x\}$ is inductive. Now show that if $\natn\cap(n, n+1)=\varnothing$ and $x \in\natn\cap(n +1, n +2)$ then $\natn\setminus\{x\}$ is inductive.
\end{hproof}
\begin{thm}[Archimedian property]
\[
\forall x \in\realn ~\exists n \in\natn: ~~x<n
\]
\end{thm}
\begin{proof}
If $x < 1$ there is nothing to prove, so let $x \ge1$. Define the set
\begin{equation}
A = \{n \in\natn\setvert n \le x\}
\end{equation}
$A$ is bounded above by definition. There exists the supremum $s =\sup A$. By definition, $s-1$ is not an upper bound of $A$, i.e. $\exists m \in A: ~~s-1 < m$. Therefore $s \le m +1$.
\begin{equation}
m \in A \subset\natn\implies m + 1 \in\natn
\end{equation}
Since $s$ is an upper bound of $A$, this implies that $m+1\not\subset A$, so therefore $m +1 > x$.
\end{proof}
\begin{cor}\label{cor:minimum}
Every non-empty subset of $\natn$ has a minimum, and every non-empty subset of $\natn$ that is bounded above has a maximum.
\end{cor}
\begin{proof}
Let $A \subset\natn$. Propose that $A$ has no minimum. Define the set
\begin{equation}
\tilde{A} := \{n \in\natn\setvert\forall m \in A: ~n < m\}
\end{equation}
$1$ is a lower bound of $A$, but according to the proposition $A$ has no minimum, so therefore $1\notin A$. This implies that $1\in\tilde{A}$.
\begin{equation}
n \in\tilde{A}\implies n < m ~\forall m \in A
\end{equation}
But since there exists no natural number between $n$ and $n+1$, this means that $n+1$ is also a lower bound of $A$, and therefore
\begin{equation}
n+1 \le m ~\forall m \in A \implies n+1 \in\tilde{A}
\end{equation}
So $\tilde{A}$ is an inductive set, hence $\tilde{A}=\natn$. Therefore $A =\varnothing$.
\end{proof}
\begin{defi}
We define the following new sets:
\begin{align*}
&\intn := \{x \in\realn\setvert x \in\natn_0 \vee (-x) \in\natn_0\}\\
\exists q \in\natn: ~~ \frac{1}{y-x} < q \left( \iff\frac{1}{q} < y - x \right)
\end{equation}
Let $p \in\intn$ be the greatest integer that is smaller than $y \cdot q$. The existence of $p$ is ensured by corollary \Cref{cor:minimum}. Then $\frac{p}{q} < y$ and
\begin{equation}
p + 1 \ge y \cdot q \implies y \le\frac{p}{q} + \frac{1}{q} < \frac{p}{q} + (y - x)
\end{equation}
\begin{equation}
\implies x < \frac{p}{q} < y
\end{equation}
\end{proof}
\begin{defi}[Absolute values]
We define the following function
\begin{align*}
|\cdot|: \realn&\longrightarrow [0, \infty) \\
x &\longmapsto
\begin{cases}
x &, x \ge 0 \\
-x &, x < 0
\end{cases}
\end{align*}
\end{defi}
\begin{thm}
\[
x, y \in\realn\implies |xy| = |x||y|
\]
\end{thm}
\begin{proof}
\reader
\end{proof}
\begin{defi}[Complex numbers]
Complex numbers are defined as the set $\cmpln=\realn^2$. Addition and multiplication are defined as mappings $\cmpln\times\cmpln\rightarrow\cmpln$. Let $(x, y), (\tilde{x}, \tilde{y})\in\cmpln$.
$\cmpln$ is a field. Let $z =(x, y)\in\cmpln$. We define
\begin{align*}
\real(z) = \Re(z) = x& ~~\text{ the real part}\\
\imag(z) = \Im(z) = y& ~~\text{ the imaginary part}
\end{align*}
\end{defi}
\begin{rem}\leavevmode
\begin{enumerate}[(i)]
\item We will not prove that $\cmpln$ fulfils the field axioms here, this can be left as an exercise to the reader. However, we will note the following statements
\item The imaginary unit is defined as $i =(0, 1)$
\[
(0, 1) \cdot (x, y) = (-y, x)
\]
Especially
\[
i^2 = (0, 1)^2 = (-1, 0) = -(1, 0) = -1
\]
\end{enumerate}
We also introduce the following notation
\[
(x, y) = (x, 0) + i\cdot(y, 0) = x + iy
\]
\end{rem}
\begin{thm}[Fundamental theorem of algebra]
Every non-constant, complex polynomial has a complex root. I.e. for $n \in\natn$, $\alpha_0, \cdots, \alpha_n \in\cmpln$, $\alpha_n \ne0$ there is some $x \in\cmpln$ such that
