\item$x$ is the supremum of $A$ if $x$ 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 $A$.
$A$ 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.
\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 $\forall n \in\natn: ~~\Phi(n)$.
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 $A \subset\natn$ follows that $\natn= A$.
\item
Let $n \in\natn$. Define $A =\set[mn \in\natn]{m \in\natn}$. Then $1\in A$ because of M2.
Now let $m \in A$, therefore $nm \in\natn$
\begin{align}
\implies&(m + 1)n = mn + n \in\natn\\
\iff&m + 1 \in A
\end{align}
Hence $A$ is inductive, so $\natn\subset A$. From $A \subset\natn$ follows that $\natn= A$.
\item
Define $A =\set[1\le n]{n \in\natn}$. Then $1\in A$ since $1\le1$. Now let $n \in A$.
\begin{align}
&0 \le 1 \implies n \le n + 1 \implies 1 \le n + 1
\iff&n + 1 \in A
\end{align}
Hence $A$ is inductive, so $\natn\subset A$. But since $A \subset\natn$ it must follow that $\natn= A$.
Let $x \in\natn\cap(1, 2)$, and define $A =\natn\setminus\set{x}=\set[n \ne x]{n \in\natn}$. Obviously, $1\in A$ and $2\in A$. Let $n \in A$, then
\begin{align}
&1 \le n \implies 2 \le n + 1 \\
\iff&n + 1 \in A
\end{align}
Thus $A$ is inductive. Since $A \subset\natn$ we have $A =\natn$, so there are no natural numbers between $1$ and $2$.
Now assume $\natn\cap(n, n+1)=\emptyset$, and consider $x \in\natn\cap(n +1, n +2)$. We will take another look at $A$ with this new $x$. Obviously $1\in A$.
Let $m \in A$, we want to show that $m +1\in A$. To do this, assume that $m +1\not\in A$, i.e.
\begin{equation}
m + 1 \in\natn\cap (n + 1, n + 2) \implies m \in\natn\cap (n, n + 1)
\end{equation}
However since $\natn\cap(n, n +1)=\emptyset$ this $m$ can't exist, so $m +1\in A$.
So $A$ is still inductive, and $A =\natn$ still holds.
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}
Let $p \in\intn$ be the greatest integer that is smaller than $y \cdot q$. The existence of $p$ is ensured by \Cref{cor:minimum}. Then $\frac{p}{q} < y$ and
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
