\documentclass[../main_text.tex]{subfiles}
\begin{document}
\setcounter{exercise}{0}

\part{Assignment 3}

\exercise*
\begin{tcolorbox}
    Suppose $x,y \in \R$ and $x < y$. Prove that there exists $i \in \R\backslash\Q$ such that $x < i < y$.
\end{tcolorbox}

If either $x$ or $y$ (or both) are not rational numbers, we can simply take the average like so: $\frac{x+y}{2}$,
in a similar way we did for the rationals. Since $x$ or $y$ isn't rational, the resulting fraction will also not be
a rational and this proves the statement.

Now if $x,y \in \Q$, we cannot use this average trick, because the resulting fraction will be a rational itself and
so it doesn't satisfy the restriction that it must be in $\R\backslash\Q$. So we have to take a different approach.

Let $x,y \in \Q$ with $x < y$ and $m := \frac{x+y}{2}$, so $x < m < y$. Then, let $X = \{a \in \R : x < a < m\}$ and
let $Y = \{b \in \R : m < b < y\}$. Since $x < m$ and $m < y$, these are nonempty and they are bounded, because
of the restrictions $x < a < m$ and $m < b < y$. So, there exists $k \in X$, that is not rational such that
$x < k < m$ and there exists $h \in Y$ that is not rational such that $m < h < y$. Pick either $k$ or $h$ as $i$,
since $x < k < m < h < y$. \qed

\exercise*
\begin{tcolorbox}
    Let $E \subset (0,1)$ be the set of all real numbers with decimal representation using only the digits 1 and 2:
    \begin{equation}
        E := \{x \in (0,1) : \forall j \in \N, \exists d_j \in \{1,2\} \text{ such that } x = 0.d_1d_2...\}
    \end{equation}
    Prove that $|E| = |\mathcal{P}(\N)|$.
\end{tcolorbox}

\end{document}
