assignments/main_text.tex

@@ -1,6 +1,5 @@
 \documentclass{template/homework}
 
-\setcounter{secnumdepth}{0} % Makes sections unnumbered
 \usepackage{enumitem} % Gives access to better enumeration items
 \usepackage{tcolorbox} % Gives boxes
 \usepackage{subfiles} % Makes subfiles easier (and you want those)

assignments/week1/main.tex

@@ -1,7 +1,8 @@
 \documentclass[../main_text.tex]{subfiles}
 \begin{document}
+\setcounter{exercise}{0}
 
-\section{Assignment 1}
+\part{Assignment 1}
 
 \exercise*[0.3.6]
 % For some reason I can't put a fitted tcbox here and I really don't like it
@@ -132,7 +133,7 @@ So each set $A_i := \{i\}  \forall  i \in \N$. Since $\N$ is countably infinite,
 Each set is definitely finite, because they all contain just one element.
 Finally, the union of the collection of sets is equal to $\N$, which is not a finite set.
 
-\exercise[6]
+\exercise*
 \begin{tcolorbox}
     \begin{enumerate}[label=\emph{\alph*)}, wide]
         \item Compute $f(4/15)$. Find $q$ such that $f(q) = 108$.

assignments/week2/main.tex

@@ -1,7 +1,8 @@
 \documentclass[../main_text.tex]{subfiles}
 \begin{document}
+\setcounter{exercise}{0}
 
-\section{Assignment 2}
+\part{Assignment 2}
 
 \exercise*[1.1.1]
 \begin{tcolorbox}
@@ -17,7 +18,7 @@ Working out the right side with the distributive law, gives $0 < (-x*z)-(-x*y)$.
 Using $-1*-1 = 1$, gives $0 < (-xz)-(-xy)$, thus $0 < xy - xz$. The right part can be split again: $xz < xy$.
 Then, the < can be flipped, which gives $xy > xz$. \qed
 
-\exercise[1.1.2]
+\exercise*[1.1.2]
 \begin{tcolorbox}
     Let $S$ be an ordered set. Let $A \subset S$ be a non-empty finite subset. Then A is bounded. Furthermore,
     $\text{inf} A$ exists and in in $A$ and $\sup A$ exists and is in $A$.
@@ -45,7 +46,7 @@ of $A$ in $A$ \footnote{It might even be that I have already proven that $A$ has
 Then that's also good enough to show that $A$ is bounded, since in order for $A$ to have a supremum,
 it must also be bounded.}. This will be left to the reader.
 
-\exercise[1.1.5]
+\exercise*[1.1.5]
 \begin{tcolorbox}
     Let $S$ be an ordered set. Let $A \subset S$ and suppose $b$ is an upper bound for $A$. Suppose $b \in A$.
     Show that $b = \sup A$.
@@ -63,7 +64,7 @@ not an upper bound of $A$ and thus also not its supremum. If $c > b$, then $b$ i
 $c$ and therefore $c$ cannot be the supremum. The only option left is that $c=b$, and therefore $b = \sup A$.
 \qed
 
-\exercise[1.1.6]
+\exercise*[1.1.6]
 \begin{tcolorbox}
     Let $S$ be an ordered set. Let $A \subset S$ be nonempty and bounded above. Suppose $\sup A$ exists
     and $\sup A \notin A$. Show that $A$ contains a countably infinite subset.
@@ -79,7 +80,7 @@ $a < b$ and $x < a $ $\forall x \in A$. So $b$ is in fact not the least upper bo
 which is in contradiction with our assumption earlier. Ergo, $|X| \geq |\N|$. Since $X$ then is at least of the same
 cardinality as $\N$, it must also contain a countably infinite subset. \qed
 
-\exercise[1.2.7]
+\exercise*[1.2.7]
 \begin{tcolorbox}
     Prove the arithmetic-geometric mean inequality. That is, for two positive real numbers $x, y$, we have
     \begin{equation*}
@@ -103,7 +104,7 @@ Now to prove the second statement. We assume $x = y$ is a positive real number.
     \sqrt{xy}=\sqrt{x^2}=x=\frac{2x}{2}=\frac{x + y}{2}. \qed
 \end{equation*}
 
-\exercise[1.2.9]
+\exercise*[1.2.9]
 \begin{tcolorbox}
     Let $A$ and $B$ be two nonempty bounded sets of real numbers. Define the set $C := \{a + b : a \in A,
     b \in B\}$. Show that $C$ is a bounded set and that
@@ -133,7 +134,7 @@ $y$ can be given by the supremum; $x \leq \sup A$ and $y \leq \sup B$. So, $\sup
 
 A similar argument can be given for the infimum, which is left to the reader.
 
-\exercise[7]
+\exercise*
 \begin{tcolorbox}
     Let
     \begin{equation*}

assignments/week3/main.tex

@@ -1,8 +1,34 @@
 \documentclass[../main_text.tex]{subfiles}
 \begin{document}
+\setcounter{exercise}{0}
 
-\section{Assignment 3}
+\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}