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+++ b/assignments/week1/1.tex
@@ -4,6 +4,7 @@
 \section*{Week 1}
 
 \exercise*[0.3.6]
+% For some reason I can't put a fitted tcbox here and I really don't like it
 \begin{tcolorbox}
     Prove:
     \begin{enumerate}[label=\emph{\alph*)}]
@@ -48,10 +49,8 @@
 \end{enumerate}
 
 
-\exercise[0.3.11]
-\begin{tcolorbox}
-    Prove by induction that $n < 2^n$ for all $n \in \N$.
-\end{tcolorbox}
+\exercise*[0.3.11]
+\tcbox{Prove by induction that $n < 2^n$ for all $n \in \N$.}
 
 For this proof we will use induction. For this, we have to prove the base case, i.e. $n = 1$,
 and the inductive step, $n < 2^n \implies n + 1 < 2^{n+1}$.
@@ -65,10 +64,8 @@ Since $m \geq 1$, $m + 1 < 2m$, and thus $m + 1 < 2m < 2^{m+1}$.
 Since both the base case and inductive step hold, we can close the induction, proving the proposition. \qed
 
 
-\exercise[0.3.12]
-\begin{tcolorbox}
-    Show that for a finite set $A$ of cardinality $n$, the cardinality of $\mathcal{P}(A)$ is $2^n$.
-\end{tcolorbox}
+\exercise*[0.3.12]
+\tcbox{Show that for a finite set $A$ of cardinality $n$, the cardinality of $\mathcal{P}(A)$ is $2^n$.}
 
 The power set of a set $A$, $\mathcal{P}(A)$, is defined as the set of all possible subsets of $A$.
 This is very similar to an inclusion/exclusion problem.
@@ -92,4 +89,49 @@ Since this doubles the number of subsets, the cardinality of $\mathcal{P}(C)$ is
 
 Both the base case and the inductive step hold, which closes the induction and proves the proposition. \qed
 
+\exercise*[0.3.15]
+\tcbox{Prove that $n^3 + 5n$ is divisible by 6 for all $n \in \N$.}
+
+In order to prove this proposition, we will use induction. To do this, we need to prove the following lemma,
+of which we will see the usefulness later:
+
+\begin{lemma}
+    \label{lem:div6}
+    $3n^2 + 3n + 6$ is divisible by 6 for all $n \in \N$.
+\end{lemma}
+
+This lemma we will also prove by induction. For this, we prove the base case and the inductive step.
+First, for the base case we have $n = 1$, yielding $3 * 1^2 + 3 * 1 + 6 = 12$, which is divisibly by 6.
+
+Then, for the inductive step we assume that the lemma holds for a certain $m \in \N$.
+So, $3m^2 + 3m + 6$ is divisible by 6. 
+Substituting $m$ with $m+1$ gives $3(m+1)^2 + 3(m+1) + 6$, which can be expanded to
+$3m^2 + 9m + 12$. Rewriting this with our assumption in mind gives the following:
+$(3m^2 + 3m + 6) + (6m + 6)$. We know from our assumption that the first part is divisible by 6, and since $m \in \N$,
+$6m + 6$ is also divisible by 6, and so the whole expression is as well. \qed
+
+Now for the original proposition. We will prove this by induction. First we prove the base case, where $n=1$.
+Then, $1^3 + 5*1 = 6$, which is definitely divisible by 6.
+
+For the inductive step, we assume that the proposition holds for a certain $m \in \N$. So, $m^3 + 5m$ is divisible by 6.
+When we increase $m$ by 1, we get: $(m+1)^3 + 5(m+1)$.
+Expanded, this is the same as $m^3 + 3m^2 + 8m + 6$. When we rearrange the terms we can get the following expression:
+$(m^3 + 5m) + (3m^2 + 3m + 6)$. From Lemma \ref{lem:div6}, we know that the latter part is divisible by 6.
+The prior part is divisible by 6 because of the assumption of the inductive step. So together, this expression
+is also divisible by 6. \qed
+
+
+\exercise*[0.3.19]
+\begin{tcolorbox}
+    Give an example of a countably infinite collection of finite sets $A_1, A_2,...$,
+    whose union is not a finite set.
+\end{tcolorbox}
+
+The easiest example is simply the collection of singleton sets containing a natural number.
+So each set $A_i := \{i\}  \forall  i \in \N$. Since $\N$ is countably infinite, so the collection of sets.
+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*
+
 \end{document}