Merge pull request 'Finished assignment 2' (#13) from assignment2 into master

Reviewed-on: #13

assignments/week1/main.tex

@@ -132,7 +132,7 @@ 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*
+\exercise*[6]
 \begin{tcolorbox}
     \begin{enumerate}[label=\emph{\alph*)}, wide]
         \item Compute $f(4/15)$. Find $q$ such that $f(q) = 108$.

assignments/week2/main.tex

@@ -4,6 +4,187 @@
 \section{Week 2}
 
 \exercise*[1.1.1]
-Exercise 1.1.1 comes here.
+\begin{tcolorbox}
+    Let $F$ be an ordered field and $x,y,z \in F$. If $x < 0$ and $y < z$, then $xy > xz$.
+\end{tcolorbox}
 
+So let's assume the premise. $F$ is an ordered field and $x,y,z \in F$,
+and we choose $x,y$ and $z$ such that $x < 0$ and $y < z$.
+
+From $x < 0$ it follows that $(-x) > 0$. From $y < z$ it follows that $0 < z - y$.
+From both of these, we can conclude that $0 < (-x)(z-y)$.
+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]
+\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 $A$.
+\end{tcolorbox}
+
+In order to prove that $A$ is bounded, we have to prove that it has an upper and a lower bound. Let us prove
+that $A$ is bounded above first. 
+
+In particular, we have to prove that $\exists a \in A$ such that $x \leq b$ for all $x \in E$.
+Since $A$ is non-empty and finite, we can use induction on the cardinality of $A$, since that will always
+be some natural number $n$. So, we have to prove two cases: the base case, where $|A|=1$, and the inductive step,
+where we will assume that when $A$ has an upper bound when it has cardinality $m$,
+then it also has an upper bound when its cardinality is equal to $m+1$.
+
+The base case is quite simple; if $A = \{x\}$, then $x$ is the greatest element and $A$ has an upper bound.
+Now for the inductive step. We assume that for some set $B \subset S$ with cardinality $m$,
+$B$ is bounded above. Thus, there is some $b \in B$ such that $b$ is greater than all other
+elements in $B$. Now, let's add a new element $h \in S$ to $B$, such that $h$ is distinct from all elements already in
+$B$ and the cardinality of $B$ is now $m+1$. Then, since $S$ is well ordered, we can compare $h$ also to $b$.
+Either $h$ is greater than this $b$, in which case $h$ is the new greatest element, or it is less than $b$, in which
+case $b$ stays the greatest element of $B$. In both cases however, $B$ remains bounded above. \qed
+
+A similar argument can be made to prove the existence of the lower bound, the supremum of $A$ in $A$ and the infimum
+of $A$ in $A$ \footnote{It might even be that I have already proven that $A$ has a supremum present in $A$.
+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]
+\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 $b = \sup A$.
+\end{tcolorbox}
+
+So, let $S$ be an ordered set, with $A \subset S$ and $b \in A$ being an upper bound for $A$. Since $b$ is an upper
+bound, $a \leq b$ for all $a \in A$. Since $b \in A$ as well, we know that there is some element in $A$ which is
+the greatest element of them all, and all other elements are smaller. 
+
+Now let's assume that $b \neq \sup A$. Then either some other element of $A$ is the supremum,
+which would imply that $b$ is not larger than this element, which is a contradiction.
+The other possibility is that there is an element $c \in S\backslash A$ that is the
+supremum. Because $S$ is ordered, $c$ must either be greater than, smaller than or equal to $b$. If $c < b$, c is 
+not an upper bound of $A$ and thus also not its supremum. If $c > b$, then $b$ is an upper bound that is smaller than
+$c$ 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]
+\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.
+\end{tcolorbox}
+
+Let $S$ be an ordered set, with $A \subset S$ nonempty and bounded above. We assume that $b = \sup A$ exists and
+$b \notin A$ $(\implies b \in S\backslash A)$. We are asked to show this then implies that $\exists X \subset A$
+such that $|X| \geq |\N|$. We will prove this with a proof by contradiction.
+
+We assume that no such set $X$ exists, i.e. $|X| < |\N|$. So, $A$ also doesn't have to countably infinite anymore.
+Since $b \notin A$ and $A$ is ordered, finite and nonempty, there is a greatest element $a \in A$ such that
+$a < b$ and $x < a $ $\forall x \in A$. So $b$ is in fact not the least upper bound of $A$,
+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]
+\begin{tcolorbox}
+    Prove the arithmetic-geometric mean inequality. That is, for two positive real numbers $x, y$, we have
+    \begin{equation*}
+        \sqrt{xy} \leq \frac{x + y}{2}.
+    \end{equation*}
+    Furthermore, equality occurs if and only if $x = y$.
+\end{tcolorbox}
+
+Let us prove the first statement first. So we let $x,y \in \R$ such that $x,y > 0$. Then we will prove the statement
+by contradiction. Hence, we assume that
+\begin{equation*}
+    \sqrt{xy} > \frac{x + y}{2}.
+\end{equation*}
+
+We can multiply both sides with 2. This results in $2\sqrt{xy} > x + y$. We can pull the left part into the right,
+so we get $0 > x - 2\sqrt{xy} + y$. We can restructure the right side to $0 > (\sqrt{x} - \sqrt{y})^2$.
+We know that $0 \leq z^2$, $\forall z \in \R$, so this is a contradiction. \qed
+
+Now to prove the second statement. We assume $x = y$ is a positive real number. Then,
+\begin{equation*}
+    \sqrt{xy}=\sqrt{x^2}=x=\frac{2x}{2}=\frac{x + y}{2}. \qed
+\end{equation*}
+
+\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
+    \begin{align*}
+        \sup C &= \sup A+\sup B \; \text{and}\\
+        \inf C &= \inf A+\inf B.
+    \end{align*}
+\end{tcolorbox}
+
+First, let us show that $C$ is a bounded set. Since $A$ and $B$ are both subsets of $\R$, which is an ordered field, 
+all elements of $C$ must also be real numbers. Let $a$ be an upper bound for $A$ and $b$ be an upper bound for $B$.
+So $x \leq a \; \forall x \in A$ and $y \leq b \; \forall y \in B$. Since $C$ is defined as the sum of any element in
+$A$ with any element in $B$, an upper bound of $C$, $c$, can be found as $c \leq a + b$. A similar argument can be
+made for the lower bound of $C$, which makes $C$ bounded. \qed
+
+To prove that $\sup C = \sup A + \sup B$, we will show that $\sup C \geq \sup A + \sup B$ and
+$\sup C \leq \sup A + \sup B$. 
+
+Let $a = \sup A$ and $b = \sup B$. So $x \leq a$ for all $x \in A$ and $y \leq b$ for all $y \in B$. Then,
+$x + y \leq a + b$. Since $z \leq x + y$ for all $z \in C$ because of the definition of $C$,
+$z \leq a + b$. In other words, $\sup C \leq \sup A + \sup B$.
+
+Now to prove the other direction. Let $c = \sup C$. So $z \leq c$ for all $c \in C$. Since all elements in $C$ are
+the sum of an element $x \in A$ and $y \in B$, $x + y \leq c$ for all $x,y$. The least upper bound for these $x$ and
+$y$ can be given by the supremum; $x \leq \sup A$ and $y \leq \sup B$. So, $\sup A + \sup B \leq c \implies
+\sup A + \sup B \leq \sup C$, completing the equality. \qed
+
+A similar argument can be given for the infimum, which is left to the reader.
+
+\exercise*[7]
+\begin{tcolorbox}
+    Let
+    \begin{equation*}
+        E = \{ x \in \R : x > 0 \; \text{and} \; x^3 < 2\}.
+    \end{equation*}
+
+    \begin{enumerate}[label=\emph{\alph*)}]
+        \item Prove that $E$ is bounded above.
+        \item Let $r = \sup E$ (which exists by part a)). Prove that $r > 0$ and $r^3=2$.
+
+            \textit{Hint:} Adapt the proof used in Example 1.2.3.
+    \end{enumerate}
+\end{tcolorbox}
+
+So, let $E$ and $r$ be defined as in the exercise statement. Then:
+\begin{enumerate}[label=\emph{\alph*)}]
+    \item $x \leq 2$ is an upper bound for $E$, as $2 * 2 * 2 = 8$. So, $E$ is bounded above.
+    \item As $1 \in E$, $r \geq 1 > 0$, so the first part of the statement holds. In order to show that $r^3=2$,
+        we want to show that $r^3 \leq 2$ and $r^3 \geq 2$ hold.
+
+        First, let's show that $r^3 \geq 2$. We will take a similar approach as in Example 1.2.3 from the textbook.
+        So, take a positive number $s$ such that $s^3 < 2$. We wish to find an $h > 0$ such that $(s + h)^3 < 2$.
+        As $2 - s^3 > 0$, we have $\frac{2 - s^3}{3s^2 + 3s + 1} > 0$. Choose an $h \in \R$ such that $0 < h <
+        \frac{2 - s^3}{3s^2+3s+1}$. Furthermore, assume $h < 1$. Estimate,
+        
+        \begin{align*}
+            (s+h)^3 - s^3 &= h(3s^2 + 3sh + h^2) \\
+                          &< h(3s^2 + 3s + 1) \quad &\text{(since } h < 1) \\
+                          &< 2 - s^3 \quad &\text{(since } h < \frac{2 - s^3}{3s^2 + 3s + 1}).
+        \end{align*}
+
+        Therefore, $(s+h)^3 < 2$. Hence $s + h \in E$, but as $h > 0$, we have $s + h > s$. So $s < r = \sup E$.
+        As $s$ was an arbitrary positive number such that $s^3 < 2$, it follows that $r^3 \geq 2$.
+
+        Now take an arbitrary positive number $s$ such that $s^3 > 2$. We wish to find an $h > 0$ such that 
+        $(s-h)^3 > 2$ and $s - h$ is still positive. As $s^3 - 2 > 0$, we have that $\frac{s^3 - 2}{3s^2+1} > 0$.
+        Let $h := \frac{s^3-2}{3s^2+1}$, and check that $s - h = s - \frac{s^3-2}{3s^2+1} = 
+        \frac{2s^3+s+2}{3s^2+1} > 0$. Assume that $h < 1$. Estimate,
+
+        \begin{align*}
+            s^3 - (s-h)^3 &= h(3s^2 - 3sh + h^2) \\
+                          &< h(3s^2+h^2) \quad &\text{(since } s>0 \; \text{ and } h > 0) \\
+                          &< h(3s^2+1) \quad &\text{(since } h < 1) \\
+                          &= s^3 - 2 &\text{(because of the definition of } h).
+        \end{align*}
+
+        By subtracting $s^3$ from both sides and multiplying by -1, we find $(s-h)^3>2$. Therefore, $s-h \notin E$.
+        Moreover, if $x \geq s-h$, then $x^3 \geq (s-h)^3 > 2$ (as $x > 0$ and $s-h > 0$) and so $x \notin E$. Thus,
+        $s - h$ is an upper bound for $E$. However, $s-h < s$, or in other words, $s > r = \sup E$. 
+        Hence, $r^3 \leq 2$.
+
+        Together, $r^3 \geq 2$ and $r^3 \leq 2$ imply $r^3 = 3$. \qed
+\end{enumerate}
 \end{document}