diff --git a/assignments/main_text.pdf b/assignments/main_text.pdf index 1615538..279e02f 100644 Binary files a/assignments/main_text.pdf and b/assignments/main_text.pdf differ diff --git a/assignments/main_text.tex b/assignments/main_text.tex index 442519d..ad5bebd 100644 --- a/assignments/main_text.tex +++ b/assignments/main_text.tex @@ -1,6 +1,7 @@ \documentclass{template/homework} \usepackage{enumitem} +\usepackage{tcolorbox} \usepackage{subfiles} \title{MIT OCW Real Analysis} diff --git a/assignments/template b/assignments/template index 7b85d98..de9419d 160000 --- a/assignments/template +++ b/assignments/template @@ -1 +1 @@ -Subproject commit 7b85d987143e4d5c71a12cd116811ce6d0be0bf5 +Subproject commit de9419d4cc22f8204a8d49376bd5e507efa01a43 diff --git a/assignments/week1/1.pdf b/assignments/week1/1.pdf index 41e6977..257c441 100644 Binary files a/assignments/week1/1.pdf and b/assignments/week1/1.pdf differ diff --git a/assignments/week1/1.tex b/assignments/week1/1.tex index a012809..3cada10 100644 --- a/assignments/week1/1.tex +++ b/assignments/week1/1.tex @@ -1,22 +1,239 @@ \documentclass[../main_text.tex]{subfiles} \begin{document} +\section*{Week 1} + \exercise*[0.3.6] -\begin{enumerate}[label=\emph{\alph*)}] - \item Wanting to show: - $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ +% 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*)}] + \item $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ + \item $A \cup (B \cap C) = (A \cup B) \cap (a \cup C)$ + \end{enumerate} +\end{tcolorbox} - In order to prove this equivalence, we have to prove the implication both ways. We use two lemmas for this. +\begin{enumerate}[label=\emph{\alph*)}, wide] + \item In order to prove this equivalence, we have to prove the implication both ways. We use two lemmas for this. - \begin{lemma}[test] + \begin{lemma} + \label{lem:set1} $A \cap (B \cup C) \implies (A \cap B) \cup (A \cap C)$ - - Let $x \in A \cap (B \cup C)$. \end{lemma} - \begin{definition} - Test. - \end{definition} - \item + + Let $x \in A \cap (B \cup C)$. + By the definition of set intersection, $x \in A$ and $x \in B \cup C$. + By the definition of set union, $x \in A$ and $(x \in B$ or $x \in C)$. + From propositional logic we know that for propositions $P$, $Q$ and $R$ the following holds: + $P \land (Q \lor R) \iff (P \land Q) \lor (P \land R)$. + So, substituting for this particular case yields $(x \in A$ and $x \in B)$ or $(x \in A$ and $x \in C)$. + Using the definition of set intersection again gets $x \in A \cap B$ or $x \in A \cap C$. + Using the definition of set union again gives $x \in (A \cap B) \cup (A \cap C)$. \qed + + \begin{lemma} + \label{lem:set2} + $(A \cap B) \cup (A \cap C) \implies A \cap (B \cup C)$ + \end{lemma} + + Let $x \in (A \cap B) \cup (A \cap C)$. + By the definition of set union, $x \in (A \cap B)$ or $x \in (A \cap C)$. + By the definition of set intersection, $(x \in A$ or $x \in B)$ and $(x \in A$ or $x \in B)$. + Using the same propositional logical equivalence as in Lemma \ref{lem:set1}, this gives + $x \in A$ and $(x \in B$ or $x \in C)$. + Wrapping up, we use the definition of set union to get $x \in A$ and $x \in B \cup C$ and + the definition of intersection to get $x \in A \cap (B \cup C)$. \qed + + Using Lemma \ref{lem:set1} and \ref{lem:set2}, we get the desired equivalence of + $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$. \qed + \item This proof is so similar to a) that it feels like a waste of time and will therefore be left to the reader. +\end{enumerate} + + +\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, $n < 2^n \implies n + 1 < 2^{n+1}$. + +First, let's prove the base case. When $n = 1$, we get $1 < 2^1$, which is certainly true. + +Then, for the inductive step. We assume that the proposition holds for any $m \in \N$. +So, $m < 2^m$. Multiplying both sides with 2 gives $2m < 2^{m+1}$. +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] +\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. +It is built up by all the possible combinations of the different elements being either inside a certain subset or not. +For all possible subsets of $A$, we have that for every element $x \in A$ there are 2 possibilities, +either $x$ is in the subset or it isn't. +This means that for every additional element, the number of subsets increases by a factor of 2, with a minimum of 1, +in case of $A = \emptyset$. We will prove this formally now, using induction. + +For this, the base case is a set of 1 element (but the theorem also holds for the empty set, where $n=0$). +Let us assume that $A := \{ \pi \}$. +Then the cardinality of $\mathcal{P}(A)$ is $2^1$, with $\mathcal{P}(A)= \{ \emptyset, \{ \pi \}\}$. + +For the inductive step, we assume that for any set $B$ of cardinality $m$, the cardinality of the power set of $B$ +is $2^m$. Then, we will add an element $x \notin B$ to $B$ to increase its cardinality by 1, to $m + 1$, +creating a new set $C$. +Note that all the possible subsets of $B$ are still viable subsets of $C$, since $B \subset C$. +In order to create the new subsets, we can simply keep all the subsets of $B$, duplicate them and take the union with +the new element $x$, so now we also have all combinations of the old sets with possibly $x$ being in them. +Since this doubles the number of subsets, the cardinality of $\mathcal{P}(C)$ is $2^{m+1}$. + +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* +\begin{tcolorbox} + \begin{enumerate}[label=\emph{\alph*)}, wide] + \item Compute $f(4/15)$. Find $q$ such that $f(q) = 108$. + \item Use the \textbf{Theorem} to prove that $f$ is a bijection. + \end{enumerate} +\end{tcolorbox} + +See the assignment PDF for the full assignment specification and theorem. + +\begin{enumerate}[label=\emph{\alph*)}, wide] + \item $\frac{4}{15}$, if written as a product of prime factors, is equal to $\frac{2^2}{3^1*5^1}$. + Since this fraction is not a natural number, we have to use the second part of the definition of $f$. + So, $f(q) = 2^{2*2} * 3^{2 * 1 - 1} * 5^{2 * 1 - 1} = 240$. + + For the inverse of $f$, it is still necessary to compute the factorization in prime numbers. + Using the powers of the primes we can deduce whether the prime present is, if applicable, + part of either the numerator or the denominator. + $180 = 2^2 * 3^2 * 5^1$. Because of the way $f$ is defined, we know that all the prime factors with an even + power are part of the numerator and all prime factors with an odd power are part of the denominator (except 1, + which just maps to itself). When we backtrack using this information, we then get the following fraction: + $\frac{2^1 * 3^1}{5^1} = \frac{6}{5}$. + \item In order to prove that $f$ is a bijection, we have to prove that $f$ is injective and surjective. + \begin{description} + \item[Injectivity:] We want to show that $f$ is 1-1, + i.e. $f(x_1)=f(x_2) \implies x_1 = x_2$. + + So, let's assume that for any $x_1, x_2 \in \{ q > 0 : q \in \Q \}$, $f(x_1) = f(x_2)$. + Since the function $f$ has 3 parts, based on the input, we have to prove this statement for those 3 + parts separately as well. First, the easiest case, where the input set is $\{1\}$. + Then, $f(x) = 1 \forall x$, so $f$ is injective. + + For the case where $x \in \N\backslash\{1\}$, $f(x) := p_1^{2r_1}***p_N^{2r_N}$. + We know from the \textbf{Theorem} that any fraction can be uniquely written as a product of prime + factors with exponents, so when we assume $f(x_1) = f(x_2)$, we can also assume that $x_1$ and $x_2$ + have a unique prime factorization associated with them. So let's assume that $f(x_1) = f(x_2)$. + This means that $p_1^{2r_1}***p_N^{2r_N} = q_1^{2s_1}***q_M^{2s_M}$, where $p_i^{r_i}$ and $q_j^{s_j}$ + denote the prime factors for both sides. We can further expand this expression into: + + \begin{align} + p_1^{r_1} * p_1^{r_1} *** p_N^{r_N} * p_N^{r_N} &= + q_1^{s_1} * q_1^{s_1} *** q_M^{s_M} * q_M^{s_M} \implies \\ + p_1^{r_1}***p_N^{r_N} * p_1^{r_1}***p_N^{r_N} &= + q_1^{s_1}***q_M^{s_1} * q_1^{s_1}***q_N^{s_M} \implies \\ + x_1 * x_1 &= x_2 * x_2 + \end{align} + + Because we know that each fraction constitutes a unique prime factorization, we also know that $x_1$ and + $x_2$ are uniquely derived. This is why the implications in the equation above hold. + Because both $x_1$ and $x_2 > 0$, $x_1 = x_2$. + + Now for the case where $x \in \Q\backslash\N$. + Then $f(x) := p_1^{2r_1}***p_N^{2r_N}q_1^{2s_1-1}***q_M^{2s_M-1}$, using the unique factorization + derived from the \textbf{Theorem}. So again, we assume that for any $x_1, x_2 \in \Q\backslash\N$, + $f(x_1) = f(x_2)$. + Using the definition of $f$, we get: + $p_1^{2r_1}***p_N^{2r_N}q_1^{2s_1-1}***q_M^{2s_M-1} = + v_1^{2t_1}***v_n^{2t_n}w_1^{2u_1-1}***w_m^{2u_m-1}$.\footnote{The super- and subscripts become a bit + abracadabra, but I think everything is unique and readable this way.} + Expanding this expression further, we get: + + \begin{align} + \frac{p_1^{2r_1}}{p_1}***\frac{p_N^{2r_N}}{p_N}\frac{q_1^{2s_1}}{q_1}***\frac{q_M^{2s_M}}{q_M} &= + \frac{v_1^{2t_1}}{v_1}***\frac{v_n^{2t_n}}{v_n}\frac{w_1^{2u_1}}{w_1}***\frac{w_m^{2u_m}}{w_m} + \implies \\ + \frac{p_1^{r_1}*p_1^{r_1}}{p_1}***\frac{p_N^{r_N}*p_N^{r_N}}{p_N} + \frac{q_1^{s_1}*q_1^{s_1}}{q_1}***\frac{q_M^{s_M}*q_M^{s_M}}{q_M} &= + \frac{v_1^{t_1}*v_1^{t_1}}{v_1}***\frac{v_n^{t_n}*v_n^{t_n}}{v_n} + \frac{w_1^{u_1}*w_1^{u_1}}{w_1}***\frac{w_m^{u_m}*w_m^{u_m}}{w_m} \implies \\ + \frac{x_1*x_1}{p_1***p_N*q_1***q_M} &= \frac{x_2*x_2}{v_1***v_n*w_1***w_m} + \end{align} + + \textit{I'm kinda stuck at this point. + I see that this is definitely injective, since the way the exponents + are defined, you will always know which prime factors belong to the numerator or to the denominator. + But I fail to prove this using the direct definition of $f$ like we could do for the natural numbers. + This is because the products of the denominators in the last equation are not unique. + So maybe I simplified them too much and shouldn't try and write them in terms of $x_1$ and $x_2$ like + we did earlier, and try and focus more on just the exponents, but I feel it becomes really hard + to show that $x_1 = x_2$ that way.} + + \item[Surjectivity:] We want to show that $f$ is onto, i.e. $f(\{ q > 0 : q \in \Q \}) = \N$. + + In order to prove this, we will take an arbitrary $y \in \N$, and show that $\exists x : f(x)=y$. + We know from the \textbf{Theorem} that $y$ can be written as a product of unique prime factors, + $p_1^{r_1}***p_N^{r_N}$. From the definition of $f$ we know that if the exponents of the prime factors + $r$ are even, they belong to the numerator of $x$ and if the exponents are + odd, they belong to the denominator of $x$. If there are no prime factors with odd exponents, + $x$ will be a natural number. If $y=1$, $x=1$. + + We will now only consider the case that $y$ is a prime factorization with factors with odd exponents + \footnote{The case for a prime factorization with solely even exponents can be backtracked + in a similar fashion, just without the case for odd exponents and making $x$ a fraction.}. + Then, we can find $x$ in the following way: we multiply each prime factor $p_i^{2r_i-1}$ with $p_i$ + and take the square root. We know that the square root of $p_i^{2r_i}$ is defined, since the exponent + is multiplied by a factor 2, which the root negates. This will yield a prime factorization that we will + put in the denominator of a fraction. We do the same for the prime factors with even exponent, but + without multiplying with $p_i$. The prime factors we gain like that we put as a product in the + numerator of the fraction. So, we gain a fraction with both the numerator and the denominator + consisting of products of prime numbers, which are natural numbers, and so the fraction is positive + and in fact a fraction. \qed + \end{description} \end{enumerate} \end{document}