Merge pull request 'Finished assignment 1' (#12) from assignment1 into master
Reviewed-on: #12
This commit is contained in:
Philippe Zwietering
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\documentclass{template/homework}
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\documentclass{template/homework}
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\usepackage{enumitem}
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\usepackage{enumitem}
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\usepackage{tcolorbox}
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\usepackage{subfiles}
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\usepackage{subfiles}
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\title{MIT OCW Real Analysis}
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\title{MIT OCW Real Analysis}
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\documentclass[../main_text.tex]{subfiles}
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\documentclass[../main_text.tex]{subfiles}
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\begin{document}
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\begin{document}
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\section*{Week 1}
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\exercise*[0.3.6]
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\exercise*[0.3.6]
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\begin{enumerate}[label=\emph{\alph*)}]
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% For some reason I can't put a fitted tcbox here and I really don't like it
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\item Wanting to show:
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\begin{tcolorbox}
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$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$
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Prove:
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\begin{enumerate}[label=\emph{\alph*)}]
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\item $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$
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\item $A \cup (B \cap C) = (A \cup B) \cap (a \cup C)$
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\end{enumerate}
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\end{tcolorbox}
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In order to prove this equivalence, we have to prove the implication both ways. We use two lemmas for this.
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\begin{enumerate}[label=\emph{\alph*)}, wide]
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\item In order to prove this equivalence, we have to prove the implication both ways. We use two lemmas for this.
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\begin{lemma}[test]
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\begin{lemma}
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\label{lem:set1}
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$A \cap (B \cup C) \implies (A \cap B) \cup (A \cap C)$
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$A \cap (B \cup C) \implies (A \cap B) \cup (A \cap C)$
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\end{lemma}
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Let $x \in A \cap (B \cup C)$.
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Let $x \in A \cap (B \cup C)$.
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By the definition of set intersection, $x \in A$ and $x \in B \cup C$.
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By the definition of set union, $x \in A$ and $(x \in B$ or $x \in C)$.
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From propositional logic we know that for propositions $P$, $Q$ and $R$ the following holds:
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$P \land (Q \lor R) \iff (P \land Q) \lor (P \land R)$.
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So, substituting for this particular case yields $(x \in A$ and $x \in B)$ or $(x \in A$ and $x \in C)$.
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Using the definition of set intersection again gets $x \in A \cap B$ or $x \in A \cap C$.
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Using the definition of set union again gives $x \in (A \cap B) \cup (A \cap C)$. \qed
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\begin{lemma}
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\label{lem:set2}
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$(A \cap B) \cup (A \cap C) \implies A \cap (B \cup C)$
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\end{lemma}
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\end{lemma}
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\begin{definition}
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Test.
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Let $x \in (A \cap B) \cup (A \cap C)$.
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\end{definition}
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By the definition of set union, $x \in (A \cap B)$ or $x \in (A \cap C)$.
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\item
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By the definition of set intersection, $(x \in A$ or $x \in B)$ and $(x \in A$ or $x \in B)$.
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Using the same propositional logical equivalence as in Lemma \ref{lem:set1}, this gives
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$x \in A$ and $(x \in B$ or $x \in C)$.
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Wrapping up, we use the definition of set union to get $x \in A$ and $x \in B \cup C$ and
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the definition of intersection to get $x \in A \cap (B \cup C)$. \qed
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Using Lemma \ref{lem:set1} and \ref{lem:set2}, we get the desired equivalence of
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$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$. \qed
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\item This proof is so similar to a) that it feels like a waste of time and will therefore be left to the reader.
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\end{enumerate}
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\exercise*[0.3.11]
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\tcbox{Prove by induction that $n < 2^n$ for all $n \in \N$.}
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For this proof we will use induction. For this, we have to prove the base case, i.e. $n = 1$,
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and the inductive step, $n < 2^n \implies n + 1 < 2^{n+1}$.
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First, let's prove the base case. When $n = 1$, we get $1 < 2^1$, which is certainly true.
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Then, for the inductive step. We assume that the proposition holds for any $m \in \N$.
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So, $m < 2^m$. Multiplying both sides with 2 gives $2m < 2^{m+1}$.
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Since $m \geq 1$, $m + 1 < 2m$, and thus $m + 1 < 2m < 2^{m+1}$.
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Since both the base case and inductive step hold, we can close the induction, proving the proposition. \qed
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\exercise*[0.3.12]
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\tcbox{Show that for a finite set $A$ of cardinality $n$, the cardinality of $\mathcal{P}(A)$ is $2^n$.}
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The power set of a set $A$, $\mathcal{P}(A)$, is defined as the set of all possible subsets of $A$.
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This is very similar to an inclusion/exclusion problem.
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It is built up by all the possible combinations of the different elements being either inside a certain subset or not.
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For all possible subsets of $A$, we have that for every element $x \in A$ there are 2 possibilities,
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either $x$ is in the subset or it isn't.
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This means that for every additional element, the number of subsets increases by a factor of 2, with a minimum of 1,
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in case of $A = \emptyset$. We will prove this formally now, using induction.
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For this, the base case is a set of 1 element (but the theorem also holds for the empty set, where $n=0$).
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Let us assume that $A := \{ \pi \}$.
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Then the cardinality of $\mathcal{P}(A)$ is $2^1$, with $\mathcal{P}(A)= \{ \emptyset, \{ \pi \}\}$.
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For the inductive step, we assume that for any set $B$ of cardinality $m$, the cardinality of the power set of $B$
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is $2^m$. Then, we will add an element $x \notin B$ to $B$ to increase its cardinality by 1, to $m + 1$,
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creating a new set $C$.
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Note that all the possible subsets of $B$ are still viable subsets of $C$, since $B \subset C$.
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In order to create the new subsets, we can simply keep all the subsets of $B$, duplicate them and take the union with
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the new element $x$, so now we also have all combinations of the old sets with possibly $x$ being in them.
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Since this doubles the number of subsets, the cardinality of $\mathcal{P}(C)$ is $2^{m+1}$.
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Both the base case and the inductive step hold, which closes the induction and proves the proposition. \qed
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\exercise*[0.3.15]
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\tcbox{Prove that $n^3 + 5n$ is divisible by 6 for all $n \in \N$.}
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In order to prove this proposition, we will use induction. To do this, we need to prove the following lemma,
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of which we will see the usefulness later:
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\begin{lemma}
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\label{lem:div6}
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$3n^2 + 3n + 6$ is divisible by 6 for all $n \in \N$.
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\end{lemma}
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This lemma we will also prove by induction. For this, we prove the base case and the inductive step.
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First, for the base case we have $n = 1$, yielding $3 * 1^2 + 3 * 1 + 6 = 12$, which is divisibly by 6.
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Then, for the inductive step we assume that the lemma holds for a certain $m \in \N$.
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So, $3m^2 + 3m + 6$ is divisible by 6.
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Substituting $m$ with $m+1$ gives $3(m+1)^2 + 3(m+1) + 6$, which can be expanded to
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$3m^2 + 9m + 12$. Rewriting this with our assumption in mind gives the following:
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$(3m^2 + 3m + 6) + (6m + 6)$. We know from our assumption that the first part is divisible by 6, and since $m \in \N$,
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$6m + 6$ is also divisible by 6, and so the whole expression is as well. \qed
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Now for the original proposition. We will prove this by induction. First we prove the base case, where $n=1$.
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Then, $1^3 + 5*1 = 6$, which is definitely divisible by 6.
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For the inductive step, we assume that the proposition holds for a certain $m \in \N$. So, $m^3 + 5m$ is divisible by 6.
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When we increase $m$ by 1, we get: $(m+1)^3 + 5(m+1)$.
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Expanded, this is the same as $m^3 + 3m^2 + 8m + 6$. When we rearrange the terms we can get the following expression:
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$(m^3 + 5m) + (3m^2 + 3m + 6)$. From Lemma \ref{lem:div6}, we know that the latter part is divisible by 6.
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The prior part is divisible by 6 because of the assumption of the inductive step. So together, this expression
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is also divisible by 6. \qed
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\exercise*[0.3.19]
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\begin{tcolorbox}
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Give an example of a countably infinite collection of finite sets $A_1, A_2,...$,
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whose union is not a finite set.
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\end{tcolorbox}
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The easiest example is simply the collection of singleton sets containing a natural number.
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So each set $A_i := \{i\} \forall i \in \N$. Since $\N$ is countably infinite, so the collection of sets.
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Each set is definitely finite, because they all contain just one element.
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Finally, the union of the collection of sets is equal to $\N$, which is not a finite set.
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\exercise*
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\begin{tcolorbox}
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\begin{enumerate}[label=\emph{\alph*)}, wide]
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\item Compute $f(4/15)$. Find $q$ such that $f(q) = 108$.
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\item Use the \textbf{Theorem} to prove that $f$ is a bijection.
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\end{enumerate}
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\end{tcolorbox}
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See the assignment PDF for the full assignment specification and theorem.
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\begin{enumerate}[label=\emph{\alph*)}, wide]
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\item $\frac{4}{15}$, if written as a product of prime factors, is equal to $\frac{2^2}{3^1*5^1}$.
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Since this fraction is not a natural number, we have to use the second part of the definition of $f$.
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So, $f(q) = 2^{2*2} * 3^{2 * 1 - 1} * 5^{2 * 1 - 1} = 240$.
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For the inverse of $f$, it is still necessary to compute the factorization in prime numbers.
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Using the powers of the primes we can deduce whether the prime present is, if applicable,
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part of either the numerator or the denominator.
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$180 = 2^2 * 3^2 * 5^1$. Because of the way $f$ is defined, we know that all the prime factors with an even
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power are part of the numerator and all prime factors with an odd power are part of the denominator (except 1,
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which just maps to itself). When we backtrack using this information, we then get the following fraction:
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$\frac{2^1 * 3^1}{5^1} = \frac{6}{5}$.
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\item In order to prove that $f$ is a bijection, we have to prove that $f$ is injective and surjective.
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\begin{description}
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\item[Injectivity:] We want to show that $f$ is 1-1,
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i.e. $f(x_1)=f(x_2) \implies x_1 = x_2$.
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So, let's assume that for any $x_1, x_2 \in \{ q > 0 : q \in \Q \}$, $f(x_1) = f(x_2)$.
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Since the function $f$ has 3 parts, based on the input, we have to prove this statement for those 3
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parts separately as well. First, the easiest case, where the input set is $\{1\}$.
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Then, $f(x) = 1 \forall x$, so $f$ is injective.
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For the case where $x \in \N\backslash\{1\}$, $f(x) := p_1^{2r_1}***p_N^{2r_N}$.
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We know from the \textbf{Theorem} that any fraction can be uniquely written as a product of prime
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factors with exponents, so when we assume $f(x_1) = f(x_2)$, we can also assume that $x_1$ and $x_2$
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have a unique prime factorization associated with them. So let's assume that $f(x_1) = f(x_2)$.
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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 $q_j^{s_j}$
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denote the prime factors for both sides. We can further expand this expression into:
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\begin{align}
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p_1^{r_1} * p_1^{r_1} *** p_N^{r_N} * p_N^{r_N} &=
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q_1^{s_1} * q_1^{s_1} *** q_M^{s_M} * q_M^{s_M} \implies \\
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p_1^{r_1}***p_N^{r_N} * p_1^{r_1}***p_N^{r_N} &=
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q_1^{s_1}***q_M^{s_1} * q_1^{s_1}***q_N^{s_M} \implies \\
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x_1 * x_1 &= x_2 * x_2
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\end{align}
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Because we know that each fraction constitutes a unique prime factorization, we also know that $x_1$ and
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$x_2$ are uniquely derived. This is why the implications in the equation above hold.
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Because both $x_1$ and $x_2 > 0$, $x_1 = x_2$.
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Now for the case where $x \in \Q\backslash\N$.
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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
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derived from the \textbf{Theorem}. So again, we assume that for any $x_1, x_2 \in \Q\backslash\N$,
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$f(x_1) = f(x_2)$.
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Using the definition of $f$, we get:
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$p_1^{2r_1}***p_N^{2r_N}q_1^{2s_1-1}***q_M^{2s_M-1} =
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v_1^{2t_1}***v_n^{2t_n}w_1^{2u_1-1}***w_m^{2u_m-1}$.\footnote{The super- and subscripts become a bit
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abracadabra, but I think everything is unique and readable this way.}
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Expanding this expression further, we get:
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\begin{align}
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\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} &=
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\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}
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\implies \\
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\frac{p_1^{r_1}*p_1^{r_1}}{p_1}***\frac{p_N^{r_N}*p_N^{r_N}}{p_N}
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\frac{q_1^{s_1}*q_1^{s_1}}{q_1}***\frac{q_M^{s_M}*q_M^{s_M}}{q_M} &=
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\frac{v_1^{t_1}*v_1^{t_1}}{v_1}***\frac{v_n^{t_n}*v_n^{t_n}}{v_n}
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\frac{w_1^{u_1}*w_1^{u_1}}{w_1}***\frac{w_m^{u_m}*w_m^{u_m}}{w_m} \implies \\
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\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}
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\end{align}
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\textit{I'm kinda stuck at this point.
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I see that this is definitely injective, since the way the exponents
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are defined, you will always know which prime factors belong to the numerator or to the denominator.
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But I fail to prove this using the direct definition of $f$ like we could do for the natural numbers.
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This is because the products of the denominators in the last equation are not unique.
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So maybe I simplified them too much and shouldn't try and write them in terms of $x_1$ and $x_2$ like
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we did earlier, and try and focus more on just the exponents, but I feel it becomes really hard
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to show that $x_1 = x_2$ that way.}
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\item[Surjectivity:] We want to show that $f$ is onto, i.e. $f(\{ q > 0 : q \in \Q \}) = \N$.
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In order to prove this, we will take an arbitrary $y \in \N$, and show that $\exists x : f(x)=y$.
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We know from the \textbf{Theorem} that $y$ can be written as a product of unique prime factors,
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$p_1^{r_1}***p_N^{r_N}$. From the definition of $f$ we know that if the exponents of the prime factors
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$r$ are even, they belong to the numerator of $x$ and if the exponents are
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odd, they belong to the denominator of $x$. If there are no prime factors with odd exponents,
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$x$ will be a natural number. If $y=1$, $x=1$.
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We will now only consider the case that $y$ is a prime factorization with factors with odd exponents
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\footnote{The case for a prime factorization with solely even exponents can be backtracked
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in a similar fashion, just without the case for odd exponents and making $x$ a fraction.}.
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Then, we can find $x$ in the following way: we multiply each prime factor $p_i^{2r_i-1}$ with $p_i$
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and take the square root. We know that the square root of $p_i^{2r_i}$ is defined, since the exponent
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is multiplied by a factor 2, which the root negates. This will yield a prime factorization that we will
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put in the denominator of a fraction. We do the same for the prime factors with even exponent, but
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without multiplying with $p_i$. The prime factors we gain like that we put as a product in the
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numerator of the fraction. So, we gain a fraction with both the numerator and the denominator
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consisting of products of prime numbers, which are natural numbers, and so the fraction is positive
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and in fact a fraction. \qed
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\end{description}
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\end{enumerate}
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\end{enumerate}
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\end{document}
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\end{document}
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Reference in New Issue
Block a user