User:Marty/UCSC Math 110 Fall 2008/Homework 3

This homework assignment has three problems, and your solutions should be typed using LaTex (which you have installed previously. Follow these steps to complete the homework assignment:
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 * Save and compile the file, and view the PDF (this will help you see what you are trying to do).
 * The PDF file should look like this: [[Image:Math110Fall2008HW3.pdf‎]]
 * Edit the template to complete the computations and proofs.
 * Make sure to change the name and e-mail addresses to your own (you are not Student X).
 * Follow all of the directions in the template, which are given on lines that begin with a % symbol in the body of the text.
 * Save and compile the file now and then, to view the PDF results.
 * After you finish, save and compile the PDF, print the file, and turn it in.
 * There are three sections to complete.
 * The first section is worth 3 points.
 * The second section is worth 3 points.
 * The third section is worth 4 points.

% % AMS-LaTeX Paper ************************************************ % **** --- \documentclass{amsart} \usepackage{graphicx} \usepackage{amsfonts} \usepackage{amscd} \usepackage{amssymb} \usepackage{xy} % \vfuzz2pt % Don't report over-full v-boxes if over-edge is small \hfuzz2pt % Don't report over-full h-boxes if over-edge is small % THEOREMS --- \newtheorem{thm}{Theorem}[section] \newtheorem{cor}[thm]{Corollary} \newtheorem{lem}[thm]{Lemma} \newtheorem{prop}[thm]{Proposition} \theoremstyle{definition} \newtheorem{defn}[thm]{Definition} \theoremstyle{remark} \newtheorem{rem}[thm]{Remark} \numberwithin{equation}{section} % MATH --- \newcommand{\Matrix}[4]{ \left( \begin{array}{cc} #1 & #2 \\  #3 & #4 \\ \end{array} \right) } \newcommand{\norm}[1]{\left\Vert#1\right\Vert} \newcommand{\abs}[1]{\left\vert#1\right\vert} \newcommand{\set}[1]{\left\{#1\right\}}

\newcommand{\NN}{\mathbb N} \newcommand{\ZZ}{\mathbb Z} \newcommand{\QQ}{\mathbb Q} \newcommand{\RR}{\mathbb R} \newcommand{\CC}{\mathbb C} \newcommand{\isom}{\cong} % \begin{document}

\title[]{Math 110, Homework 3}% \author{Student X}% \address{Dept. of Mathematics, University of California, Santa Cruz, CA 95064}% \email{studentx AT ucsc DOT edu}%

%\date{}% %\dedicatory{}% %\commby{}% % \maketitle % \section{Solving a linear Diophantine equation} In this section, we analyze the linear Diophantine equation: $$4862 x + 6171y = 374.$$ We begin by applying the Euclidean algorithm to find the greatest common divisor of $6171$ and $4862$. \begin{eqnarray*} 6171 & = & 1 (4862) + 1309, \\ 4862 & = & 3 (1309) + 935, \\ % Fill out the rest of the Euclidean algorithm, copying the format of the above lines precisely! % The eqnarray* environment tells Tex that you are writing a series of lines of math. % The double-backslash \\ tells Tex to start a new line. % The ampersands (&) make the equal signs line up, one above the next. % Observe the punctuation at the end of lines (commas). % On the last line, end the line with a period instead of a comma. Do not put a \\ at the end of the last line. \end{eqnarray*} Using the above Euclidean algorithm, we find that the greatest common divisor of $6171$ and $4862$ is $g$. % Replace the letter g by the greatest common divisor that you find. % Observe that even simple numbers, like 6171 are placed between dollar signs. % Tex italicizes "stuff" that it thinks is math. % So, by always putting dollar signs around math (including simple numbers), you will get a consistent appearance.

Recall that $GCD(4862,6171) \cdot LCM(4862,6171) = 4862 \cdot 6171$. It follows that the least common multiple of $6171$ and $4862$ can be computed: $$LCM(6171, 4862) = \frac{4862 \cdot 6171}{g} = \ell.$$ %Replace the letters g and l by the greatest common divisor and least common multiple that you find. Now, we can follow the Euclidean algorithm backwards to find a solution to the Diophantine equation: \begin{eqnarray*} g & = & u - w(v), \\ & = & u - w(t - r(u)), \\ & = & u + wr(u) - w(t), \\ & = & z(u) - w(t), \\ \end{eqnarray*} % Replace all of the letters above with numbers, with g replaced by the greatest common divisor. % Add lines with exactly the same format as needed. Remember to end lines with \\, and put & = & at the beginning of each new line. % The last line should look like 4862x + 6171y, for some numbers x and y which you have to find. % Remember to end the last line with a period, and no \\ From the above computation, we have found a solution to the Diophantine equation $4862 x + 6171 y = 374$. Our solution is $(u,v)$. % Replace u and v by the solution you have found. % You definitely should check that your solution works!

Every solution of the Diophantine equation $4862 x + 6171 y = 374$ has the form: $$(u + \frac{l n}{4862}, v - \frac{l n}{6171}),$$ for some integer $n$. % Replace u, v, and l by the appropriate numbers.

\section{Fibonacci numbers} In this section, we prove the following theorem, using induction. \begin{thm} Let $F_n$ be the Fibonacci sequence, defined recursively as follows: \begin{itemize} \item $F_0 = 1$. \item $F_1 = 1$. \item For all natural numbers $n$, if $n > 1$ then $F_n = F_{n-1} + F_{n-2}$. \end{itemize} Then, for every positive natural number $i$, $GCD(F_i, F_{i-1}) = 1$. In other words, consecutive Fibonacci numbers are relatively prime. \end{thm} \proof %Write a proof of this theorem, using induction on i, and the definition of greatest common divisor. %Make sure that your proof is written here, after the \proof line, and before the \qed line. %Write your proof in paragraphs. Remember to put any mathematics between dollar signs. %Notice that subscripts are created with an "underscore", i.e., $F_{n-1}$ will look like F, with a subscript of n-1. \qed

\section{Irrationality} In this section, we prove the following theorem. \begin{thm} Let $n$ be a natural number, and suppose that $n \geq 2$. Then, there does not exist a rational number $q$, such that $q^n = n!$. \end{thm} \proof We argue by contradiction. Suppose that $n$ is a natural number, $n \geq 2$, $q$ is a rational number, and $q^n = n!$. % First, use a theorem from class to prove that q must be an integer.

Since $q$ is an integer, $q$ has a canonical decomposition: $$q = \prod_{p \in P} p^{e_p},$$ for some natural numbers $e_p$ (where $P$ denotes the set of prime numbers). The canonical decomposition of $q^n$ is given by: % Imitate the canonical decomposition of q, given two lines above, to write the canonical decomposition of q^n. % Observe that \prod creates a "big pi", the symbol for a product. % Observe that \in creates an "element" symbol. % Always remember to put "clumps" between { and }. % This is especially important when there are subscripts within superscripts, as in p^{e_p}.

Also, the natural number $n!$ has a canonical decomposition: $$n! = \prod_{p \in P} p^{f_p}.$$ Recall from class that the exponents $f_p$ are given by: $$f_p = \lfloor \frac{n}{p} \rfloor + \lfloor \frac{n}{p^2} \rfloor + \cdots + \lfloor \frac{n}{p^t} \rfloor,$$ where $t = \lfloor log_p(n) \rfloor$. % Observe that \lfloor and \rfloor produce the left and right "floor" brackets around a quantity. % Also, recall that \frac{x}{y} makes a fraction with x on top, y on bottom.

Since $q^n = n!$, the uniqueness of canonical decomposition implies that: $$n e_p = f_p$$ % Do you understand why the left side is n e_p? This should be clear from something you wrote above.

Using a geometric series, it is known that: $$\sum_{i=1}^\infty \frac{1}{p^i} = \frac{1}{p} + \frac{1}{p^2} + \frac{1}{p^3} + \cdots = \frac{1}{p - 1} \leq 1.$$ % The code \leq gets made into a "less than or equal to" sign. % Explain why f_p is less than n, using the above fact, and the previous expression for f_p. % This step might be a bit difficult. Here are a few tips. % Observe that taking "floors" only decreases the size of numbers. % Therefore, you can bound f_p by the sum \frac{n}{p} + \frac{n}{p^2} + \cdots + \frac{n}{p^t} % This finite sum can be bounded (strictly, with a < instead of a \leq) by the infinite sum \frac{n}{p} + \frac{n}{p^2} + \cdots. % This sum can be evaluated using the above geometric series.

Since $f_p < n$, and $f_p = n e_p$, it follows that: $$n e_p < n.$$ % Explain why this implies that e_p is equal to zero.

Since $e_p = 0$, for all $p \in P$, it follows that $q = 1$. But $1^n = 1 \neq n!$, since $n \geq 2$. This contradicts the assumption that $q^n = n!$.

Hence there cannot exist a rational number $q$, such that $q^n = n!$. \qed

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