Homework 7

.tex

School

Texas A&M University *

*We aren’t endorsed by this school

Course

222

Subject

Computer Science

Date

Feb 20, 2024

Type

tex

Pages

Uploaded by PrivateSteelKomodoDragon37

% Comment lines start with % % LaTeX commands start with \ % This template was provided by Jennifer Welch for CSCE 222-200, Honors, Spring 2015 \documentclass[12pt]{article} % This is an article with font size 12-point % Packages add features \usepackage{times} % font choice \usepackage{amsmath} % American Mathematical Association math formatting \usepackage{amsthm} % nice formatting of theorems \usepackage{amssymb} % provides some symbols \usepackage{latexsym} % provides some more symbols \usepackage{fullpage} % uses most of the page (1-inch margins) \usepackage{enumitem} \usepackage{hyperref} \setlength{\parskip}{.1in} % increase the space between paragraphs \renewcommand{\baselinestretch}{1.1} % increase the space between lines % Convenient renaming of symbols for logic formulas \newcommand{\NOT}{\neg} \newcommand{\AND}{\wedge} \newcommand{\OR}{\vee} \newcommand{\XOR}{\oplus} \newcommand{\IMPLIES}{\rightarrow} \newcommand{\IFF}{\leftrightarrow} \providecommand{\myceil}[1]{$\left \lceil #1 \right \rceil$} \providecommand{\myfloor}[1]{$\left \lfloor #1 \right \rfloor$} \newcommand{\powerset}[1]{\mathbb{P}(#1)} % Actual content starts here. \begin{document} \begin{center} % center all the material between begin and end {\large % use larger font CSCE 222 (Carlisle), Discrete Structures for Computing \\ % \\ is line break Spring 2022 \\ Homework 7} \end{center} \rule{6in}{.1pt} % horizontal line 6 inches long and .1 point high \begin{center} {\large Type your name below the pledge to sign\\ On my honor, as an Aggie, I have neither given nor received unauthorized aid on this academic work.\\ **HUY QUANG LAI**} \end{center} % blank line separates paragraphs. First line of a paragraph is automatically % indented. \rule{6in}{.1pt} % horizontal line 6 inches long and .1 point high \noindent % don't indent {\bf Instructions:} % \bf makes text boldface % \em makes text emphasized (italics)

\begin{itemize} % makes an itemized list \item The exercises are from the textbook. You are encouraged to work extra problems to aid in your learning; remember, the solutions to the odd-numbered problems are in the back of the book. \item Grading will be based on correctness, clarity, and whether your solution is of the appropriate length. \item Always justify your answers. \item Don't forget to acknowledge all sources of assistance in the section below, and write up your solutions on your own. \item {\em Turn in .pdf file to Gradescope by the start of class on Monday, March 7, 2022.} It is simpler to put each problem on its own page using the LaTeX clearpage command. \end{itemize} \rule{6in}{.1pt} % horizontal line 6 inches long and .1 point high {\bf Help Received:} % \bf makes text boldface \begin{itemize} \item Rosen, Kenneth H. \emph{Discrete Mathematics and Its Applications}. McGraw- Hill, 2019. \end{itemize} \rule{6in}{.1pt} % horizontal line 6 inches long and .1 point high %--------------------------------------------------------------------- % \subsection makes a subsection heading; * leaves it unnumbered. % (Usually subsections are inside sections, but the \section command % used a font that was larger than I wanted.) \subsection*{Exercises for Section 5.1:} \noindent{\bf{6: (2 point).}} \noindent{Prove that $1\cdot1!+2\cdot2!+\cdots+n\cdot n!=(n+1)!-1$ whenever $n$ is a non-negative integer.} \noindent{\textit{Basis Step}:} \noindent\\ $P(n)=n\cdot n!$\\ $P(0)=0$\\ $(0+1)!-1=1-1=0$\\ $n\cdot n!=(n+1)!-1$ when $n=0$ \noindent{\textit{Induction Hypothesis}:} \noindent\\ $\displaystyle\sum_{i=0}^{n}(n\cdot n!)=(n+1)!-1$ \noindent{\textit{Induction Step}:} \noindent\\ $\displaystyle\sum_{i=0}^{n+1}(n\cdot n!)$\\ $\displaystyle=\sum_{i=0}^{n}(n\cdot n!)+(n+1)(n+1)!$\\ $=(n+1)!-1+(n+1)(n+1)!$\\ $=(n+1)!(1-1+(n+1))$\\ $=(n+1)(n+1)!$ \clearpage

\noindent{\bf{18(a-f): (3 points).}}\\ Let $P(n)$ be the statement that $n!<n^n$, where $n$ is an integer greater than 1. \begin{enumerate} \item What is the statement $P(2)$?\\ $P(2)\equiv2!<2^2$ \item Show that $P(2)$ is true, completing the basis step of a proof by mathematical induction that $P(n)$ is true for all integers $n$ greater than 1.\\ $2<4$\\ $P(2)\equiv$ True \item What is the inductive hypothesis of a proof by mathematical induction that $P(n)$ is true for all integers $n$ greater than 1?\\ Let $P(k)$ be True for $k\geq2$.\\ $k!<k^k$ \item What do you need to prove in the inductive step of a proof by mathematical induction that $P(n)$ is true for all integers $n$ greater than 1?\\ Prove that $P(k+1)$ is also True.\\ $(k+1)!<(k+1)^{k+1}$ \item Complete the inductive step of a proof by mathematical induction that $P(n)$ is true for all integers $n$ greater than 1.\\ $(k+1)!=(k+1)k$\\ $(k+1)k!<(k+1)^k<(k+1)(k+1)^k$\\ $(k+1)(k+1)^k=(k+1)^{k+1}$\\ Therefore, $(k+1)!<(k+1)^{k+1}$ \item Explain why these steps show that this inequality is true whenever $n$ is an integer greater than 1.\\ Since we have completed the base and inductive steps, by the principle of mathematical induction, the inequality is true for any $n\geq2$. If we had shown $P(3)$ as our basis step, then the inequality would only be proven for $n\geq3$. \end{enumerate} \clearpage \noindent{\bf 32:{(3 points).}}\\ Prove that 3 divides $n^3+2n$ whenever $n$ is a positive integer. \noindent{\emph{Base Case}:}\\ If $n=1$, $1^3+2(1)=3$. So it is divisible by 3. \noindent{\emph{Inductive Step}:}\\ $(n+1)^3+2(n+1)$\\ $=n^3+3n^2+3n+1+2n+2$\\ $=(n^3+2n)+(3n^2+3n+3)$\\ $=(n^3+2n)+3(n^2+n+1)$ \noindent{$3(n^2+n+1)$ is divisible by 3.}\\ Therefore, $n^3+2n$ is divisible by 3 whenever $n$ is a positive integer. \clearpage \subsection*{Exercises for Section 5.2:} \noindent{\bf{6(a-c): (3 points)}} \begin{enumerate} \item Determine which amounts of postage can be formed using just 3-cent and 10-cent stamps.\\ We can form the following amounts of postage as indicated:\\

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