Problem_Set_04_Solutions

Problem_Set_04_Solutions - %&LaTeX...

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\documentclass{article} \usepackage{fullpage} \usepackage{amsmath} \newcommand{\fr}[2]{\frac{#1}{#2}} \ \begin{document} \ \begin{center} \textbf{ArsDigita University} \ \textbf{Month 2: Discrete Mathematics - Professor Shai Simonson} \ \textbf{Problem Set 4 -- Induction and Recurrence Equations} \ \textbf{Thanks to Jeffrey Radcliffe and Joe Rizzo for many of the solutions.\\ Pasted together and modified by Dimitri Kountourogiannis} P \end{center} \ \begin{enumerate} \item What's wrong with the following proofs by induction? W \begin{enumerate} \item All binary strings are identical. The proof is by induction on the size of the string. For \textit{n=}0 all binary strings are empty and therefore identical. Let \textit{X = b}$_{\mathit{n}}$\textit{b}$_{\mathit{n-1}}$\textit{\dots b}$_{\mathit{1}}$\textit{b}$_{\mathit{0}}$ be an arbitrary binary string of length \textit{n+}1. Let \textit{Y = b}$_{\mathit{n}}$\textit{b}$_{\mathit{n-1}}$\textit{\dots b}$_{\mathit{1}}$ and \textit{Z =} \textit{b}$_{\mathit{n-1}}$\textit{\dots b}$_{\mathit{1}}$\textit{b}$_{\mathit{0}}$. Since both \textit{Y} and \textit{Z} are strings of length less than \textit{n+}1, by induction they are identical. Since the two strings overlap, \textit{X} must also be identical to each of them. \\ \textbf{ANSWER} \\ This is not a subtly flawed proof. It is a terribly flawed proof. The problem is with the induction step. It is never true that the pasting together of two identical strings is identical to the strings. Furthermore, the $n=1$ case fails for another reason: A string of length 1 is not the pasting together of two strings of length 0. \\ \ \item Any amount of change greater than or equal to twenty can be gotten with a combination of five cent and seven cent coins. The proof is by induction on the amount of change. For twenty cents use four five-cent coins. Let \textit{n} \texttt{>} 20 be the amount of change. Assume that \textit{n} = 7\textit{x +} 5\textit{y} for some non-negative integers \textit{x} and \textit{y.} For any \textit{n \texttt{>}} 20, either \textit{x \texttt{>}} 1, or
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\textit{y \texttt{>}} 3. If \textit{x \texttt{>}} 1, then since 3(5) -- 2(7) = 1, \textit{n+}1 = 5(\textit{y}+3)+7(\textit{x}--2). If \textit{y \texttt{>}} 3, then since 3(7) -- 4(5) = 1, \textit{n+}1 = 7(\textit{x}+3)+5(\textit{y}--4). In either case, we showed that \textit{n+}1 = 7\textit{u +} 5\textit{v} where \textit{u} and \textit{v} are non-negative integers. \\ \textbf{ANSWER} \\ For any $n > 20$, either $x > 1$ or $y > 3$ is false. For example, if $n=21$, then $21 = 0 \cdot 5 + 3 \cdot 7$. \end{enumerate} \ \item Prove by induction that: P \begin{enumerate} \item The \textit{n}th Fibonacci number, $F_n$ equals \[ G_n = \fr{1}{\sqrt{5}}\left[(\fr{1 +\sqrt{5}}{2})^{n} - (\fr{1 + \sqrt{5} }{2})^n \right] \] where \textit{F}$_{\mathit{0}}$ = 0 and \textit{F}$_{\mathit{1}}$ = 1.
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Problem_Set_04_Solutions - %&LaTeX...

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