From 89862ae6a0554870a7708ae73112f86d2d21fc8d Mon Sep 17 00:00:00 2001
From: Holden Rohrer <hr@hrhr.dev>
Date: Thu, 10 Feb 2022 01:12:45 -0500
Subject: new teachers, new work

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+\newfam\bbold
+\def\bb#1{{\fam\bbold #1}}
+\font\bbten=msbm10
+\font\bbsev=msbm7
+\font\bbfiv=msbm5
+\textfont\bbold=\bbten
+\scriptfont\bbold=\bbsev
+\scriptscriptfont\bbold=\bbfiv
+\font\bigbf=cmbx12 at 24pt
+
+\def\answer{\smallskip\bgroup}
+\def\endanswer{\egroup\medskip}
+\def\section#1{\medskip\goodbreak\noindent{\bf #1}}
+\let\impl\Rightarrow
+
+\headline{\vtop{\hbox to \hsize{\strut Math 2106 - Dr. Gupta\hfil Due Thursday
+2022-01-27 at 11:59 pm}\hrule height .5pt}}
+
+\centerline{\bigbf Homework 3 - Holden Rohrer}
+\bigskip
+
+\noindent{\bf Collaborators:} None
+
+\section{Hammack 2.7: 2, 9, 10}
+
+\item{2.} Write the following as an English sentence:
+$\forall x\in\bb R, \exists n\in \bb N, x^n\geq 0.$
+
+\answer
+For all real numbers $x,$ there is a natural number $n$ such that $x^n$
+is nonnegative.
+This statement is true because, for all real numbers, $x^2 \geq 0$ and
+$2\in\bb N.$
+\endanswer
+
+\item{9.} Write the following as an English sentence:
+$\forall n\in\bb Z, \exists m\in\bb Z, m = n+5.$
+
+\answer
+For all integers $n,$ there is an integer $m$ which is 5 greater than
+$n.$
+This statement is true because the integers are closed under addition.
+\endanswer
+
+\item{10.} Write the following as an English sentence:
+$\exists m\in\bb Z, \forall n\in\bb Z, m = n + 5.$
+
+\answer
+There is an integer $m$ such that for all integers $n,$ $m$ is 5 greater
+than $n.$
+This statement is false because $m$ cannot equal $0+5$ and $1+5$ at the
+same time.
+\endanswer
+
+\section{Hammack 2.9: 1, 7, 10}
+
+\item{1.} Translate the following sentence into symbolic logic: ``If $f$
+is a polynomial and its degree is greater than 2, then $f'$ is not
+constant.
+\answer
+Where $P$ is the set of polynomials, and $\mathop{\rm degree}(p)$ is the
+degree of a polynomial $p,$
+$$\forall p\in P, \left(\mathop{\rm degree}(p) > 2\right) \impl \exists
+a,b\in\bb R, f'(a) \neq f'(b).$$
+\endanswer
+
+\item{7.} Translate the following sentence into symbolic logic: ``There
+exists a real number $a$ for which $a+x = x$ for every real number $x.$
+\answer
+$$\exists a\in\bb R, \forall x\in\bb R, a+x = x.$$
+\endanswer
+
+\item{10.} Translate the following sentence into symbolic logic: ``If
+$\sin(x) < 0,$ then it is not the case that $0\leq x\leq\pi.$
+\answer
+$$\forall x\in\bb R, \sin(x) < 0 \impl \lnot(0\leq x\leq\pi).$$
+\endanswer
+
+\section{Hammack 2.10: 2, 5, 10}
+\item{2.} Negate the following sentence: ``If $x$ is prime, then $\sqrt
+x$ is not a rational number.''
+
+\answer
+There is a prime number $x$ such that $\sqrt x$ is a rational number.
+\endanswer
+
+\item{5.} Negate the following sentence: ``For every positive number
+$\epsilon,$ there is a positive number $M$ for which $|f(x)-b|<\epsilon$
+whenever $x > M.$
+
+\answer
+There is a positive number $\epsilon$ such that for all $M$ there is an
+$x > M$ such that $|f(x)-b|>\epsilon$
+\endanswer
+
+\item{10.} If $f$ is a polynomial and its degree is greater than 2, then
+$f'$ is not constant.
+
+\answer
+There is a polynomial with degree greater than 2 such that $f'$ is
+constant.
+\endanswer
+
+\section{Hammack 4: 4, 12, 20}
+
+\item{4.} Prove ``Suppose $x,y\in\bb Z.$ If $x$ and $y$ are odd, then
+$xy$ is odd'' with direct proof.
+\answer
+Suppose $x,y\in\bb Z$ and that $x$ and $y$ are odd.
+Since $x$ is odd, there exists $j\in\bb Z$ such that $x = 2j+1.$
+Since $y$ is odd, there exists $k\in\bb Z$ such that $y = 2k+1.$
+$xy = (2j+1)(2k+1) = 4jk + 2j + 2k + 1 = 2(2jk + j + k) + 1.$
+Because $2jk + j + k$ is an integer, $xy$ is odd because it is one more
+than two times an integer.
+\endanswer
+
+\item{12.} Prove ``If $x\in\bb R$ and $0<x<4,$ then ${4\over x(4-x)}\geq
+1.$'' with direct proof.
+\answer
+Let $x\in\bb R$ and $0<x<4.$
+$$(x-2)^2 \geq 0 \to 4 - (x-2)^2\leq 4\to 4x-x^2 = x(4-x) \leq 4 \to
+{4\over x(4-x)}\geq {4\over 4} = 1.$$
+\endanswer
+
+\item{20.} Prove ``If $a$ is an integer, and $a^2|a,$ then
+$a\in\{-1,0,1\}.$'' with direct proof.
+
+\answer
+$a^2|a$ requires $a^2 \leq |a|.$ For $|a| > 1,$ $a^2 = |a|^2 > |a|,$ so
+we will check $a^2|a$ for the remaining cases $\{-1,0,1\}.$
+
+$n|m$ iff there is a $k\in\bb Z,$ $k\neq 0,$ $m = nk.$
+For $0,$  $0^2 = 1(0),$ so $0^2|0.$
+For $1,$  $1^2 = 1(1),$ so $1^2|1.$
+For $-1,$ $(-1)^2 = -1(-1),$ so $(-1)^2|1.$
+\endanswer
+
+\section{Problem not from the textbok}
+
+\item{1.} Prove that for all positive real numbers $x,$ the sum of $x$ and its
+reciprocal is greater than or equal to 2.
+
+\answer
+Let $x$ be a positive real number.
+For all real numbers $y,$ $y^2 \geq 0,$ so $(x-1)^2 \geq 0.$
+This is equal to
+$$x^2 - 2x + 1 \geq 0 \to x^2 + 1 \geq 2x \to x + 1/x \geq 2,$$
+since dividing by $x > 0$ is a valid algebraic operation.
+\endanswer
+
+\bye
-- 
cgit