\documentclass[12pt,a4paper]{amsart}
%%%%%%%%%%% Here is where you tell LaTeX to use packages. Each
%%%%%%%%%%% package gives you access to different commands and
%%%%%%%%%%% functionality. Most of them are unnecessary for the
%%%%%%%%%%% mathematics in this document.

%%%%%%%%%%% If you hadn't figured out, the % symbol automatically
%%%%%%%%%%% nulls out all subsequent characters on that line. So
%%%%%%%%%%% it's used for commenting code.
\usepackage{hyperref}
\usepackage{tikz}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{framed}
\usepackage{multicol}
\usepackage[OT2,T1]{fontenc}
\usepackage{graphicx}
\usepackage[all]{xy}
\usepackage{color}
\usepackage{mathrsfs}
\usepackage{mathtools}

%%%%%%%%%%% Determining margins and spacing.
\setlength\topmargin{-0.3in}
\addtolength{\oddsidemargin}{-.625in}
	\addtolength{\evensidemargin}{-.625in}
	\addtolength{\textwidth}{1.25in}
\setlength\footskip{1in}

%%%%%%%%%%% I think this command enables `sections'
\numberwithin{equation}{section}

%%%%%%%%%%% This command allows you to label things as `theorem',
%%%%%%%%%%% `conjecture', etc.
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{question}{Question}[theorem]
\newtheorem{example}{Example}[theorem]


\usepackage[utf8]{inputenc}
\title{Math 220.201 Class 9 Questions}
\begin{document}
\maketitle

\begin{enumerate}
%\item [1.] Prove that no odd integer can be written as the sum of three even integers.
%\vskip 0.6in
\item [1.] Prove that the equation $x^6+x^4+2x^2+1=0$ has no real solutions.
\vskip 0.2in
\begin{proof} [Direct Proof]
We wish to show that $\not\exists x\in\mathbb{R}, x^6+x^4+2x^2+1=0$. We will prove the equivalent statement that $\forall x\in\mathbb{R}, x^6+x^4+2x^2+1 \neq 0$.

For any real number $x$,
$$x^6+x^4+2x^2+1=(x^3)^2+(x^2)^2+2x^2+1 \ge 0+0+2(0)+1=1$$
Therefore, $x^6+x^4+2x^2+1 \ge 1$. So $x^6+x^4+2x^2+1 \neq 0$, as desired. 
\end{proof}
\begin{proof} [Proof by Contradiction]
Suppose for a contradiction that there exists a real number $x$ such that $x^6+x^4+2x^2+1=0$. Then
$$-1=x^6+x^4+2x^2=(x^3)^2+(x^2)^2+2x^2\ge 0+0+2(0)=0$$
because $x$ is a real number. So $-1 \ge 0$, which is a contradiction! Therefore, there is no such $x$.
\end{proof}
\vskip 0.2in
\item [2.] Let $x$ be a nonzero real number. If $x+\frac{1}{x}<2$, then $x<0$.\footnote{You can adapt your argument to prove the following well-known theorem.
\begin{center}
{\bf Arithmetic Mean - Geometric Mean Inequality:} For any positive real numbers $x$ and $y$, $\sqrt{xy} \le \frac{x+y}{2}$.
\end{center}
\begin{proof}
Since $\sqrt{x}-\sqrt{y}$ is a real number,
$$(\sqrt{x}-\sqrt{y})^2\ge 0 \implies x-2\sqrt{xy}+y\ge 0 \implies x+y \ge 2\sqrt{xy}\implies \frac{x+y}{2}\ge \sqrt{xy}$$
\end{proof}}
\vskip 0.2in
Note that in symbols, the statement is
$$\forall x\in\mathbb{R} \text{ s.t. }x\neq 0, ((x+\frac{1}{x}<2) \implies (x<0))$$
\vskip 0.2in
\begin{proof} [Proof by Contradiction]
Suppose that there exists $x \neq 0$ such that $x+\frac{1}{x}<2$ and $x\ge 0$. Then
$$\begin{aligned}
x+\frac{1}{x}<2 &\implies x^2+1<2x & \text{because }x\ge 0\\
&\implies x^2-2x+1<0 &\\
&\implies (x-1)^2<0 &
\end{aligned}$$
But $(x-1)^2\ge 0$, because $x-1\in\mathbb{R}$. Therefore we have a contradiction!
\end{proof}
\begin{proof} [Proof by Contrapositive]
We show that if $x\ge 0$, then $x+\frac{1}{x}\ge 2$. Suppose that $x\ge 0$. First, since $x$ is assumed to be nonzero, it follows that $x>0$. Then
$$\begin{aligned}
(x-1)^2\ge 0 &\implies x^2-2x+1\ge 0 & \\
&\implies x^2+1\ge 2x &\\
&\implies x+\frac{1}{x} \ge 2 & \text{because }x>0
\end{aligned}$$
which completes the proof.
\end{proof}
\vskip 0.2in
\item [3.] Let $x$ be an irrational number. Then there is no {\bf largest} rational number $y$ with the following property: $y\le x$.
\vskip 0.2in
In symbols, the statement is saying that
$$\forall x\in \mathbb{R}-\mathbb{Q}, \not\exists y\in\mathbb{Q}, (y\le x \land \forall z\in\mathbb{Q}, (z\le x \implies z\le y))$$
\vskip 0.2in
\begin{proof} [Proof by Contradiction]
Suppose for a contradiction that there is such an irrational number $x$ and a rational number $y$ such that $y\le x$ and $y$ is the largest among all rational numbers with this property. Since $x$ is irrational and $y$ is rational, they are unequal, so $y<x$. Thus, $x-y>0$ Then there is some positive integer $n$ such that $x-y>\frac{1}{n}$.\footnote{The proof of this is delving into the axiomatic construction of the real numbers, as limits of \emph{Cauchy sequences} of rational numbers. For now I will justify it as follows: if $x-y\le \frac{1}{n}$ for \emph{every} natural number $n$, then this would mean that $x-y\le \lim\limits_{n\to\infty}\frac{1}{n}=0$. However, we've assumed that $x-y>0$, so this is a contradiction.} Then $x>y+\frac{1}{n}$, and $y+\frac{1}{n}$ is rational. So by the assumption, $y+\frac{1}{n} \le y$. Thus, $\frac{1}{n}\le 0$. This is clearly false, so we have a contradiction!
\end{proof}
\begin{proof} [Direct Proof]
It suffices to show that for any irrational $x$ and for any rational $y$ such that $y\le x$, there exists some rational $z$ such that $y < z \le x$. That is, we are proving the equivalent statement
$$\forall x\in\mathbb{R}-\mathbb{Q}, \forall y\in\mathbb{Q}, (y\le x \implies \exists z\in\mathbb{Q}, (z\le x \land z>y))$$
Suppose that $y$ is a rational number such that $y\le x$. Since $x$ is irrational and $y$ is irrational, they are unequal, and so $y<x$. Thus, $x-y>0$. Then, there is some positive integer $n$ such that $x-y>\frac{1}{n}$. Now take $z=y+\frac{1}{n}$. Clearly $z>y$, $z$ is rational (because it is the sum of two rational numbers), and it is also true that $x>z$ because $x-y>\frac{1}{n}$. Thus, we have constructed a valid $z$, as desired.
\end{proof}
\vskip 0.2in
Use the following theorem in questions 4 and 5:
\begin{center}
{\bf Intermediate Value Theorem:} For every continuous function $f$ on the closed interval $[a,b]$, and for every number $k$ between $f(a)$ and $f(b)$, there is some $c\in [a,b]$ such that $f(c)=k$.
\end{center}
\item [4.] \begin{itemize}
\item The equation $x^5+2x-5=0$ has a solution on the interval $[1,2]$.
\vskip 0.2in
\begin{proof}
Let $f(x)=x^5+2x-5$. The Intermediate Value Theorem tells us that
\begin{center}
For every number $k$ between $f(1)$ and $f(2)$, there is some $c\in [1,2]$ such that $f(c)=k$.
\end{center}
$f(1)=1^5+2(1)-5=-3$ and $f(2)=2^5+2(2)-5=31$. The Intermediate Value Theorem therefore implies
\begin{center}
For every number $k$ between -3 and 31, there is some $c\in [1,2]$ such that $f(c)=k$.
\end{center}
Since 0 is between -3 and 31, it follows that there is some $c\in [1,2]$ such that $f(c)=0$. This is the desired statement.
\end{proof}
\vskip 0.2in
\item The equation $x^5-2x-5=0$ has \emph{exactly one} solution on the interval $[1,2]$.
\end{itemize}
\vskip 0.2in
\begin{proof}
Suppose there are two distinct real numbers $c, d\in [1,2]$ such that $c^5+2c-5=d^5+2d-5=0$. Then $c^5+2c=d^5+2c$. Either $c>d$ or $d>c$, so without loss of generality, $c>d$. Then $c^5>d^5$ and $2c>2d$. Thus, $c^5+2c-5>d^5+2d-5$. We have a contradiction!
\end{proof}
\vskip 0.2in
\item [5.] Any polynomial equation $f(x)=0$ of odd degree has a real number solution. (Note: this proof is much more involved than anything you'll be expected to prove at this point in the course.)
\begin{proof}
We begin by proving a lemma.
\begin{lemma}
Let $f(x)$ be an odd-degree polynomial. Then there is some real number $a$ such that $f(a)>0$, and some real number $b$ such that $f(b)<0$.
\end{lemma}
\begin{proof} [Proof of Lemma]
Since $f$ has odd degree, its degree is equal to $2k+1$ for some nonnegative integer $k$. Then, by the definition of a polynomial, we can write
$$f(x)=a_{2k+1}x^{2k+1}+a_{2k}x^{2k}+\ldots+a_1x+a_0=\sum\limits_{i=0}^{2k+1}a_ix^i$$
for some real numbers $a_0, a_1, \ldots, a_{2k+1}$ such that $a_{2k+1}\neq 0$. If the required conclusion holds for a polynomial $f(x)$, then it holds for the polynomial $t\cdot f(x)$ for any nonzero real number $t$. Therefore, we may assume without loss of generality that $a_{2k+1}=1$. Now consider plugging in $x=1+|a_{2k}|+|a_{2k-1}|+\ldots+|a_1|+|a_0|$. Then
$$\begin{aligned}
x=1+|a_{2k}|+\ldots+|a_0| &\ge 1+|a_{2k}|+|\frac{a_{2k-1}}{x}|+\ldots+|\frac{a_0}{x^{2k}}| & \text{because }x\ge 1\\
&> |a_{2k}|+|\frac{a_{2k-1}}{x}|+\ldots+|\frac{a_0}{x^{2k}}| &\\
x^{2k+1} &> |a_{2k}x^{2k}|+|a_{2k-1}x^{2k-1}|+\ldots+|a_0| & \text{ because }x\text{ is positive}\\
x^{2k+1} &> |a_{2k}x^{2k}+a_{2k-1}x^{2k-1}+\ldots+a_0| & \text{(Triangle inequality)}
\end{aligned}$$
Thus, it follows that $x^{2k+1}+a_{2k}x^{2k}+a_{2k-1}x^{2k-1}+\ldots+a_0>0$. This is our value `$a$'. To get our value `$b$', take $x=-1-|a_{2k}|-\ldots-|a_0|$, and use similar reasoning.
\end{proof}
By the lemma, there exist real numbers $a,b$ such that $f(a)>0$ and $f(b)<0$. $0$ is therefore between $f(a)$ and $f(b)$, so by the Intermediate Value Theorem, there is some real number $c$ in the interval between $a$ and $b$ such that $f(c)=0$. This completes the proof.
\end{proof}

\end{enumerate}
\end{document}