new teachers, new work

6 files changed, 1008 insertions, 0 deletions

diff --git a/gupta/Makefile b/gupta/Makefile
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+++ b/gupta/Makefile
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+.POSIX:
+.SUFFIXES: .tex .pdf
+
+PDFTEX = pdftex
+PDFLATEX = pdflatex
+
+.tex.pdf:
+	$(PDFTEX) $<
+
+all: hw1.pdf hw2.pdf hw3.pdf hw4.pdf hw5.pdf
+
+clean:
+	rm -f *.pdf *.log *.aux
+
+hw2.pdf: hw2.tex
+	$(PDFLATEX) hw2.tex
+	$(PDFLATEX) hw2.tex
+	$(PDFLATEX) hw2.tex
diff --git a/gupta/hw1.tex b/gupta/hw1.tex
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+++ b/gupta/hw1.tex
@@ -0,0 +1,18 @@
+\noindent{\it (a)}
+
+I have access to eduroam and I live on campus, so my internet is stable.
+I have a scanner app for pictures of paper work, and I have a mic and
+camera for online lectures. I don't have any other technological
+concerns.
+
+\noindent{\it (b)}
+
+I'm a first-year Math and CS double major, and I'm taking this class
+because I'm interested in abstract algebra and analysis, which have this
+class as a prerequisite.
+
+\noindent{\it (c)}
+
+I don't have any other concerns about this class.
+
+\bye
diff --git a/gupta/hw2.tex b/gupta/hw2.tex
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+++ b/gupta/hw2.tex
@@ -0,0 +1,386 @@
+\documentclass[10pt,twoside]{article}
+
+\usepackage{amssymb,amsmath,amsthm,amsfonts, epsfig, graphicx, dsfont,
+  bbm, bbold, url, color, setspace, multirow,pinlabel,tikz,pgfplots}
+\usepackage[all]{xy}
+
+\usepackage{fancyhdr} \setlength{\voffset}{-1in}
+\setlength{\topmargin}{0in} \setlength{\textheight}{9.5in}
+\setlength{\textwidth}{6.5in} \setlength{\hoffset}{0in}
+\setlength{\oddsidemargin}{0in} \setlength{\evensidemargin}{0in}
+\setlength{\marginparsep}{0in} \setlength{\marginparwidth}{0in}
+\setlength{\headsep}{0.25in} \setlength{\headheight}{0.5in}
+\pagestyle{fancy}
+
+\onehalfspace
+
+\fancyhead[LO,LE]{Math 2106 - Dr. Gupta} \fancyhead[RO,RE]{Due Thursday 1/20/2022 at 11:59 pm}
+\chead{\textbf{}} \cfoot{}
+\fancyfoot[LO,LE]{} \fancyfoot[RO,RE]{Page \thepage\ of
+  \pageref{LastPage}} \renewcommand{\footrulewidth}{0.5pt}
+\parindent 0in
+
+\makeatletter
+\def\old@comma{,}
+\catcode`\,=13
+\def,{%
+  \ifmmode%
+    \old@comma\discretionary{}{}{}%
+  \else%
+    \old@comma%
+  \fi%
+}
+\makeatother
+
+\let\implies\Rightarrow
+\def\lnot{{\sim}}
+
+%% ------------------------------------------------------%%
+%% -------------------Begin Document---------------------%%
+%% ------------------------------------------------------%%
+\begin{document}
+
+\begin{center}
+    \huge{\bf{Homework 2} - Holden Rohrer}
+\end{center}
+
+\medskip
+
+\noindent \large{\textbf{Collaborators:}}
+
+\medskip
+
+\begin{itemize}
+	\item Hammack 1.1: 16, 28, 52
+	\begin{itemize}
+	    \item[16.] Write the set $\{6a + 2b ~|~ a, b \in \mathbb Z\}$ by listing its elements between curly braces.
+	    \begin{proof}[Answer]
+            If and only if $x$ is in this set, $x+6$ and $x+2$ are in
+            this set. $0$ is in this set with $a=0$ and $b=0.$ This is
+            the set of even numbers $\{\ldots,-4,-2,0,2,4,\ldots\}.$
+	    \end{proof}
+        \item[28.] Write the set $\{\ldots, -\frac32, -\frac34, 0,
+            \frac34, \frac32, \frac94, 3, \frac{15}4, \frac92\}$ in
+            set-biulder notation.
+        \begin{proof}[Answer]
+            This set is $\{\frac34 x : x \in \mathbb Z\}.$
+        \end{proof}
+        \item[52.] Sketch the set of points $\{(x,y)\in\mathbb R^2 :
+            (y-x^2)(y+x^2) = 0\}.$
+        \begin{proof}[Answer]
+            This set is the union of the two graphs $y = x^2$ and $y =
+            -x^2.$
+            \begin{figure}[h!]
+            \centering
+            \begin{tikzpicture}[scale=3]
+                \draw[<->] (-1,0) -- (1,0) node[right] {$x$};
+                \draw[<->] (0,-1) -- (0,1) node[above] {$y$};
+                \draw (0,0) parabola (1,1);
+                \draw (0,0) parabola (-1,1);
+                \draw (0,0) parabola (1,-1);
+                \draw (0,0) parabola (-1,-1);
+            \end{tikzpicture}
+            \end{figure}
+            %%TO DRAW
+        \end{proof}
+	\end{itemize}
+	\item Hammack 1.2: 2
+    \begin{itemize}
+        \item[2.] Suppose $A = \{\pi,e,0\}$ and $B = \{0,1\}.$
+        \begin{itemize}
+            \item[(a)] Write out $A\times B.$
+            \begin{proof}[Answer]
+                $A\times B = \{(\pi,0), (\pi, 1), (e,0), (e,1), (0,0),
+                (1,0)\}.$
+            \end{proof}
+            \item[(b)] Write out $B\times A.$
+            \begin{proof}[Answer]
+                $B\times A = \{(0,\pi), (1,\pi), (0,e), (1,e), (0,0),
+                (1,0)\}.$
+            \end{proof}
+            \item[(c)] Write out $A\times A.$
+            \begin{proof}[Answer]
+                $A\times A = \{(\pi,\pi), (\pi,e), (\pi, 0), (e,\pi),
+                (e,e), (e,0), (0,\pi), (0,e), (0,0)\}.$
+            \end{proof}
+            \item[(d)] Write out $B\times B.$
+            \begin{proof}[Answer]
+                $B\times B = \{(0,0), (0,1), (1,0), (1,1)\}.$
+            \end{proof}
+            \item[(e)] Write out $A\times \emptyset.$
+            \begin{proof}[Answer]
+                $A\times\emptyset = \emptyset.$
+            \end{proof}
+            \item[(f)] Write out $(A\times B)\times B.$
+            \begin{proof}[Answer]
+                $(A\times B)\times B = \{((\pi,0),0), ((\pi, 1),0),
+                ((e,0),0), ((e,1),0), ((0,0),0), ((0,1),0),
+                ((\pi,0),1), ((\pi, 1),1),
+                ((e,0),1), ((e,1),1), ((0,0),1), ((0,1),1)\}.$
+            \end{proof}
+            \item[(g)] Write out $A\times(B\times B).$
+            \begin{proof}[Answer]
+                $A\times(B\times B) = \{(\pi,(0,0)), (\pi,(1,0)),
+                (e,(0,0)), (e,(1,0)), (0,(0,0)), (0,(1,0)),
+                (\pi,(0,1), (\pi,(1,1)),
+                (e,(0,1)), (e,(1,1)), (0,(0,1)), (0,(1,1))\}.$
+            \end{proof}
+            \item[(h)] Write out $A\times B\times B.$
+            \begin{proof}[Answer]
+                $A\times B\times B = \{(\pi,0,0), (\pi, 1,0),
+                (e,0,0), (e,1,0), (0,0,0), (0,1,0),
+                (\pi,0,1), (\pi, 1,1),
+                (e,0,1), (e,1,1), (0,0,1), (0,1,1)\}.$
+            \end{proof}
+        \end{itemize}
+    \end{itemize}
+	\item Hammack 1.3: 2, 8, 10, 16
+    \begin{itemize}
+        \item[2.] List all subsets of $\{1,2,\emptyset\}.$
+        \begin{proof}[Answer]
+            The subsets are $\emptyset, \{1\}, \{2\}, \{\emptyset\},
+            \{1,2\}, \{1,\emptyset\}, \{2,\emptyset\},
+            \{1,2,\emptyset\}.$
+        \end{proof}
+        \item[8.] List all subsets of $\{\{0,1\}, \{0,1,\{2\}\},
+            \{0\}\}.$
+        \begin{proof}[Answer]
+            The subsets are $\emptyset, \{\{0,1\}\}, \{\{0,1,\{2\}\}\},
+            \{\{0\}\}, \{\{0,1\},\{0\}\}, \{\{0,1\},\{0,1,\{2\}\}\},
+            \{\{0\},\{0,1,\{2\}\}\}, \{\{0,1\},\{0\},\{0,1,\{2\}\}\}.$
+        \end{proof}
+        \item[10.] Write out $\{X\subseteq N : |X| \leq 1\}.$
+        \begin{proof}[Answer]
+            This set is $\{-1,0,1\}.$
+        \end{proof}
+        \item[16.] Decide if $\{(x,y) : x^2 - x = 0\} \subseteq
+            \{(x,y) : x-1 = 0\}$ is true or false.
+        \begin{proof}[Answer]
+            False. $(0,0)$ is in the first set but not in the second
+            set, so it cannot be a subset.
+        \end{proof}
+    \end{itemize}
+	\item Hammack 1.4: 6, 14, 16, 18, 20
+    \begin{itemize}
+        \item[6.] Find $\mathcal P(\{1,2\})\times\mathcal P(\{3\}).$
+        \begin{proof}[Answer]
+            This is $\{(\emptyset,\emptyset), (\{1\},\emptyset),
+            (\{2\},\emptyset), (\{1,2\},\emptyset), (\emptyset, \{3\}),
+            (\{1\}, \{3\}), (\{2\}, \{3\}), (\{1,2\}, \{3\}).$
+        \end{proof}
+        \item[14.] Suppose $|A| = m$ and $|B| = n.$ Find
+            $|\mathcal P(\mathcal P(A))|.$
+        \begin{proof}[Answer]
+            $|\mathcal P(\mathcal P(A))| = 2^{|\mathcal P(A)|} =
+            2^{2^m}.$
+        \end{proof}
+        \item[16.] Suppose $|A| = m$ and $|B| = n.$ Find
+            $|\mathcal P(A)\times \mathcal P(B)|.$
+        \begin{proof}[Answer]
+            This is $|\mathcal P(A)|\cdot|\mathcal P(B)| = 2^{|\mathcal
+            P(A)|}2^{|\mathcal P(B)|} = 2^{nm}$
+        \end{proof}
+        \item[18.] Suppose $|A| = m$ and $|B| = n.$ Find
+            $|\mathcal P(A\times \mathcal P(B))|.$
+        \begin{proof}[Answer]
+            By similar techniques, this set has cardinality $2^{m2^n}.$
+        \end{proof}
+        \item[20.] Suppose $|A| = m$ and $|B| = n.$ Find $|\{X\subseteq\mathcal P(A) : |X|\leq 1\}|.$
+        \begin{proof}[Answer]
+            This is $\emptyset$ and every one-element subset of
+            $\mathcal P(A),$ so it has cardinality $2^m + 1.$
+        \end{proof}
+    \end{itemize}
+	\item Hammack 1.6: 2
+    \begin{itemize}
+        \item[2.] Let $A = \{0,2,4,6,8\}$ and $B = \{1,3,5,7\}$ have
+            universal set $U = \{0,1,2,\ldots,8\}.$ Find:
+        \begin{itemize}
+            \item[(a)] $\bar A = B.$
+            \item[(b)] $\bar B = A.$
+            \item[(c)] $A\cap\bar A = \emptyset.$
+            \item[(d)] $A\cup\bar A = U.$
+            \item[(e)] $A - \bar A = A.$
+            \item[(f)] $\bar{A\cup B} = \emptyset.$
+            \item[(g)] $\bar A \cap \bar B = \emptyset.$
+            \item[(h)] $\bar{A\cap B} = U.$
+            \item[(i)] $\bar A \times B = B^2.$
+        \end{itemize}
+        %\begin{proof}[Answer]
+        %\end{proof}
+    \end{itemize}
+	\item Hammack 1.8: 4, 10, 12, 14
+    \begin{itemize}
+        \item[4.] For each $n\in \mathbb N,$ let $A_n = \{-2n,0,2n\}.$
+        \begin{itemize}
+            \item[(a)] $\bigcup_{i\in N} A_i = \{2n:n\in\mathbb N\}$
+            \item[(b)] $\bigcap_{i\in N} A_i = \{0\}$
+        \end{itemize}
+        \item[10.]
+        \begin{itemize}
+            \item[(a)] $\bigcup_{i\in [0,1]} [x,1]\times[0,x^2] =
+                \{(x,y)\in \mathbb [0,1]^2 : y \leq x^2\}.$
+            \item[(b)] $\bigcap_{i\in [0,1]} [x,1]\times[0,x^2] =
+                \{(1,0)\},$ as can be seen from the intersection of the
+                $[0,1]\times[0,0]$ and $[1,1]\times[0,1]$ elements.
+        \end{itemize}
+        \item[12.] If $\bigcap_{\alpha\in I} A_\alpha =
+            \bigcup_{\alpha\in I} A_\alpha,$ what do you think can be
+            said about the relationships between the sets $A_{\alpha}?$
+        \begin{proof}[Answer]
+            For any $\alpha,\beta \in I,$ $A_\alpha = A_\beta.$
+        \end{proof}
+        \item[14.] If $J\neq\emptyset$ and $J\subseteq I,$ does it
+            follow that $\bigcap_{\alpha\in I} A_\alpha \subseteq
+            \bigcap_{\alpha\in J} A_\alpha?$ Explain.
+        \begin{proof}[Answer]
+            Yes, this does follow because
+            $\bigcap_{\alpha\in I} A_\alpha = \bigcap_{\alpha\in J}
+            A_\alpha \cap \bigcap_{\beta\in I - J} A_\beta
+            \subseteq \bigcap_{\alpha\in J} A_\alpha.$
+        \end{proof}
+    \end{itemize}
+	\item Hammack 2.3: 2, 6
+    \begin{itemize}
+        \item[2.] Convert the following sentence into the form ``{\em If
+            P, then Q}.'' ``For a function to be continuous, it is
+            sufficient that it is differentiable.''
+        \begin{proof}[Answer]
+            ``If a function is differentiable, that function is
+            continuous.''
+        \end{proof}
+        \item[6.] Convert the following sentence into the form ``{\em If
+            P, then Q}.'' ``Whenever a surface has only one side, it is
+            non-orientable.''
+        \begin{proof}[Answer]
+            ``If a surface has one side, that surface is
+            non-orientable.''
+        \end{proof}
+    \end{itemize}
+	\item Hammack 2.5: 10
+    \begin{itemize}
+        \item[10.] Suppose the statement $((P\land Q)\lor R) \implies
+            (R\lor S)$ is false. Find the truth values of $P$, $Q,$ $R,$
+            and $S.$
+        \begin{proof}[Answer]
+            $P=Q={\rm True}$ and $R=S={\rm False}$ make this statement
+            false.
+        \end{proof}
+    \end{itemize}
+	\item Hammack 2.6: 14
+    \begin{itemize}
+        \item[14.] Decide whether or not $P\land(Q\lor\lnot Q)$ and
+            $(\lnot P) \implies (Q\land\lnot Q)$ are logically
+            equivalent.
+        \begin{proof}[Answer]
+            These are logically equivalent. $Q\lor ~Q \equiv T$ and
+            $Q\land ~Q \equiv F,$ giving us simplified expressions
+            $P\land T \equiv P$ and $\lnot P \implies F \equiv P,$ so
+            these are logically equivalent expressions.
+        \end{proof}
+    \end{itemize}
+	\item Problems not from the textbook
+  	\begin{enumerate}
+         \item Use truth tables to prove that each of the following compound propositions is \emph{not} a tautology. These implication are common logical fallacies (errors in reasoning) since the conclusion does not follow from the hypotheses.
+         \begin{enumerate}
+              \item $[(P\implies Q)\land Q]\implies P$
+              \begin{proof}[Answer]
+                  \begin{tabular}{|c|c|c|c|c|c|}
+                      \hline
+                      $P$ & $Q$ & $P \implies Q$ & $(P \implies Q)\land
+                      Q$ & $[(P\implies Q)\land Q] \implies P$ \\\hline
+
+                      T & T & T & T & T\\\hline
+                      T & F & F & F & T\\\hline
+                      F & T & T & T & F\\\hline
+                      F & F & T & F & T\\\hline
+                  \end{tabular}
+              \end{proof}
+  			  \item $[(P\implies Q)\land\lnot{P}]\implies\lnot{Q}$
+              \begin{proof}[Answer]
+                  \begin{tabular}{|c|c|c|c|c|c|}
+                      \hline
+                      $P$ & $Q$ & $P \implies Q$ & $(P \implies Q)\land
+                      \lnot P$ & $[(P\implies Q)\land \lnot P] \implies
+                      \lnot Q$ \\\hline
+
+                      T & T & T & F & T\\\hline
+                      T & F & F & F & T\\\hline
+                      F & T & T & T & F\\\hline
+                      F & F & T & T & T\\\hline
+                  \end{tabular}
+              \end{proof}
+ 		\end{enumerate}
+		\item Use truth tables to prove that each of the following compound propositions is a tautology. These are the four most important ``rules of inference'' in propositional logic. Each rule gives a conclusion that follows from a set of hypotheses and thus give building blocks for correct proofs.
+ 		 \begin{enumerate}
+             \item $[P\land(P\implies Q)]\implies Q$
+             \begin{proof}[Answer]
+                  \begin{tabular}{|c|c|c|c|c|c|}
+                      \hline
+                      $P$ & $Q$ & $P \implies Q$ &
+                      $P\land (P \implies Q)$ &
+                      $[P\land (P\implies Q)] \implies Q$ \\\hline
+
+                      T & T & T & T & T\\\hline
+                      T & F & F & F & T\\\hline
+                      F & T & T & F & T\\\hline
+                      F & F & T & F & T\\\hline
+                  \end{tabular}
+             \end{proof}
+			 \item $[\lnot Q\land(P\implies Q)]\implies\lnot{P} $
+             \begin{proof}[Answer]
+                  \begin{tabular}{|c|c|c|c|c|c|}
+                      \hline
+                      $P$ & $Q$ & $P \implies Q$ & $(P \implies Q)\land
+                      \lnot P$ & $[(P\implies Q)\land \lnot P] \implies
+                      \lnot Q$ \\\hline
+
+                      T & T & T & F & T\\\hline
+                      T & F & F & F & T\\\hline
+                      F & T & T & T & F\\\hline
+                      F & F & T & T & T\\\hline
+                  \end{tabular}
+             \end{proof}
+			 \item $[(P\implies Q)\land(Q\implies R)]\implies(P\implies R)$
+             \begin{proof}[Answer]
+                 Let $(1)$ be $(P\implies Q)\land(Q\implies R).$
+
+                  \begin{tabular}{|c|c|c|c|c|c|c|c|}
+                      \hline
+                      $P$ & $Q$ & $R$ & $P \implies Q$ &
+                      $Q \implies R$ &
+                      $(1)$ &
+                      $P \implies R$ &
+                      $(1) \implies (P\implies R)$ \\\hline
+
+                      T & T & T & T & T & T & T & T\\\hline
+                      T & T & F & T & F & F & F & T\\\hline
+                      T & F & T & F & T & F & T & T\\\hline
+                      T & F & F & F & T & F & F & T\\\hline
+                      F & T & T & T & T & T & T & T\\\hline
+                      F & T & F & T & F & F & T & T\\\hline
+                      F & F & T & T & T & T & T & T\\\hline
+                      F & F & F & T & T & T & T & T\\\hline
+                      % to break up lol
+                  \end{tabular}
+             \end{proof}
+             \item $[(P\lor Q)\land\lnot{P}]\implies Q$
+             \begin{proof}[Answer]
+                  \begin{tabular}{|c|c|c|c|c|c|}
+                      \hline
+                      $P$ & $Q$ & $P \lor Q$ & $(P \lor Q)\land
+                      \lnot P$ & $[(P\lor Q)\land \lnot P] \implies Q$
+                      \\\hline
+
+                      T & T & T & F & T\\\hline
+                      T & F & T & F & T\\\hline
+                      F & T & T & T & T\\\hline
+                      F & F & F & F & T\\\hline
+                  \end{tabular}
+             \end{proof}
+		\end{enumerate}
+	\end{enumerate}
+\end{itemize}
+
+\label{LastPage}
+\end{document}
diff --git a/gupta/hw3.tex b/gupta/hw3.tex
new file mode 100644
index 0000000..1961ce3
--- /dev/null
+++ b/gupta/hw3.tex
@@ -0,0 +1,151 @@
+\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
diff --git a/gupta/hw4.tex b/gupta/hw4.tex
new file mode 100644
index 0000000..fe80e3f
--- /dev/null
+++ b/gupta/hw4.tex
@@ -0,0 +1,218 @@
+\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{\bf Answer.}\par}
+\def\endproof{\leavevmode\hskip\parfillskip\vrule height 6pt width 6pt
+depth 0pt{\parfillskip0pt\medskip}}
+\let\endanswer\endproof
+\def\section#1{\medskip\goodbreak\noindent{\bf #1}}
+\let\impl\Rightarrow
+\def\nmid{\not\hskip2.5pt\mid}
+
+\headline{\vtop{\hbox to \hsize{\strut Math 2106 - Dr. Gupta\hfil Due Thursday
+2022-02-03 at 11:59 pm}\hrule height .5pt}}
+
+\centerline{\bigbf Homework 4 - Holden Rohrer}
+\bigskip
+
+\noindent{\bf Collaborators:} None
+
+\section{Hammack 4: 26}
+
+\item{26.} Prove the following with direct proof: every odd integer is a
+difference of two squares.
+
+\answer
+Let $n$ be an odd integer.
+We will show that there are two perfect squares $a^2$ and $b^2$ (with
+$a,b\in\bb Z$) such that $n$ is their difference.
+
+Because it is odd, there exists an integer $k$ such that $n = 2k+1.$
+$$(k+1)^2-k^2 = k^2+2k+1-k^2 = 2k+1 = n,$$
+so any odd integer can be written as the difference of two squares.
+\endanswer
+
+\section{Hammack 5: 6, 12, 18, 20, 24, 28}
+
+\item{6.} Prove the following with contrapositive proof: suppose
+$x\in\bb R.$ If $x^3-x>0,$ then $x>-1.$
+
+\answer
+For contrapositive proof, let $x \leq -1.$
+We will prove that $x^3-x\leq 0.$
+
+We obtain $x+1 \leq 0$ and $x-1 \leq -2.$
+$$x(x-1)(x+1) \leq 0,$$
+because the product of three non-positive numbers is non-positive.
+% is this sufficient??
+\endanswer
+
+\item{12.} Prove the following with contrapositive proof: suppose
+$a\in\bb Z.$ If $a^2$ is not divisible by 4, then $a$ is odd.
+
+\answer
+For contrapositive proof, let $a$ be not odd (even). We will show that
+$a^2$ is divisible by 4.
+
+By the definition of even, there exists $k$ such that $a = 2k.$
+$a^2 = 4k^2,$ and $k^2\in\bb Z,$ so $a^2$ is divisible by 4.
+\endanswer
+
+\item{18.} Prove the following with either direct or contrapositive
+proof: for any $a,b\in\bb Z,$ it follows that $(a+b)^3\equiv a^3+b^3
+\pmod{3}$
+
+\answer
+Let $a,b\in\bb Z.$
+We will prove that $(a+b)^3\equiv a^3+b^3\pmod{3}.$
+
+$$(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 = a^3 + b^3 + 3(a^2b+ab^2),$$
+so $(a+b)^3\equiv a^3 + b^3 \pmod{3}$ by definition of modular
+equivalence.
+\endanswer
+
+\item{20.} Prove the following with either direct or contrapositive
+proof: if $a\in\bb Z$ and $a\equiv 1\pmod{5},$ then $a^2\equiv
+1\pmod{5}.$
+
+\answer
+Let $a\in\bb Z$ and $a\equiv 1\pmod{5}.$
+We will prove that $a^2\equiv 1\pmod{5}.$
+
+There exists $k\in\bb Z$ such that $a = 5k+1.$
+$$a^2 = (5k+1)^2 = 25k^2 + 10k + 1 = 5(5k^2+2k) + 1,$$
+so $a^2 \equiv 1\pmod{5}.$
+\endanswer
+
+\item{24.} Prove the following with either direct or contrapositive
+proof: if $a\equiv b \pmod{n}$ and $c\equiv d \pmod{n},$ then $ac\equiv
+bd \pmod{n}.$
+
+\answer
+Let $a,b,c,d\in\bb R.$
+Let $a\equiv b \pmod{n}$ and $c\equiv d \pmod{n}.$
+We will show that $ac\equiv bd\pmod{n}.$
+
+Therefore, there is $k\in\bb Z$ such that $a = b + nk,$ and there is
+$j\in\bb Z$ such that $c = d + nj.$
+$$ac = (b+nk)(d+nj) = bd + n(jb+dk+njk),$$
+so $ac \equiv bd \pmod{n},$ because $jb+dk+njk\in\bb Z.$
+\endanswer
+
+\item{28.} Prove the following with either direct or contrapositive
+proof: if $n\in\bb Z,$ then $4\nmid (n^2-3).$
+
+\smallskip
+{\bf Lemma.}
+Let $n\in\bb Z.$ We will show that, for some $k\in\bb Z,$ $n^2 = 4k$ or
+$n^2 = 4k+1.$
+$n$ can be written as exactly one of $4j,$ $4j+1,$ $4j+2,$ and $4j+3,$
+where $j\in\bb Z.$
+
+In the first case $n = 4j,$ $n^2 = 16j^2 = 4(4j^2),$ so with $k=4j^2,$
+$n^2 = 4k.$
+
+In the second case $n = 4j+1,$ $n^2 = 16j^2 + 8j + 1 = 4(4j^2+2j) + 1,$
+so with $k = 4j^2+2j,$ $n^2 = 4k + 1.$
+
+In the third case $n = 4j+2,$ $n^2 = 16j^2 + 16j + 4 = 4(4j^2+4j+1),$ so
+with $k = 4j^2+4j+1,$ $n^2 = 4k.$
+
+In the fourth case $n = 4j+3,$ $n^2 = 16j^2 + 24j + 9 = 4(4j^2+6j+2)+1,$
+so with $k = 4j^2+6j+2,$ $n^2 = 4k+1.$
+
+All of these values of $k$ are integers by integer closure.
+
+$n^2\neq 4k+2$ and $n^2\neq 4k+3$ for any integer $k$ because $n^2$ is
+an integer and it can be written as exactly one of $4k,$ $4k+1,$ $4k+2,$
+and $4k+3.$
+\endproof
+
+\answer
+Let $n\in\bb Z.$
+By the lemma, $n^2 = 4k$ or $n^2 = 4k+1.$
+In the case $n^2 = 4k,$ $n^2 - 3 = 4k - 3 = 4(k-1) + 1,$
+which is not divisible by $4.$
+In the case $n^2 = 4k+1,$ $n^2 - 3 = 4k - 2 = 4(k-1) + 2,$
+which is not divisible by $4.$
+\endanswer
+
+\section{Hammack 6: 4, 6, 8}
+
+\item{4.} Prove the following by contradiction: $\sqrt 6$ is irrational.
+
+\answer
+For the sake of contradiction, assume that $\sqrt 6$ is rational.
+Therefore, there exist coprime $p,q\in\bb Z$ such that $\sqrt 6 =
+{p\over q}.$
+
+$$p = q\sqrt 6 \to p^2 = 6q^2.$$
+$p^2$ is even only if $p$ is even ($2\mid p$), so $4\mid p^2,$ so $2\mid
+q^2,$ and thus $2\mid q.$
+Therefore, $(q,p) = 2\neq 1,$ meaning they're not coprime.
+\endanswer
+
+\item{6.} Prove the following by contradiction: if $a,b\in\bb Z,$ then
+$a^2-4b-2\neq 0.$
+
+\answer
+Let $a,b\in\bb Z,$ and for the sake of contradiction, let $a^2-4b-2 =
+0.$
+$$a^2-4b-2 \equiv 0 \equiv a^2-2 \to a^2 \equiv 2 \bmod{4}.$$
+Therefore, there exists $k\in\bb Z$ such that $a^2 = 4k+2.$
+By the lemma, $a^2 \neq 4k + 2.$
+\endanswer
+
+\item{8.} Prove the following by contradiction: suppose $a,b,c\in\bb Z.$
+If $a^2+b^2=c^2,$ then $a$ or $b$ is even.
+
+\answer
+Let $a,b,c\in\bb Z$ such that $a^2+b^2=c^2.$
+Suppose, for the sake of contradiction, $a$ and $b$ are odd.
+There exists $j$ and $k$ such that $a=2k+1$ and $b=2j+1.$
+
+Therefore, $a^2 + b^2 = (2k+1)^2 + (2j+1)^2 = 4k^2 + 4k + 1 + 4j^2 + 4j
++ 1 = 4(k^2+k+j^2+j) + 2 = c^2.$
+By the lemma, $c^2 \neq 4k + 2.$
+\endanswer
+
+\section{Problems not from the textbok}
+
+\item{1.} A perfect square is an integer $n$ for which there exists an
+integer $k$ such that $n = k^2.$ Prove that if $n$ is a positive integer
+such that $n\equiv 2\bmod 4$ or $n\equiv 3\bmod 4,$ then $n$ is not a
+perfect square.
+
+\answer
+This has already been proven in the above lemma.
+\endanswer
+
+\item{2.} After a grueling slog through the snow to reach Ponce City
+Market, you decide to reward yourself by buying three boxes of candy
+from Collier’s. One box contains mint candies, one chocolate candies,
+and the other is mixed. Unfortunately, all three boxes were incorrectly
+labeled! What is the smallest number of candies that you need to remove
+and sample to be able to correctly label all three boxes? Carefully
+justify your reasoning.
+
+\answer
+We only need to sample one candy.
+We sample the box labeled ``mixed,'' and without loss of generality we
+get a chocolate candy. This is the chocolate box.
+This box cannot be ``mixed'' because it is labeled incorrectly, and this
+box cannot be ``mint'' because the mint box doesn't have chocolate
+candy.
+Now, the box labeled ``mint'' must be the mixed box because it cannot be
+the chocolate box (we only have one of those) and it cannot be the mixed
+box because it is incorrectly labeled.
+By elimination, the last box is the mixed box.
+\endanswer
+
+\bye
diff --git a/gupta/hw5.tex b/gupta/hw5.tex
new file mode 100644
index 0000000..bb1bd43
--- /dev/null
+++ b/gupta/hw5.tex
@@ -0,0 +1,217 @@
+\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{\bf Answer.}\par}
+\def\endproof{\leavevmode\hskip\parfillskip\vrule height 6pt width 6pt
+depth 0pt{\parfillskip0pt\medskip}}
+\let\endanswer\endproof
+\def\section#1{\medskip\vskip0pt plus 1in\goodbreak\vskip 0pt plus -1in%
+\noindent{\bf #1}}
+\let\impl\to
+\def\nmid{\hskip-3pt\not\hskip2.5pt\mid}
+\def\problem#1{\par\penalty-100\item{#1}}
+
+\headline{\vtop{\hbox to \hsize{\strut Math 2106 - Dr. Gupta\hfil Due Thursday
+2022-02-17 at 11:59 pm}\hrule height .5pt}}
+
+\centerline{\bigbf Homework 5 - Holden Rohrer}
+\bigskip
+
+\noindent{\bf Collaborators:} None
+
+\section{Hammack 7: 6, 9, 12}
+
+\problem{6.} Suppose $x,y\in\bb R.$ Then $x^3+x^2y = y^2+xy$ if and only if
+$y = x^2$ or $y = -x.$
+
+$x^2(x+y) = y(x+y).$
+\answer
+$(\Rightarrow)$
+
+Let $x^3 + x^2y = y^2+xy.$
+We then have $x^2(x+y) = y(x+y).$
+If $y = -x,$ $x+y = 0 \impl x^2(x+y) = y(x+y).$
+Otherwise, we can divide by $x+y$ because it is nonzero, giving
+$y = x^2.$
+Therefore, $y=-x$ or $y=x^2.$
+
+$(\Leftarrow)$
+
+Let $y = -x$ or $y = x^2.$
+We will first consider the case $y = -x,$ then the case $y = x^2.$
+
+With $y = -x,$ $x^3 + x^2y = x^3 - x^3 = 0 = x^2 - x^2 = y^2 + xy.$
+
+If $y = x^2,$ $x^3 + x^2y = x^3 + x^4 = x^3 + x^4 = xy + y^2.$
+\endanswer
+
+\problem{9.} Suppose $a\in\bb Z.$ Prove that $14\mid a$ if and only if
+$7\mid a$ and $2\mid a.$
+
+\answer
+$(\Rightarrow)$
+
+Let $14\mid a.$
+This gives $a = 14k$ for some integer $k.$
+$a = 2(7k),$ so with $7k\in\bb Z,$ we get $2\mid a.$
+Similarly, $a = 7(2k),$ so with $2k\in\bb Z,$ we get $7\mid a.$
+
+$(\Leftarrow)$
+
+Let $7\mid a$ and $2\mid a.$
+These give $a = 7j$ for some $j$ and $a = 2k$ for some integers $j$ and
+$k.$
+A product of odd number $7$ and odd number $j$ cannot be even (and $a$
+is even because $2\mid a$), so $j$ must be even.
+Thus, there exists $l\in\bb Z$ such that $j = 2l.$
+This gives $a = 7(2l) = 14l \to 14\mid a.$
+\endanswer
+
+\problem{12.} There exist a positive real number $x$ for which $x^2 < \sqrt
+x.$
+
+\answer
+Observe that $x = 1/4$ gives $x^2 = 1/16$ and $\sqrt x = 1/2,$ so $1/16
+< 1/2.$
+\endanswer
+
+\section{Hammack 8: 12, 22, 28}
+
+\problem{12.} If $A,$ $B,$ and $C$ are sets, then $A-(B\cap C) =
+(A-B)\cup(A-C).$
+
+\answer
+$(\subseteq)$
+
+Let $x\in A - (B\cap C).$
+This gives us $x\in A$ and $x\not\in B\cap C.$
+We get $x\not\in B$ or $x\not\in C.$
+WLOG, let $x\not\in B.$
+$x\in A$ and $x\not\in B,$ so $x\in A-B,$ so $x\in (A-B)\cup(A-C).$
+
+$(\supseteq)$
+
+Let $x\in (A-B)\cup (A-C).$
+This gives $x\in A-B$ or $x\in A-C.$
+WLOG, let $x\in A-B.$
+Therefore, $x\in A$ and $x\not\in B.$
+This implies $x\not\in B\cap C,$ so $x\in A-(B\cap C).$
+
+Since we have $A-(B\cap C) \subseteq (A-B)\cup(A-C)$ and $(A-B)\cup(A-C)
+\subseteq A-(B\cap C),$
+we obtain $$A-(B\cap C) = (A-B)\cup(A-C).$$
+\endanswer
+
+\problem{22.} Let $A$ and $B$ be sets. Prove that $A\subseteq B$ if and
+only if $A\cap B = A.$
+
+\answer
+$(\Rightarrow)$
+
+Let $A\subseteq B.$
+Let $x\in A.$
+By subset, $x\in B.$
+And if and only if $x\in A$ and $x\in B,$ $x\in A\cap B,$ so $A = A\cap
+B.$
+
+$(\Leftarrow)$
+
+Let $A\cap B = A.$
+This implies $A\subseteq A\cap B,$ or $x\in A\impl x\in A\cap B\impl
+x\in B.$
+$x\in A\impl x\in B$ is the definition of $A\subseteq B.$
+\endanswer
+
+\problem{28.} Prove $\{12a+25b:a,b\in\bb Z\} = \bb Z.$
+
+\answer
+Let $A = \{12a+25b:a,b\in\bb Z\}.$ If $x\in A,$ $x\in\bb Z$ because the
+integers are closed under addition and mulitplication.
+
+If $x\in\bb Z,$ let $b = x$ and $a = -2x,$ giving us $12(-2x) + 25x =
+x\in A.$
+
+Therefore, since $A\subseteq\bb Z$ and $\bb Z\subseteq A,$ these two
+sets are equal.
+\endanswer
+
+\section{Hammack 9: 6, 28, 30, 34}
+Prove or disprove each of the following statements.
+
+\problem{6.} If $A,$ $B,$ $C,$ and $D$ are sets, then $(A\times
+B)\cup(C\times D) = (A\cup C)\times(B\cup D).$
+
+\answer
+{\bf Disproof.}
+
+Let $A = B = \{1\}$ and $C = D = \{2\}.$
+The set $A\times B = \{(1,1)\}.$
+The set $C\times D = \{(2,2)\}.$
+And the set $A\cup C = B\cup D = \{1,2\}.$
+
+$$(A\times B)\cup(C\times D) = \{(1,1),(2,2)\} \neq
+\{(1,1),(1,2),(2,1),(2,2)\} = (A\cup C)\times(B\cup D).$$
+\endanswer
+
+\problem{28.} Suppose $a,b\in\bb Z.$ If $a\mid b$ and $b\mid a,$ then $a =
+b.$
+
+\answer
+We will show this by contrapositive.
+Let $a\neq b.$ We will show that $a\nmid b$ or $b\nmid a.$
+
+WLOG, $a > b.$
+Immediately, $a\nmid b.$
+\endanswer
+
+\problem{30.} There exist integers $a$ and $b$ for which $42a + 7b = 1.$
+
+\answer
+{\bf Disproof.}
+
+Let $a,b\in\bb Z.$
+For the sake of contradiction, assume $42a + 7b = 1.$
+Dividing by 7, $6a + b = 1/7.$
+By closure of the integers, $6a+b\in\bb Z,$ and $1/7\not\in\bb Z,$
+giving a contradiction.
+\endanswer
+
+\problem{34.} If $X\subseteq A\cup B,$ then $X\subseteq A$ or $X\subseteq
+B.$
+
+\answer
+{\bf Disproof.}
+
+Let $A = \{1\},$ $B = \{2\},$ and $X = A\cup B.$
+Immediately, $X\subseteq A\cup B.$
+And then, $2\in X,$ but $2\not\in A,$ so $X\not\subseteq A.$
+Also, $1\in X,$ but $1\not\in B,$ so $X\not\in B.$
+\endanswer
+
+\section{Problem not from the textbok}
+
+\problem{1.} Let $A,$ $B,$ and $C$ be arbitrary sets. Prove that if
+$A-C\not\subseteq A-B,$ then $B\not\subseteq C.$
+
+\answer
+We will prove this by contrapositive.
+Let $B\subseteq C.$ We will show that $A-C\subseteq A-B.$
+
+If $x\in B,$ $x\in C,$ so by contrapositive, if $x\not\in C,$ $x\not\in
+B.$
+
+Let $y\in A-C.$
+By definition of setminus, $y\in A$ and $y\not\in C.$
+As established, this implies $y\not\in B.$
+Therefore, with $y\in A$ and $y\not\in B,$ $y\in A-B,$ so
+$A-C\subseteq A-B.$
+\endanswer
+
+\bye