aboutsummaryrefslogtreecommitdiff
path: root/gupta/hw2.tex
diff options
context:
space:
mode:
Diffstat (limited to 'gupta/hw2.tex')
-rw-r--r--gupta/hw2.tex386
1 files changed, 386 insertions, 0 deletions
diff --git a/gupta/hw2.tex b/gupta/hw2.tex
new file mode 100644
index 0000000..922703a
--- /dev/null
+++ 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}
generated by cgit (git 2.34.1) at 2024-05-18 09:24:32 +0000