I feel that my code isn't written very efficiently.
\documentclass[a4paper,11pt]{article}
\usepackage[english]{babel}
\usepackage[pdftex]{graphicx}
\usepackage{tikz}
\usepackage{wasysym}
\usepackage[all]{xy}
\usepackage{amsfonts}
\title{Discrete Mathematics -- Lecture 15}
\author{Alec Taylor}
\date{August 25, 2011}
\begin{document}
\maketitle
\section{Disjoint definition}
Sets A and B are \underline{disjoint} if A $\cap$ B = \{ c,d,e \}
\\[2mm]
But if C=\{ x,y,z \} then A and C are disjoint, B and C are disjoint.
\begin{center}
\line(1,0){250}
\end{center}
Note: A $\subseteq$ A so A $\in$ $\mathcal{P}$(A)
\\[2mm]
$\mathcal{P}$ $\subseteq$ A so $\emptyset$ $\in$ $\mathcal{P}$(A)
\\[2mm]
x $\in$ A, \{x\} A, \{x\} $\in$ $\mathcal{P}$(A)
\section{Cartesian definition}
The \underline{cartesian product} of the set $A_1$, $A_2$, ..., $A_n$ is $A_1$ $\times$ $A_2$ $\times$ ... $\times$ $A_n$
\newline
=\{($a_1$, $a_2$, ..., $a_n$)(a, $\in$$A_1$)$\land$($a_2$$\in$$A_2$)$\land$...$\land$($a_n$$\in$$A_n$)\}
\\[2mm]
The element ($a_1$, $a_2$, ..., $a_n$) is an \underline{ordered n-tuple}.
\\[2mm]
E.g.: (x,y) in cartesian plan (i.e.: the coordiantes of a point in the plane) is an ordered pair, the Cartesian plane is the product $\mathbb{R}$x$\mathbb{R}$ or $\mathbb{R}^2$.
\section{Cardinality definition}
In general if $\|$A$\|$=n, $\|$B$\|$=m then $\|$A $\times$ B$\|$ = $\|$A$\|$ $\|$B$\|$
\\[2mm]
$\|$$A^n$$\|$=$\|$A$\|$ $\|$A$\|$ (<n times>) $\|$A$\|$ = $\|$$A^n$$\|$
\end{document}
\documentclass[a4paper,11pt]{article}
\usepackage[english]{babel}
\usepackage[pdftex]{graphicx}
\usepackage{tikz}
\usepackage{wasysym}
\usepackage[all]{xy}
\usepackage{amsfonts}
\title{Discrete Mathematics -- Lecture 16}
\author{Alec Taylor}
\date{August 25, 2011}
\begin{document}
\maketitle
\section{Laws}
Let A = \{ x $\in$ $\mathbb{Z}$ $\|$ x $\le$ 5 \}
\newline
Let B = \{ x $\in$ $\mathbb{Z}$ $\|$ x $>$ 0 \}
\\[2mm]
A $\cap$ B = \{ x $\in$ $\mathbb{Z}$ $\|$ (x $\le$ 5) $\land$ (x $>$0) \}
\newline
= \{ 1,2,3,4,5 \}
\\[2mm]
A $\cup$ B = \{ x $\in$ $\mathbb{Z}$ $\|$ (x $\le$ 5) $\lor$ (x $>$ 0) \} = $\mathbb{Z}$
\\[2mm]
$\bar{A}$ = \{ x $\in$ $\mathbb{Z}$ $\|$ x $>$ 5 \}
\newline
= \{x $\in$ $\mathbb{Z}$ $\|$-(x$\le$5) \}
\\[2mm]
A $\cap$ $\mathbb{P}$ = A
\\[2mm]
A $\cap$ $\bar{A}$ = $\diameter$
\\[2mm]
A $\cap$ $\diameter$ = $\diameter$
\\[2mm]
A $\cup$ (A $\cap$ B) $\equiv$ A
\\[2mm]
A $\cap$ B
\begin{tikzpicture}
\filldraw[fill=gray] (-2,-2) rectangle (3,2);
\scope % A \cap B
\clip (0,0) circle (1);
\fill[white] (1,0) circle (1);
\endscope
% outline
\draw (0,0) circle (1)
(1,0) circle (1);
\end{tikzpicture}
\\[2mm]
$\overline{A \cap B}$ \tikz \fill[even odd rule] (0,0) circle (1) (1,0) circle (1);
\section{Equivalency}
Two expressions involving sets are equivalent if ew can represent them with the same venn diagram. This corresponds to using truthg tables to establish equivalence of logical expressions.
\subsection{Examples}
\subsubsection{Example 1}
Let $\mathbb{P}$ = \{ 1,2, ..., 10 \}
\newline
A = \{ 2,4,7,9 \}
\newline
B = \{ 1,4,6,7,10 \}
\newline
C = \{ 3,5,7,9 \}
\subsubsection{Example 2}
Find B $\cap$ $\bar{C}$
\newline
$\bar{C}$=\{1,2,4,6,8,10\}
\newline
So B $\cap$ $\bar{C}$ = \{1,4,6,10\}
\subsubsection{Example 3}
Find (A $\cap$ $\bar{B}$) $\cup$ C = (A $\cap$ \{ 2,3,5,9,9 \}) $\cup$ C = \{ 2,9 \} $\cup$ C = \{ 2,3,5,7,9 \}
\subsubsection{Example 4}
Find $\overline{B \cup C}$ $\cap$ C = \{ $\overline{1,3,4,5,6,7,8,10}$\} $\cap$ C = \{ 2,8 \} $\cap$ C = $\diameter$
\subsubsection{Example 5}
Show that (A $\cup$ $\bar{B}$) $\cap$ (A $\cup$ B) = A
\newline
(A $\cup$ $\bar{B}$) $\cap$ (A $\cup$B) = ((A $\cup$ $\bar{B}$) $\cap$ A) $\cup$ ((A $\cup$ $\bar{B}$) $\cap$ B) \underline{\bf distr}
\newline
=((A $\cup$ $\bar{B}$) $\cap$ A) $\cup$ ((A $\cap$ B) $\cup$ ($\bar{B}$ $\cap$ B)) \underline{\bf distr}
\newline
= A $\cup$ ((A$\cap$B)$\cup$($\bar{B}$ $\cap$ B)) \underline{\bf absorption}
\newline
= A $\cup$ (A $\cap$ B) \underline{\bf identity}
\newline
= {\bf A}
\end{document}
Is there a better way of writing it?
FYI: I type up these notes during the lectures.
Best Answer
Here's my nomination for how to rewrite Lecture 16. The new version makes use of constructs of the
amsmath
,amssymb
, andntheorem
packages. I've also simplified most of the math expressions, at least relative to the initial forms. Some\mathstrut
s were inserted to raise the overline parts a bit. The command\varnothing
now denotes the empty set. Finally, I've put the two diagrams intocenter
environments so that they're each centered on the page (along with their respective captions).