\documentclass[10pt,envcountsect]{beamer}

\mode<presentation> {
  \usetheme{Warsaw}
}

\usepackage[english]{babel}
\usepackage[latin1]{inputenc}
\usepackage{times}
\usepackage[T1]{fontenc}
\usepackage{pstricks}
\usepackage{multicol}
\usepackage{amssymb}

\newcommand{\N}{\textbf{N}}
\newcommand{\Z}{\textbf{Z}}
\newcommand{\D}{\textbf{D}}
\newcommand{\Q}{\textbf{Q}}
\newcommand{\R}{\textbf{R}}
\newcommand{\dis}{\displaystyle}

\begin{document}

\title{Episode 06 -- Set-builder notation}
\subtitle{European section -- Season 2}
\date{}

\begin{frame}
  \titlepage
\end{frame}


\begin{frame}
\begin{block}{$\{2k+1 : k\in\Z\}$}\pause
$-7$, $-3$, $5$, $17$
\end{block}\pause

\begin{block}{$\{n : n\in\Z\wedge \dis\frac{12}{n}\in\Z\}$}\pause
$3$, $6$, $-12$, $1$
\end{block}\pause

\begin{block}{$\{n^2 : n\in\N\cap[0,5]\}$}\pause
$0$, $1$, $9$, $16$
\end{block}\pause

\begin{block}{$\{3k : k\in\N\}$}\pause
$6$, $21$, $336$, $4272$
\end{block}
\end{frame}

\begin{frame}
\begin{block}{$\{x\in\R : 3x\in\Z\}$}\pause
$\frac73$, $-17$, $-\frac{52}3$, $0$
\end{block}\pause

\begin{block}{$\{a+bi : a\in\R\wedge b\in\R\}$}\pause
$3+5i$, $\sqrt2-i$, $17$, $-1+\pi i$
\end{block}\pause

\begin{block}{$\{7k+1 : k\in\Z\}$}\pause
$-20$, $1$, $50$, $778$
\end{block}\pause

\begin{block}{$\{n\in\N : \exists m\in\Z, m^2=n\}$}\pause
$0$, $9$, $81$, $196$
\end{block}
\end{frame}

\begin{frame}

\begin{block}{$\{x\in\R : x^2=x\}$}\pause
$0$, $1$
\end{block}\pause

\begin{block}{$\{x\in\R : \exists p,q\in\Z, q\neq 0 \wedge xq=p\}$}\pause
$\frac{12}{7}$, $-\frac{2}{3}$, $42$, $\frac{-28}{2}$
\end{block}\pause

\begin{block}{$\{r\in\R : \exists k\in\Z, r^2=k\}$}\pause
$5$, $\sqrt2$, $-13$, $\sqrt{13}$
\end{block}\pause

\begin{block}{$\{n\in\Z : |n|\leq 5\}$}\pause
$-4$, $-2$, $0$, $3$
\end{block}
\end{frame}

\begin{frame}

\begin{block}{$\{x\in\R : |x|\leq 5\}$}\pause
$\sqrt2$, $-\pi$, $2.256$, $-4.99$
\end{block}\pause

\begin{block}{$\{x\in\R : \forall y\in\N, x\leq y\}$}\pause
$-5$, $-2\pi$, $-\frac47$, $-0.0001$
\end{block}\pause

\begin{block}{$\{(x,y) : x\in\N \wedge y\in\N \wedge\ x+y=7\}$}\pause
$(2,5)$ ; $(1,6)$ ; $(4,3)$ ; $(3,4)$
\end{block}\pause

\begin{block}{$\{x\in\R : x\leq -3 \vee\ x\geq 5\}$}\pause
$-5$, $-\sqrt{19}$, $17.25$, $5$
\end{block}
\end{frame}

\begin{frame}

\begin{block}{$\{n\in\Z : \forall m\in\Z, n>m\}$}\pause
$\varnothing$
\end{block}\pause

\begin{block}{$\{n\in\N : \nexists m\in\N-\{1,n\}, \dis\frac{n}{m}\in\N  \}$}\pause
$17$, $5$, $43$, $199$
\end{block}\pause

\begin{block}{$\{(x,y) : x\in\N \wedge y\in\N \wedge\ xy\leq 6\}$}\pause
$(2,2)$ ; $(1,5)$ ; $(3,1)$ ; $(1,1)$
\end{block}\pause

\begin{block}{$\{(x,y) : x\in\N \wedge y\in\N \wedge\ y=2x\}$}\pause
$(3,6)$ ; $(12,24)$ ; $(5,10)$ ; $(131,262)$
\end{block}

\end{frame}

\begin{frame}{Set-builder notation}

\begin{description}
\item[$\{\ldots\}$ :] A set.
\item[$:$ or $|$ :]  ``such that''.
\end{description}

\end{frame}

\begin{frame}{Sets of numbers}

\begin{description}
\item[$\N$ :] The whole numbers (natural numbers and zero).
\item[$\Z$ :] The integers.
\item[$\D$ :] The decimal numbers.
\item[$\Q$ :] The rational numbers
\item[$\R$ :] The real numbers.
\item[$\R^\star$ :] The non-zero real numbers.
\item[$\R^+$ :] The positive real numbers.
\item[$\R^-$ :] The negative real numbers.
\end{description}

\end{frame}

\begin{frame}{Set, subsets and elements}

\begin{description}
\item[$\cap$ :] The intersection of two sets.
\item[$\cup$ :] The union of two sets.
\item[$\in$ :] ``is an element of the set''
\item[$\subset$ :] ``is a subset of the set''
\end{description}

\end{frame}

\begin{frame}{Logical operators}

\begin{description}
\item[$\neg$ :] The negation of a proposition, ``not''.
\item[$\wedge$ :] The conjunction of two propositions, ``and''.
\item[$\vee$ :] The disjunction of two propositions, ``or''.
\item[$\forall$ :] ``For all''
\item[$\exists$ :] ``There exists''
\item[$\nexists$ :] ``There doesn't exist''
\end{description}

\end{frame}


\begin{frame}{Russell's paradox}
Let $R$ be the set of all sets that don't contain themselves, that
is\pause $$R=\{X:X\mbox{ is a set and }X\not\in X\}.$$\pause\ 

Then $R$ either is or is not an element of itself.\pause\ 

\begin{itemize}
\item If $R$ is not an element of
itself,\pause\
then $R$ is a set that doesn't contain itself,\pause\ and so $R$ is an
element of $R$,\pause\ which is precisely the contrary of what we
supposed.\pause
\item Conversely, if $R$ is an element of itself,\pause\ then $R$ does not
contain itself,\pause\ so we also get to a contradiction.\pause
\end{itemize}

So $R$ cannot be a set, or our definition of a set is not
satisfactory.\pause
   
\end{frame}


\end{document}
