% This is a demonstration file distributed with the
% Lecturer package (see lecturer-doc.pdf).
%
% You can recompile the file with a basic TeX implementation,
% using pdfTeX or LuaTeX with the plain format.
%
% The reusable part ends somewhere around line 250.
%
% Author: Paul Isambert.
% Date: July 2010.
\input lecturer
% Uncomment these to show the grid on which the presentation is built.
%\showgrid{1mm}
%\showgrid[-.5cm,-.5cm]{1cm}[black]
%\showgrid{1cm}[red][.4pt]
\font\maintitlefont = cmdunh10 at 23pt
\font\slidetitlefont = cmss12 at 20pt
\font\subtitlefont = cmss8
\font\titlefont = cmss10 at 20pt
\newcolor{darkred}{rgb}{.6 0 0}
\setparameter job:
mode = presentation
author = "Pete Agoras"
title = "The square root of 2 ain't rational"
date = "Some centuries B.C\rlap." % \rlap creates protrusion
fullscreen = true
autofullscreen = true
font = \tenrm
%
% Slides for the main matter.
%
\setparameter slide:
everyslide = \everyslide
top = 3cm
topskip = 1cm
left = 3.1cm
right = 5cm
baselineskip = .5cm
areas* = "title by author date"
bookmarkstyle = italic
font = \tenrm
vpos = top
\def\everyslide{%
\position{top}\slidetitle
\position{left}{\Author\par\Date}
\position{menu}{\quitvmode\showbookmarks\Bookmarks}
\position{menu}[0pt,0pt]{\the\numexpr\slidenumber-1\relax}
}
%
% Left and top areas.
%
\setarea{left}
hshift = 0pt
width = 3cm
right = .1cm
hpos = rf
bottom = .5cm
vpos = bottom
foreground = darkred
font = \subtitlefont
\setarea{top}
left = 3.1cm
vshift = 0pt
height = 2cm
vpos = bottom
font = \titlefont
\setarea{vline hline}
frame = "width = .15pt, color = black"
\setarea{vline}
hshift = 3cm
width = 0pt
\setarea{hline}
vshift = 2cm
height = 0pt
%
% The circle with the slide number.
%
\setarea{menu}
width height = 1cm
vshift* hshift* topskip = .5cm
vpos = bottom
hpos = rr
\newsymbol\Bookmarks[.35cm,padding=.1pt]
{pen .05,
move 1 -.6,
circle lu 1,
circle ur 1,
circle rd 1,
circle dl 1,
}
%
% The maths on the right.
%
\setarea{math}
width = 2cm
hshift* = 1.1cm
vshift* = 5.35cm
vshift = 5.25cm
topskip = .8cm
foreground = white
background = black
frame = width=.1cm,corner=round % No space => no need for quotes or braces. No readability either.
hpos = rr
%
% The title slide.
%
\setslide{titleslide}
areas = "title by author date"
bookmark = false
hpos = rr
everyslide = {}
\setarea{title}
hshift* = 1cm
hpos = rr
vshift = 2cm
height = 1.3cm
topskip = 1cm
background = darkred
foreground = white
frame = "width = .5em, corner = round, color = black"
font = \maintitlefont
\setarea{by}
hshift* = 5.8cm
hpos = rr
vshift = 4.5cm
height = .8cm
topskip = 10pt
top = 3pt
background = black
foreground = white
frame = "width = .5em, corner = round"
font = \it
\setarea{author}
hshift* = 5cm
hpos = rr
vshift = 6.3cm
height = 25pt
topskip = 15pt
top = 4pt
background = black
foreground = white
frame = "width = .5em, corner = round, color = darkred"
font = \slidetitlefont
\setarea{date}
hshift* = 5.5cm
vshift* = 1.7cm
height = .8cm
bottom = .3cm
topskip = 0pt
vpos = bottom
background = white
foreground = black
frame = "width = .5em, corner = bevel, color = darkred"
hpos = rr
%
% For the appendices.
% We remove the math area instead
% of redefining \everyslide.
%
\setslide{appslide}
top = 2.5cm
areas*= "math title by author date"
\abovedisplayshortskip = 0cm
\belowdisplayshortskip = 0cm
\abovedisplayskip = 0pt
\belowdisplayskip = 0pt
%
% Steps.
%
\setparameter step:
vskip = .5cm
% With this macro, the item symbol is always on the screen
% (since the step is "visible").
\def\Step{%
\step[visible=true]\quitvmode\llap{\stepsym\kern.2cm}%
\step}
%
% The maths.
%
\setstep\emptystep
everyvstep everyhstep = {}
vskip hskip = 0pt
\def\domath#1 #2 #3{%
\emptystep[on=#1,off=#2]
\domathstep{#3}%
}
\def\Domath#1 #2{%
\emptystep[off=#1,visible=true]
\domathstep{#2}%
}
\def\domathstep#1{%
\position{math}[0pt,0pt]{$#1$}%
}
%
% Item symbols.
%
\newsymbol\stepsym[2mm,padding=0pt]
{color darkred,
+ 1 .5, + -1 .5, fill,}
\newsymbol\backsym[2mm,padding=0pt]
{color darkred, move 1 0, + -1 .5, + 1 .5, fill,}
%
% Navigation to the appendices.
%
\def\Back#1{%
\presentationonly{\usecolor{darkred}{\gotoB{#1}{\subtitlefont\backsym\kern.3em Back}}}%
}
\def\To#1{%
\usecolor{darkred}{\gotoA{#1}{\subtitlefont\stepsym\kern.3em See why}}%
}
%
% And some structure.
%
\def\section#1{%
\def\sectiontitle{#1}%
\createbookmark[nosubmenutext,open]{.5}{#1}%
}
% \endinput
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% %
% UNCOMMENT THE PREVIOUS LINE TO USE THIS FILE AS A TEMPLATE, %
% OR REMOVE EVERYTHING BELOW. %
% %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\titleslide
\position{title}\Title
\position{by}{A Casual Talk By}
\position{author}\Author
\position{date}\Date
\endtitleslide
\section{Demonstration}
\slide[A simple assumption,font=]
\step[on=A]\position{math}[0cm,0cm]{}
\domath A B {{a\over b}=\sqrt2}
\domath B C {({a\over b})^2=2}
\domath C D {{a^2\over b^2}=2}
\domath D {} {a^2=2b^2}
\Step Some say the square root of 2 isn't rational.
\step Suppose it were.
\Step[A] Then we could write it as this, where $a$ and $b$ are
integers without a common factor.
\To{app1}
\Step[B] But then we can also write this.
\step[C] And this.
\step[D] And finally this.
\step Which means that $a^2$ is even.
\endslide
\slide[Its consequences]
\step[on=A]\position{math}[0cm,0cm]{}
\Domath A {a^2=2b^2}
\domath A B {(2k)^2=2b^2}
\domath B C {4k^2=2b^2}
\domath C D {2k^2=b^2}
\Step[visible=true] So what?
\step So $a$ is even. Because only even numbers produce even squares.
\To{app2}
\Step[A] Being even means being expressible in the form $2k$, where $k$ is
any integer.
\step[B] And $(2k)^2$ square gives $4k^2$.
\Step[C] Let's simplify.
\step Thus $b^2$ is even.
\step And $b$ is too.
\endslide
\slide[The problem]
\step[on=A]\position{math}[0cm,0cm]{}
\domath A {} {{a\over b}\neq\sqrt2}
\Step[visible=true] And but so we said $a$ and $b$ have no common factor.
\Step If both are even they do have a common factor: 2.
\step Which is absurd.
\Step Thus, our basic assumption is false.
\step[A] There are no such $a$ and $b$.
\Step The square root of 2 is irrational.
\step Too bad.
\endslide
\section{Appendices}
\appslide[All fractions are reducible]
\Step[visible=true]
Suppose $c\over d$ is a rational number. If c and d have no common
factor, then $a=b$ and $b=d$. If they have a common factor,
divide both by their greatest common divisor. The result
is $a\over b$, with no common factor.
\kern1em\Back{app1}
\endappslide
\appslide[An even square has an even root]
\Step[visible=true]
An even number, by definition, is expressible
in the form $2k$, where $k$ is any integer.
On the other hand, an odd number is expressible by
%
$$2k+1$$
%
Thus the square of an odd number is
%
$$(2k+1)^2$$
i.e. $$4k^2+4k+1$$
i.e. $$2\times2(k^2+k)+1$$
%
which is of the form $2k+1$
with $2(k^2+k)$ as $k$.
Hence, an odd number produces an odd square,
and thus if a square is even its root is even too.
\kern1em\Back{app2}
\endappslide
\bye