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+% 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 150.
+%
+% Author: Paul Isambert.
+% Date: July 2010.
+
+\input lecturer
+
+%\showgrid{1cm}
+
+\setparameter job:
+ fullscreen = true
+ font = \mainfont
+
+\setparameter slide:
+ width height = 12cm
+ top = 0cm
+ bottom = 3cm
+ left = 5cm
+ right = 1.6cm
+ topskip = 0pt
+ background = black % For the lines between squares.
+ foreground = white
+ vpos = center
+ hpos = ff
+
+\setarea{area1 area2 matharea}
+ width = 3cm
+
+\setarea{area1 area2}
+ height = 4.3cm
+ background = white
+
+\setarea{area2}
+ vshift = 4.7cm
+
+\setarea{matharea footnotearea}
+ height = 2.6cm
+ vshift* = 0pt
+ topskip = .1cm
+ baselineskip = .333cm
+ vpos = center
+
+\setarea{matharea}
+ background = blue
+ foreground = white
+ hpos = rr
+
+% Below the slide's text.
+\setarea{mainarea}
+ hshift = 3.4cm
+ hshift* = 0pt
+ height = 9cm
+ background = red
+
+\setarea{footnotearea}
+ hshift = 3.4cm
+ hshift* = 1.8cm
+ left right = .3cm
+ background = white
+ font = \footnotefont
+ hpos = ff
+
+\setarea{area6 area7}
+ width = 1.4cm
+ hshift* = 0pt
+ height = 1.1cm
+
+\setarea{area6}
+ vshift = 9.4cm
+ background = white
+
+\setarea{area7}
+ vshift* = 0pt
+ background = yellow
+
+
+
+\setparameter step:
+ vskip = \baselineskip
+
+\setstep\emptystep
+ vskip = 0pt
+
+% Used for the maths.
+\def\mathstep#1 #2 #3{\emptystep[on=#1,off=#2]\position{matharea}[0pt,0pt]{$#3$}\ignorespaces}
+\def\Mathstep#1 #2{\emptystep[off=#1,visible=true]\position{matharea}[0pt,0pt]{$#2$}\ignorespaces}
+
+% The footnote. Since the same name is given
+% to each footnote step, there must be at most
+% one per slide. Otherwise, a name should be given
+% manually (or automatically with a number in the name,
+% and a counter to increment it on every footnote call).
+\def\footnote#1{%
+ \showorhide{toggle=footnote}{\super{\usecolor{blue}{*}}}
+ \emptystep[footnote,on=]
+ \position{footnotearea}{\quitvmode\llap{* }#1}%
+ }
+
+% On the last equation.
+\newsymbol\cross[1.7em]{%
+ pen 0.1,
+ move 0 -.5, .5 .9, stroke,
+ move 0 .9, .5 -.5, stroke,
+ }
+% Font for text.
+\font\mainfont=cmss10
+\font\supmainfont=cmss10 at 7.5pt
+% Simple superscript (I'm not very good with
+% math font family... I'm no mathematician after all).
+\def\super#1{\raise.3em\hbox{\supmainfont#1}}
+\font\footnotefont=cmss8 at 7pt
+\mainfont
+\frenchspacing
+
+% Font for maths.
+\font\mathfont=cmss12 at 16pt
+\font\scriptmathfont=cmss12 at 10pt
+\font\scriptscriptmathfont=cmss12 at 8pt
+\textfont0=\mathfont
+\scriptfont0=\scriptmathfont
+\scriptscriptfont0=\scriptscriptmathfont
+\font\mathfont=cmss12 at 16pt
+\font\scriptmathfont=cmss12 at 12pt
+\font\scriptscriptmathfont=cmss12 at 8pt
+\textfont1=\mathfont
+\scriptfont1=\scriptmathfont
+\scriptscriptfont1=\scriptscriptmathfont
+\font\Tensy=cmsy10 at 18pt
+\font\scriptTensy=cmsy10 at 10pt
+\font\scriptscriptTensy=cmsy10 at 8pt
+\textfont2=\Tensy
+\scriptfont2=\scriptTensy
+\scriptscriptfont2=\scriptscriptTensy
+\font\Tenex=cmex10 at 18pt
+\font\scriptTenex=cmex10 at 12pt
+\font\scriptscriptTenex=cmex10 at 8pt
+\textfont3=\Tenex
+\scriptfont3=\scriptTenex
+\scriptscriptfont3=\scriptscriptTenex
+
+
+
+% \endinput
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+% %
+% UNCOMMENT THE PREVIOUS LINE TO USE THIS FILE AS A TEMPLATE, %
+% OR REMOVE EVERYTHING BELOW. %
+% %
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+
+
+
+
+\slide[A simple assumption]
+
+\mathstep A B {{a\over b}=\sqrt2}
+\mathstep B C {({a\over b})^2=2}
+\mathstep C D {{a^2\over b^2}=2}
+\mathstep 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.%
+\footnote{%
+ Suppose c/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/b, with no common factor.}
+
+\step[B] But then we can also write this.
+\step[C] And this.
+\step[D] And finally this.
+\step Which means that a\super2 is even.
+
+\endslide
+
+
+
+\slide[Its consequences]
+
+\Mathstep A {a^2=2b^2}
+\mathstep A B {(2k)^2=2b^2}
+\mathstep B C {4k^2=2b^2}
+\mathstep C D {2k^2=b^2}
+
+\step[visible=true] So what?
+\step So a is even. Because only even numbers produce even squares.%
+\footnote{%
+ 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)\super2, i.e. 4k\super2+4k+1, i.e.
+ 2x2(k\super2+k)+1, which is of the form 2k+1,
+ with 2(k\super2+k) as k.
+ Hence, an odd number produces an odd square,
+ and thus if a square is even its root is even too.}
+
+\step[A] Being even means being expressible in the form 2k, where k is
+any integer.
+\step[B] And (2k)\super2 square gives 4k\super2.
+
+\step[C] Let's simplify.
+\step Thus b\super2 is even.
+\step And b is too.
+
+\endslide
+
+
+
+\slide[The problem]
+
+\Mathstep A {{a\over b}=\sqrt2}
+\mathstep A {} {\rlap\cross{a\over b}=\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.
+
+\endslide
+
+\bye \ No newline at end of file