\documentclass[12pt  ]{article}

\usepackage{amssymb}
\usepackage{graphicx}
\usepackage{amsmath}
 \usepackage{relsize}
\usepackage{verbatim}

\usepackage{hyperref} \hypersetup{breaklinks=true,  colorlinks=true, linkcolor= black, 
hyperfootnotes= true, urlcolor= Sepia} 
\usepackage{breakurl}
%%%%Now define a  new style, rm and tiny
\makeatletter
\def\url@leostyle{%
  \@ifundefined{selectfont}{\def\UrlFont{\sf}}{\def\UrlFont{\footnotesize\rm}}}
\makeatother
%%%%% Now actually use the newly defined style.
\urlstyle{leo}

\hypersetup{breaklinks=true,
pagecolor=white,
colorlinks=true,
linkcolor= blue,
hyperfootnotes= true,
urlcolor=blue
}
\urlstyle{rm}

\usepackage{setspace} \newenvironment{bigquote}
              { \begin{quotation} \begin{small} \begin{singlespace}  }
               {  \end{singlespace}  \end{small}  \end{quotation}   }

\usepackage{wrapfig}
 \usepackage{boxedminipage} \setlength{\fboxsep}{12pt} 


\newcounter{figurecounter}
\setcounter{figurecounter}{0}
\newcounter{tablecounter}
\setcounter{tablecounter}{0}
\newcounter{sidebarcounter}
\setcounter{sidebarcounter}{0}
 \newcounter{chaptercounter}
 \setcounter{chaptercounter}{0}
 
 
 \usepackage{pictexwd}
  \usepackage{pdftricks}
  \usepackage{comment}
 

\usepackage{setspace}
\usepackage{amsfonts}
\usepackage{color}
\usepackage{palatino}
%\usepackage {urwchancal}
\usepackage {pzccal}

%\DeclareMathAlphabet{\mathcalligra}{T1}{calligra}{m}{n}
\usepackage{calligra}
\usepackage[T1]{fontenc}
% \usepackage{calligra-type1}



%\setcounter{page}{1}
%\setlength{\oddsidemargin}{0in}
 \setlength{\textwidth}{6in}
%\setlength{\textheight}{9.25in}
%\setlength{\topmargin}{0in}
%\setlength{\headheight}{0in}
%\setlength{\headsep}{0in}


\newcommand{\margincomment}[1]
  {\mbox{}\marginpar{\tiny\hspace{0pt}#1}}

\newcommand{\comments}[1]{}
\renewcommand{\baselinestretch}{1.2}
 \parindent 24pt
 \parskip 10pt
 
 

%\newcommand{\argmax}{\mathop{\rm argmax}}

\begin{document}

% \newcommand{{=}less} 
%  \mathsmaller {  \;\widetilde{\prec}\;  } 
% A ${=}less B $


\titlepage

\vspace*{12pt}

    
 
\begin{center}



{ \sc \large  Something Hilbert Got Wrong and Euclid Got Right: The Method of  Superposition  and the Side-Angle-Side Axiom in Propositions 1 to 4 of Book I of the Elements }
 
February 14, 2012

\bigskip



 
   Eric B.  Rasmusen  

{\it Abstract}


\end{center}

\begin{small}

 Add a new postulate saying we can draw a line of length pi times the reference line, adn we get interestnig new geometrya dn can square the circle. 
 
 
\noindent
Rasmusen: Dan R. and Catherine M.
Dalton
Professor, Department of Business Economics and Public Policy, Kelley
School
of Business, Indiana University. BU 438, 1309 E. 10th Street,
Bloomington,
Indiana, 47405-1701. (812) 855-9219. Fax: 812-855-3354.
\url{mailto:erasmuse@indiana.edu}{ erasmuse@indiana.edu}, \url{http://www.rasmusen.org}.  

{\small
  \noindent
 This paper:
\url{http://www.rasmusen.org/papers/euclid-plus.pdf}.   }

  
 \noindent
  Keywords: xxx  


   I  would like to thank xxx    




 \end{small}



\newpage
 
 

 
\noindent
{\sc  Notes}

 I could bring in an arc definition based on pi, but I would have to define pi rather than derive it. I should probably do that anyway. $\pi  \equiv \frac{circumference}{2*radius}$.Can I prove that this is the same for all circles? 

I can have another postulate: 

\noindent
Postulate 5C: To draw a segment equal to a circle's circumference divided by twice its radius. 

 Then I can square the circle and so forth. Probably I can trisect the angle too. 

I need a definition of circumference. 


Add pi to get Euclid+.  Add the cube root of two to get Euclid++ and double the cube. 


Proposition I-45B (quadrature) To construct a rectangle  equal to a given rectilinear figure in a given rectilinear angle  and with one side equal to the reference segment. 

 Proposition I-45C (area) To construct a segment  equal to a given rectilinear figure. (maybe make this a corollary)

Proof: 
   
   Use I-45B. The side not equal to the reference segment will equal the given rectilinear figure.   




With I-45C and Postulate 5C (the pi postulate) I can find the area of a circle. It is easy then to square the circle by constructing a square of that area. 

 The Pi Postulate is reasonable. It is like pos 1, 3. We cannot draw true segments or circles, so we postualte them. We cannot draw a true pi-segement, so we postualte it. It is no stronger a postulate.  If we draw a segement at random, almost surely t will be irrational,a nd almost surely transcndental. We can ddeduce hwo to draw a square root of two segement in Euclid. 

We can certainly square the circle now. I bet we can trisect the angle too, because pi is related to trigonometry. 


 

  
\end{document}