\documentclass[12pt]{article}
\usepackage[T1]{fontenc}
\usepackage{tikz,chessboard,comment}
\tikzset{
    pion/.style={circle,draw,ball color=gray!10},
    jeton/.style={circle,draw,ball color=gray!80}
}
\usepackage[utf8]{inputenc}
\setlength{\textwidth}{18cm}
\setlength{\textheight}{25cm}
\setlength{\hoffset}{-2.5cm}
\setlength{\voffset}{-3cm}
\begin{document}


\begin{center}
\textbf{\Huge{Abaque de Neper}}
\begin{tikzpicture}
\draw[black,thick] (0,0) grid (13,13); 
\foreach \a in {0,...,12}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (12-\a,-0.5) {\aa};
\node[rotate=45] at (13.5,1+\a) {\aa};
\draw (12-\a,-1) -- (13-\a,0);
\draw (13,\a) -- (14,1+\a);
\node at (0.5+\a,12.5-\a) {.};
}
\end{tikzpicture}
\end{center}

\section{Chapitre VII : Du mouvement des jetons dans l'abaque}

Neper fait référence à deux pièces du jeu d'échecs tel qu'il le connaissait. C'est surtout aux mouvements de ces pièces qu'il s'intéresse~: les jetons de son abaque peuvent décrire de tels mouvements.


\subsection{L'éléphant porteur de tour}

L'éléphant de Neper (\textit{elephantis turtiferi}) bouge comme la tour moderne, mais d'une seule case\footnote{Conway appelait \textit{Duke} cette pièce.}
~:


\begin{center}
\chessboard[largeboard,maxfield=e5,hlabel=false,vlabel=false,setpieces={rc3},showmover=false,markstyle=straightmove,color=gray,arrow=to,markmoves={c3-c4,c3-c2,c3-b3,c3-d3}]    
\end{center}


\subsection{Le porteur de flèches}

Le porteur de flèches (\textit{sagittiferi}, variante du vizir du jeu de Chatrang) bouge comme un fou mais d'une seule case\footnote{Les pions du jeu \emph{Alquerkonane} bougent aussi de cette manière. Les jetons du jeu de dames aussi.}
~:

\begin{center}
\chessboard[largeboard,maxfield=e5,hlabel=false,vlabel=false,setpieces={bc3},showmover=false,markstyle=straightmove,color=gray,arrow=to,markmoves={c3-b4,c3-b2,c3-d4,c3-d2}]    
\end{center}



\section{Chapitre VIII : Des axiomes et conjectures sur les mouvements de jetons}

\subsection{Axiomes}

\begin{enumerate}
    \item Axiome 1~: \textbf{Si une case n'est pas à la marge inférieure ou de droite, sa valeur est double de celle de ses voisines de droite et de dessous.}
\begin{center}
\begin{tikzpicture}[scale=3]
\draw (0,0) grid (2,2);
\node (a) at (1.5,0.5) {$x$};
\node (b) at (1.5,1.5) {$2x$};
\node (c) at (0.5,0.5) {$2x$};
\draw[->,>=stealth,very thick] (a) -- node[draw,rounded corners=3pt,fill=white,midway]{$\times 2$} (b);
\draw[->,>=stealth,very thick] (a) -- node[draw,rounded corners=3pt,fill=white,midway]{$\times 2$} (c);
\end{tikzpicture}
\end{center}

Un jeton dans une case vaut donc autant que deux jetons dans la case de droite. On peut donc, à tout instant,

\begin{itemize}
    \item enlever un jeton d'une case et remplacer ce jeton par deux jetons dans la case voisine (de droite, ou au-dessous)~:
    
\begin{center}
\begin{tikzpicture}
\draw (0,0) grid (2,1);
\draw (4,0) grid (6,1);
\draw[|->,>=stealth,very thick] (2.5,0.5) -- (3.5,0.5);
\node[jeton] at (0.5,0.5) {};
\node[jeton] at (5.3,0.3) {};
\node[jeton] at (5.7,0.7) {};
\end{tikzpicture}
\end{center}    

\item ou à l'inverse, si une case contient 2 jetons, on peut les remplacer par un jeton unique dans la case voisine (de gauche, ou au-dessus)~:

    
\begin{center}
\begin{tikzpicture}
\draw (0,0) grid (2,1);
\draw (4,0) grid (6,1);
\draw[|->,>=stealth,very thick] (2.5,0.5) -- (3.5,0.5);
\node[jeton] at (4.5,0.5) {};
\node[jeton] at (1.3,0.3) {};
\node[jeton] at (1.7,0.7) {};
\end{tikzpicture}
\end{center}    

On peut d'ailleurs réinterpréter l'axiome 1 de cette manière~:

\begin{center}
\begin{tikzpicture}[scale=3]
\draw (0,0) grid (2,2);
\node (a) at (0.5,1.5) {$x$};
\node (b) at (1.5,1.5) {$\frac{x}{2}$};
\node (c) at (0.5,0.5) {$\frac{x}{2}$};
\draw[->,>=stealth,very thick] (a) -- node[draw,rounded corners=3pt,fill=white,midway]{$/ 2$} (b);
\draw[->,>=stealth,very thick] (a) -- node[draw,rounded corners=3pt,fill=white,midway]{$/ 2$} (c);
\end{tikzpicture}
\end{center}
\end{itemize}
L'axiome 1 permet de trouver la représentation binaire d'un nombre. Par exemple si on veut savoir comment s'écrit 5 en binaire, on peut mettre 5 jetons dans la case des unités~:

\begin{center}
\begin{tikzpicture}
\draw[black,thick] (0,0) grid (5,1); 
\foreach \a in {0,...,4}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (4-\a,-0.5) {\aa};
\draw (4-\a,-1) -- (5-\a,0);
\node[jeton] at (4.5,0.5) {};
\node[jeton] at (4.25,0.25) {};
\node[jeton] at (4.75,0.25) {};
\node[jeton] at (4.25,0.75) {};
\node[jeton] at (4.75,0.75) {};
}
\end{tikzpicture}
\end{center}

L'axiome 1 permet de retirer 2 de ces jetons en ajoutant un jeton dans la case des deuxaines~:

\begin{center}
\begin{tikzpicture}
\draw[black,thick] (0,0) grid (5,1); 
\foreach \a in {0,...,4}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (4-\a,-0.5) {\aa};
\draw (4-\a,-1) -- (5-\a,0);
\node[jeton] at (4.5,0.5) {};
\node[jeton] at (4.25,0.25) {};
\node[jeton] at (4.75,0.25) {};
\node[jeton] at (3.3,0.3) {};
}
\end{tikzpicture}
\end{center}

Et comme il reste plus que 2 jetons dans la case des unités, on peut lui appliquer à nouveau l'axiome 1~:
\begin{center}
\begin{tikzpicture}
\draw[black,thick] (0,0) grid (5,1); 
\foreach \a in {0,...,4}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (4-\a,-0.5) {\aa};
\draw (4-\a,-1) -- (5-\a,0);
\node[jeton] at (4.5,0.5) {};
\node[jeton] at (3.3,0.3) {};
\node[jeton] at (3.7,0.7) {};
}
\end{tikzpicture}
\end{center}

Comme il reste moins de 2 jetons dans la case des unités, on ne peut plus lui appliquer l'axiome 1. Mais on peut appliquer l'axiome 1 à la case des deuxaines puisqu'elle contient 2 jetons~:
\begin{center}
\begin{tikzpicture}
\draw[black,thick] (0,0) grid (5,1); 
\foreach \a in {0,...,4}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (4-\a,-0.5) {\aa};
\draw (4-\a,-1) -- (5-\a,0);
\node[jeton] at (4.5,0.5) {};
\node[jeton] at (2.5,0.5) {};
}
\end{tikzpicture}
\end{center}

Comme aucune case ne contient plus d'un jeton, on ne peut plus appliquer l'axiome 1~: on a maintenant l'écriture binaire de 5 qui est \texttt{00101} (ou 4+1). 

\item Axiome 2~: \textbf{Les valeurs des cases marginales (bord droit et bord inférieur) sont les puissances de 2.}

On remarque que cet axiome pouvait se déduire d'un axiome plus simple~: la valeur de la case en bas à droite, est 1. En effet l'axiome 1 permet de remplir les bords à partir de cette valeur.
\end{enumerate}

\subsection{Corollaires}

Des deux axiomes 1 et 2, Neper déduit des résultats qu'il appelle \emph{corollaires}. En fait il s'agit de théorèmes.

\begin{enumerate}
    \item Corollaire 1~: \textbf{Un mouvement en diagonale entre sud-ouest et nord-est ne change pas la valeur d'une case.}
    
\begin{center}
\begin{tikzpicture}[scale=3]
\draw (0,0) grid (2,2);
\node (a) at (0.5,1.5) {$2x$};
\node (b) at (1.5,1.5) {$x$};
\node (c) at (0.5,0.5) {$x$};
\draw[->,>=stealth,very thick] (a) -- node[draw,rounded corners=3pt,fill=white,midway]{$/ 2$} (b);
\draw[->,>=stealth,very thick] (c) -- node[draw,rounded corners=3pt,fill=white,midway]{$\times 2$} (a);
\draw[->,>=stealth,ultra thick] (c) -- node[draw,rounded corners=5pt,fill=white,midway]{$=$} (b);
\end{tikzpicture}
\end{center}
\item Corollaire 2~: 
\begin{itemize}
    \item \textbf{Un mouvement vers le nord-ouest quadruple la valeur d'une case}~:
\begin{center}
\begin{tikzpicture}[scale=3]
\draw (0,0) grid (2,2);
\node (a) at (1.5,0.5) {$x$};
\node (b) at (1.5,1.5) {$2x$};
\node (c) at (0.5,1.5) {$4x$};
\draw[->,>=stealth,very thick] (a) -- node[draw,rounded corners=3pt,fill=white,midway]{$\times 2$} (b);
\draw[->,>=stealth,very thick] (b) -- node[draw,rounded corners=3pt,fill=white,midway]{$\times 2$} (c);
\draw[->,>=stealth,ultra thick] (a) -- node[draw,rounded corners=5pt,fill=white,midway]{$\times 4$} (c);
\end{tikzpicture}
\end{center}
\item \textbf{En avançant vers le sud-est on divise la valeur de la case par 4}~:
\begin{center}
\begin{tikzpicture}[scale=3]
\draw (0,0) grid (2,2);
\node (a) at (0.5,1.5) {$x$};
\node (b) at (1.5,1.5) {$\frac{x}{2}$};
\node (c) at (1.5,0.5) {$\frac{x}{4}$};
\draw[->,>=stealth,very thick] (a) -- node[draw,rounded corners=3pt,fill=white,midway]{$/ 2$} (b);
\draw[->,>=stealth,very thick] (b) -- node[draw,rounded corners=3pt,fill=white,midway]{$/ 2$} (c);
\draw[->,>=stealth,ultra thick] (a) -- node[draw,rounded corners=5pt,fill=white,midway]{$/ 4$} (c);
\end{tikzpicture}
\end{center}
\end{itemize}
\item Corollaire 3~: \textbf{Les valeurs sur la diagonale principale (celle qui est repérée par des points) sont les puissances de 4.}

\begin{center}
\begin{tikzpicture}[scale=1.6]
\draw[black,thick] (0,0) grid (5,5); 
\foreach \a in {0,...,4}{
\pgfmathtruncatemacro\aa{2^\a}
%\node[rotate=45] at (4-\a,-0.5) {\aa};
\node[rotate=45] at (5.5,1+\a) {\aa};
%\draw (4-\a,-1) -- (5-\a,0);
\draw (5,\a) -- (6,1+\a);
%\node at (0.5+\a,4.5-\a) {.};
};
\node (a) at (4.5,0.5) {1};
\node (b) at (3.5,1.5) {4};
\node (c) at (2.5,2.5) {16};
\node (d) at (1.5,3.5) {64};
\node (e) at (0.5,4.5) {256};
\draw[->,>=stealth,thick] (a) -- node[draw,rounded corners=3pt,fill=white,midway]{$\times 4$} (b);
\draw[->,>=stealth,thick] (b) -- node[draw,rounded corners=3pt,fill=white,midway]{$\times 4$} (c);
\draw[->,>=stealth,thick] (c) -- node[draw,rounded corners=3pt,fill=white,midway]{$\times 4$} (d);
\draw[->,>=stealth,thick] (d) -- node[draw,rounded corners=3pt,fill=white,midway]{$\times 4$} (e);
\end{tikzpicture}
\end{center}
\item Corollaire 4~: \textbf{Les cases au-dessus de la diagonale ont pour valeurs les doubles des puissances de 4.}
\begin{center}
\begin{tikzpicture}[scale=1.6]
\draw[black,thick] (0,0) grid (6,6); 
\foreach \a in {0,...,5}{
\pgfmathtruncatemacro\aa{2^\a}
%\node[rotate=45] at (5-\a,-0.5) {\aa};
\node[rotate=45] at (6.5,1+\a) {\aa};
%\draw (5-\a,-1) -- (6-\a,0);
\draw (6,\a) -- (7,1+\a);
\node at (0.5+\a,5.5-\a) {.};
};
\node (a) at (5.5,1.5) {2};
\node (b) at (4.5,2.5) {8};
\node (c) at (3.5,3.5) {32};
\node (d) at (2.5,4.5) {128};
\node (e) at (1.5,5.5) {512};
\draw[->,>=stealth,thick] (a) -- node[draw,rounded corners=3pt,fill=white,midway]{$\times 4$} (b);
\draw[->,>=stealth,thick] (b) -- node[draw,rounded corners=3pt,fill=white,midway]{$\times 4$} (c);
\draw[->,>=stealth,thick] (c) -- node[draw,rounded corners=3pt,fill=white,midway]{$\times 4$} (d);
\draw[->,>=stealth,thick] (d) -- node[draw,rounded corners=3pt,fill=white,midway]{$\times 4$} (e);
\end{tikzpicture}
\end{center}
\item Corollaire 5~: \textbf{Les cases au-dessous de la diagonale ont aussi pour valeurs les doubles des puissances de 4.}
\begin{center}
\begin{tikzpicture}[scale=1.6]
\draw[black,thick] (0,0) grid (6,6); 
\foreach \a in {0,...,5}{
\pgfmathtruncatemacro\aa{2^\a}
%\node[rotate=45] at (5-\a,-0.5) {\aa};
\node[rotate=45] at (6.5,1+\a) {\aa};
%\draw (5-\a,-1) -- (6-\a,0);
\draw (6,\a) -- (7,1+\a);
\node at (0.5+\a,5.5-\a) {.};
};
\node (a) at (4.5,0.5) {2};
\node (b) at (3.5,1.5) {8};
\node (c) at (2.5,2.5) {32};
\node (d) at (1.5,3.5) {128};
\node (e) at (0.5,4.5) {512};
\draw[->,>=stealth,thick] (a) -- node[draw,rounded corners=3pt,fill=white,midway]{$\times 4$} (b);
\draw[->,>=stealth,thick] (b) -- node[draw,rounded corners=3pt,fill=white,midway]{$\times 4$} (c);
\draw[->,>=stealth,thick] (c) -- node[draw,rounded corners=3pt,fill=white,midway]{$\times 4$} (d);
\draw[->,>=stealth,thick] (d) -- node[draw,rounded corners=3pt,fill=white,midway]{$\times 4$} (e);
\end{tikzpicture}
\end{center}
\item Corollaire 6~: \textbf{De façon analogue, les valeurs sur les diagonales parallèles à la diagonale principale, sont des produits de puissances de 2 par les puissances de 4.}
\item Corollaire 7~: \textbf{La valeur de chaque case est le produit des valeurs des cases situées en marge.}
\begin{center}
\begin{tikzpicture}
\draw[black,thick] (0,0) grid (5,5); 
\foreach \a in {0,...,4}{
\pgfmathtruncatemacro\aa{2^\a}
\pgfmathtruncatemacro\ab{2^(\a+1)}
\pgfmathtruncatemacro\ac{2^(\a+2)}
\pgfmathtruncatemacro\ad{2^(\a+3)}
\pgfmathtruncatemacro\ae{2^(\a+4)}
\node[rotate=45] at (4-\a,-0.5) {\aa};
\node[rotate=45] at (5.5,1+\a) {\aa};
\draw (4-\a,-1) -- (5-\a,0);
\draw (5,\a) -- (6,1+\a);
\node at (4.5,0.5+\a) {\aa};
\node at (3.5,0.5+\a) {\ab};
\node at (2.5,0.5+\a) {\ac};
\node at (1.5,0.5+\a) {\ad};
\node at (0.5,0.5+\a) {\ae};
}
\end{tikzpicture}
\end{center}

\end{enumerate}





\section{Chapitre IX : De la multiplication}

Le principe de la multiplication découle de la distributivité et du corollaire 7. Par exemple si on veut multiplier 5 par 9, on les écrit en binaire~:
\begin{itemize}
    \item 5 = 4+1 (\texttt{00101} en binaire~;~ dans la marge du bas)
    \item 9 = 8+1 (\texttt{01001} en binaire~;~dans la marge de droite)
\end{itemize}
Alors $5\times 9 = (4+1) \times (8+1) = 4\times 8 + 1\times 8 + 4\times 1 + 1\times 1$ et chacun des 4 produits peut être représenté par un jeton ainsi disposé\footnote{En hommage à René Descartes, on appellera \textit{produit cartésien} la pose des jetons pour une multiplication~: mettre sur l'abaque, un jeton à chaque intersection d'une ligne et d'une colonne ayant un jeton dans la marge.}~: 

\begin{center}
\begin{tikzpicture}
\draw[black,thick] (0,0) grid (6,6); 
\foreach \a in {0,...,5}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (5-\a,-0.5) {\aa};
\node[rotate=45] at (6.5,1+\a) {\aa};
\draw (5-\a,-1) -- (6-\a,0);
\draw (6,\a) -- (7,1+\a);
\node at (0.5+\a,5.5-\a) {.};
\node[jeton] at (5.5,0.5) {};
\node[jeton] at (3.5,0.5) {};
\node[jeton] at (5.5,3.5) {};
\node[jeton] at (3.5,3.5) {};
\node[pion] at (7,4.5) {};
\node[pion] at (7,1.5) {};
\node[pion] at (2.5,-1) {};
\node[pion] at (4.5,-1) {};
}
\end{tikzpicture}
\end{center}
%\newpage
Une fois la multiplication posée, on l'effectue à l'aide du corollaire 1, en bougeant les jetons en diagonale jusqu'à ce qu'ils soient alignés (ici le long de la marge du bas)~:

\begin{center}
\begin{tikzpicture}
\draw[black,thick] (0,0) grid (6,6); 
\foreach \a in {0,...,5}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (5-\a,-0.5) {\aa};
\node[rotate=45] at (6.5,1+\a) {\aa};
\draw (5-\a,-1) -- (6-\a,0);
\draw (6,\a) -- (7,1+\a);
\node at (0.5+\a,5.5-\a) {.};
\node[jeton] at (5.5,0.5) {};
\node[jeton] at (3.5,0.5) {};
\node[jeton] at (5.5,3.5) {};
\node[jeton] at (3.5,3.5) {};
\draw[->,>=stealth,ultra thick] (5.5,3.5) to (2.5,0.5);
\draw[->,>=stealth,ultra thick] (3.5,3.5) to (0.5,0.5);
}
\end{tikzpicture}
\end{center}
\bigskip
Ce qui donne l'écriture binaire du produit~: \texttt{101101} ou $32+8+4+1 = 45$~:
\begin{center}
\begin{tikzpicture}
\draw[black,thick] (0,0) grid (6,6); 
\foreach \a in {0,...,5}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (5-\a,-0.5) {\aa};
\node[rotate=45] at (6.5,1+\a) {\aa};
\draw (5-\a,-1) -- (6-\a,0);
\draw (6,\a) -- (7,1+\a);
\node at (0.5+\a,5.5-\a) {.};
\node[jeton] at (5.5,0.5) {};
\node[jeton] at (3.5,0.5) {};
\node[jeton] at (2.5,0.5) {};
\node[jeton] at (0.5,0.5) {};
}
\end{tikzpicture}
\end{center}

\bigskip

Remarque~: on peut voir la décomposition binaire de 45 avec cette figure~:

\begin{center}
\begin{tikzpicture}
\draw (0,0) -- (8,0) -- (8,4) -- (0,4) -- (0,0);
\draw (8,0) -- (9,0) -- (9,4) -- (8,4) -- (8,0);
\draw (8,4) -- (9,4) -- (9,5) -- (8,5) -- (8,4);
\draw (0,4) -- (8,4) -- (8,5) -- (0,5) -- (0,4);
\node at (4,2) {$8\times 4 = 32$};
\node at (4,4.5) {$8\times 1 = 8$};
\node[rotate=90] at (8.5,2) {$1\times 4 = 4$};
\node at (8.5,4.5) {$1$};
\draw[<->] (0,5.5) -- (9,5.5) node[sloped,midway,above] {$8+1=9$};
\draw[<->] (-0.5,0) -- (-0.5,5) node[sloped,midway,above] {$4+1=5$};
\end{tikzpicture}
\end{center}

\textbf{Exemple}

Pour multiplier $19 = 16+2+1$ par $13 = 8+4+1$, on commence par représenter en binaire, les deux nombres 19 et 13, dans les marges~:

\begin{center}
\begin{tikzpicture}
\draw[black,thick] (0,0) grid (8,8); 
\foreach \a in {0,...,7}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (7-\a,-0.5) {\aa};
\node[rotate=45] at (8.5,1+\a) {\aa};
\draw (7-\a,-1) -- (8-\a,0);
\draw (8,\a) -- (9,1+\a);
\node at (0.5+\a,7.5-\a) {.};
\node[pion] at (9,4.5) {};
\node[pion] at (9,3.5) {};
\node[pion] at (9,1.5) {};
\node[pion] at (6.5,-1) {};
\node[pion] at (5.5,-1) {};
\node[pion] at (2.5,-1) {};
}
\end{tikzpicture}
\end{center}

Puis on effectue le produit cartésien~:


\begin{center}
\begin{tikzpicture}
\draw[black,thick] (0,0) grid (8,8); 
\foreach \a in {0,...,7}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (7-\a,-0.5) {\aa};
\node[rotate=45] at (8.5,1+\a) {\aa};
\draw (7-\a,-1) -- (8-\a,0);
\draw (8,\a) -- (9,1+\a);
\node at (0.5+\a,7.5-\a) {.};
\node[jeton] at (7.5,0.5) {};
\node[jeton] at (6.5,0.5) {};
\node[jeton] at (3.5,0.5) {};
\node[jeton] at (7.5,2.5) {};
\node[jeton] at (6.5,2.5) {};
\node[jeton] at (3.5,2.5) {};
\node[jeton] at (7.5,3.5) {};
\node[jeton] at (6.5,3.5) {};
\node[jeton] at (3.5,3.5) {};
\node[pion] at (9,4.5) {};
\node[pion] at (9,3.5) {};
\node[pion] at (9,1.5) {};
\node[pion] at (6.5,-1) {};
\node[pion] at (5.5,-1) {};
\node[pion] at (2.5,-1) {};
}
\end{tikzpicture}
\end{center}

Ensuite, on utilise le corollaire 1 pour déplacer les jetons sans changer leur valeur~:


\begin{center}
\begin{tikzpicture}
\draw[black,thick] (0,0) grid (8,8); 
\foreach \a in {0,...,7}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (7-\a,-0.5) {\aa};
\node[rotate=45] at (8.5,1+\a) {\aa};
\draw (7-\a,-1) -- (8-\a,0);
\draw (8,\a) -- (9,1+\a);
\node at (0.5+\a,7.5-\a) {.};
\node[jeton] at (7.5,0.5) {};
\node[jeton] at (6.5,0.5) {};
\node[jeton] at (3.5,0.5) {};
\node[jeton] at (7.5,2.5) {};
\node[jeton] at (6.5,2.5) {};
\node[jeton] at (3.5,2.5) {};
\node[jeton] at (7.5,3.5) {};
\node[jeton] at (6.5,3.5) {};
\node[jeton] at (3.5,3.5) {};
\draw[->,>=stealth,ultra thick] (3.5,3.5) to (0.5,0.5);
\draw[->,>=stealth,ultra thick] (3.5,2.5) to (1.5,0.5);
\draw[->,>=stealth,ultra thick] (6.5,3.5) to (3.5,0.5);
\draw[->,>=stealth,ultra thick] (7.5,3.5) to (4.5,0.5);
\draw[->,>=stealth,ultra thick] (7.5,2.5) to (5.5,0.5);
}
\end{tikzpicture}
\end{center}
Alors la multiplication est effectuée, il ne reste plus qu'à utiliser l'axiome 1 pour simplifier le produit~:
\begin{center}
\begin{tikzpicture}
\draw[black,thick] (0,0) grid (8,8); 
\foreach \a in {0,...,7}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (7-\a,-0.5) {\aa};
\node[rotate=45] at (8.5,1+\a) {\aa};
\draw (7-\a,-1) -- (8-\a,0);
\draw (8,\a) -- (9,1+\a);
\node at (0.5+\a,7.5-\a) {.};
\node[jeton] at (7.5,0.5) {};
\node[jeton] at (6.5,0.5) {};
\node[jeton] at (5.5,0.5) {};
\node[jeton] at (4.3,0.3) {};
\node[jeton] at (4.7,0.7) {};
\node[jeton] at (3.3,0.3) {};
\node[jeton] at (3.7,0.7) {};
\node[jeton] at (1.5,0.5) {};
\node[jeton] at (0.5,0.5) {};
}
\end{tikzpicture}
\end{center}
À partir de
\begin{center}
\begin{tikzpicture}
\draw[black,thick] (0,0) grid (8,1); 
\foreach \a in {0,...,7}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (7-\a,-0.5) {\aa};
\draw (7-\a,-1) -- (8-\a,0);
\node[jeton] at (7.5,0.5) {};
\node[jeton] at (6.5,0.5) {};
\node[jeton] at (5.5,0.5) {};
\node[jeton] at (4.3,0.3) {};
\node[jeton] at (4.7,0.7) {};
\node[jeton] at (3.3,0.3) {};
\node[jeton] at (3.7,0.7) {};
\node[jeton] at (1.5,0.5) {};
\node[jeton] at (0.5,0.5) {};
}
\end{tikzpicture}
\end{center}
on peut faire
\begin{center}
\begin{tikzpicture}
\draw[black,thick] (0,0) grid (8,1); 
\foreach \a in {0,...,7}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (7-\a,-0.5) {\aa};
\draw (7-\a,-1) -- (8-\a,0);
\node[jeton] at (7.5,0.5) {};
\node[jeton] at (6.5,0.5) {};
\node[jeton] at (5.5,0.5) {};
\node[jeton] at (4.3,0.3) {};
\node[jeton] at (4.7,0.7) {};
\node[jeton] at (2.5,0.5) {};
\node[jeton] at (1.5,0.5) {};
\node[jeton] at (0.5,0.5) {};
}
\end{tikzpicture}
\end{center}
puis
\begin{center}
\begin{tikzpicture}
\draw[black,thick] (0,0) grid (8,1); 
\foreach \a in {0,...,7}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (7-\a,-0.5) {\aa};
\draw (7-\a,-1) -- (8-\a,0);
\node[jeton] at (7.5,0.5) {};
\node[jeton] at (6.5,0.5) {};
\node[jeton] at (5.5,0.5) {};
\node[jeton] at (3.5,0.5) {};
\node[jeton] at (2.5,0.5) {};
\node[jeton] at (1.5,0.5) {};
\node[jeton] at (0.5,0.5) {};
}
\end{tikzpicture}
\end{center}

Et on lit, en binaire, le résultat de la multiplication~: \texttt{11110111} ou $128+64+32+16+4+2+1 = 247$~: $19 \times 13 = 247$.


\begin{comment}
\begin{center}
\begin{tikzpicture}
\draw[black,thick] (0,0) grid (10,10); 
\foreach \a in {0,...,9}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (9-\a,-0.5) {\aa};
\node[rotate=45] at (10.5,1+\a) {\aa};
\draw (9-\a,-1) -- (10-\a,0);
\draw (10,\a) -- (11,1+\a);
\node at (0.5+\a,9.5-\a) {.};
}
\end{tikzpicture}
\end{center}
\end{comment}

\section{Chapitre X : De la division}

Pour effectuer une division (euclidienne), on essaye de reproduire un motif similaire aux produits cartésiens vus au chapitre précédent.

Par exemple, pour diviser 128 par 8, on fait faire ce trajet à un jeton valant 128~:

\begin{center}
\begin{tikzpicture}
\draw[black,thick] (0,0) grid (8,8); 
\foreach \a in {0,...,7}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (7-\a,-0.5) {\aa};
\node[rotate=45] at (8.5,1+\a) {\aa};
\draw (7-\a,-1) -- (8-\a,0);
\draw (8,\a) -- (9,1+\a);
\node at (0.5+\a,7.5-\a) {.};
\node[jeton] at (7.5,7.5) {};
%\node[jeton] at (4.5,3.5) {};
\draw[->,>=stealth,very thick] (7.5,7.5) to (4.5,3.5);
\draw[->,>=stealth,dotted] (4.5,0.3) to (4.5,3.3);
\node[pion] at (9,8.5) {};
\node[pion] at (3.5,-1) {};
}
\end{tikzpicture}
\end{center}

\textbf{Exemple}

Pour diviser 250 (représenté verticalement dans la marge de droite comme $128+64+32+16+8+2$) par 13 (représenté dans la marge inférieure comme $8+4+1$), on va essayer de reproduire le motif \texttt{1101} sur les lignes qui formeront le quotient~:


\begin{comment}
\begin{center}
\begin{tikzpicture}
\draw[black,thick] (0,0) grid (8,8); 
\foreach \a in {0,...,7}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (7-\a,-0.5) {\aa};
\node[rotate=45] at (8.5,1+\a) {\aa};
\draw (7-\a,-1) -- (8-\a,0);
\draw (8,\a) -- (9,1+\a);
\node at (0.5+\a,7.5-\a) {.};
\node[jeton] at (7.5,0.5) {};
}
\end{tikzpicture}
\end{center}
\end{comment}


\begin{center}
\begin{tikzpicture}
\draw[black,thick] (0,0) grid (8,8); 
\foreach \a in {0,...,7}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (7-\a,-0.5) {\aa};
\node[rotate=45] at (8.5,1+\a) {\aa};
\draw (7-\a,-1) -- (8-\a,0);
\draw (8,\a) -- (9,1+\a);
\node at (0.5+\a,7.5-\a) {.};
\node[pion] at (3.5,-1) {};
\node[pion] at (4.5,-1) {};
\node[pion] at (6.5,-1) {};
\node[pion] at (9,2.5) {};
\node[pion] at (9,4.5) {};
\node[pion] at (9,5.5) {};
\node[pion] at (9,6.5) {};
\node[pion] at (9,7.5) {};
\node[pion] at (9,8.5) {};
}
\end{tikzpicture}
\end{center}

On commence par le poids fort, en utilisant le corollaire 1 pour les déplacements~:

\begin{center}
\begin{tikzpicture}
\draw[black,thick] (0,0) grid (8,8); 
\foreach \a in {0,...,7}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (7-\a,-0.5) {\aa};
\node[rotate=45] at (8.5,1+\a) {\aa};
\draw (7-\a,-1) -- (8-\a,0);
\draw (8,\a) -- (9,1+\a);
\node at (0.5+\a,7.5-\a) {.};
\node[pion] at (3.5,-1) {};
\node[pion] at (4.5,-1) {};
\node[pion] at (6.5,-1) {};
\node[pion] at (9,2.5) {};
\node[pion] at (9,4.5) {};
\node[pion] at (9,5.5) {};
\node[pion] at (9,6.5) {};
\node[pion] at (9,7.5) {};
\node[jeton] at (7.5,7.5) {};
\draw[->,>=stealth] (7.5,7.5) to (4.5,4.5);
}
\end{tikzpicture}
\end{center}
Pour l'instant on peut continuer avec le jeton valant 64~:
\begin{center}
\begin{tikzpicture}
\draw[black,thick] (0,0) grid (8,8); 
\foreach \a in {0,...,7}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (7-\a,-0.5) {\aa};
\node[rotate=45] at (8.5,1+\a) {\aa};
\draw (7-\a,-1) -- (8-\a,0);
\draw (8,\a) -- (9,1+\a);
\node at (0.5+\a,7.5-\a) {.};
\node[pion] at (3.5,-1) {};
\node[pion] at (4.5,-1) {};
\node[pion] at (6.5,-1) {};
\node[pion] at (9,2.5) {};
\node[pion] at (9,4.5) {};
\node[pion] at (9,5.5) {};
\node[pion] at (9,6.5) {};
\node[jeton] at (7.5,6.5) {};
\node[jeton] at (4.5,4.5) {};
\draw[->,>=stealth] (7.5,6.5) to (5.5,4.5);
}
\end{tikzpicture}
\end{center}
Mais il n'y a rien à mettre dans la colonne 2 (il n'y a pas de 2 dans l'écriture binaire de 13) alors on utilise l'axiome 1~:
\begin{center}
\begin{tikzpicture}
\draw[black,thick] (0,0) grid (8,8); 
\foreach \a in {0,...,7}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (7-\a,-0.5) {\aa};
\node[rotate=45] at (8.5,1+\a) {\aa};
\draw (7-\a,-1) -- (8-\a,0);
\draw (8,\a) -- (9,1+\a);
\node at (0.5+\a,7.5-\a) {.};
\node[pion] at (3.5,-1) {};
\node[pion] at (4.5,-1) {};
\node[pion] at (6.5,-1) {};
\node[pion] at (9,2.5) {};
\node[pion] at (9,4.5) {};
\node[pion] at (9,5.5) {};
\node[jeton] at (8.7,5.2) {};
\node[jeton] at (8.3,4.8) {};
\node[jeton] at (5.5,4.5) {};
\node[jeton] at (4.5,4.5) {};
%\draw[->,>=stealth] (7.5,6.5) to (5.5,4.5);
}
\end{tikzpicture}
\end{center}
Un des jetons de valeur 16 va glisser selon le corollaire 1 pour continuer à former le quotient~:
\begin{center}
\begin{tikzpicture}
\draw[black,thick] (0,0) grid (8,8); 
\foreach \a in {0,...,7}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (7-\a,-0.5) {\aa};
\node[rotate=45] at (8.5,1+\a) {\aa};
\draw (7-\a,-1) -- (8-\a,0);
\draw (8,\a) -- (9,1+\a);
\node at (0.5+\a,7.5-\a) {.};
\node[pion] at (3.5,-1) {};
\node[pion] at (4.5,-1) {};
\node[pion] at (6.5,-1) {};
\node[pion] at (9,2.5) {};
\node[pion] at (9,4.5) {};
\node[pion] at (9,5.5) {};
\node[jeton] at (7.5,4.5) {};
\node[jeton] at (8.3,4.8) {};
\node[jeton] at (5.5,4.5) {};
\node[jeton] at (4.5,4.5) {};
\draw[->,>=stealth] (7.5,4.5) to (4.5,1.5);
}
\end{tikzpicture}
\end{center}
Un autre va se positionner pour finir le 13 d'en haut~:
\begin{center}
\begin{tikzpicture}
\draw[black,thick] (0,0) grid (8,8); 
\foreach \a in {0,...,7}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (7-\a,-0.5) {\aa};
\node[rotate=45] at (8.5,1+\a) {\aa};
\draw (7-\a,-1) -- (8-\a,0);
\draw (8,\a) -- (9,1+\a);
\node at (0.5+\a,7.5-\a) {.};
\node[pion] at (3.5,-1) {};
\node[pion] at (4.5,-1) {};
\node[pion] at (6.5,-1) {};
\node[pion] at (9,2.5) {};
\node[pion] at (9,4.5) {};
\node[pion] at (9,5.5) {};
\node[jeton] at (7.5,4.5) {};
\node[jeton] at (4.5,1.5) {};
\node[jeton] at (5.5,4.5) {};
\node[jeton] at (4.5,4.5) {};
%\draw[->,>=stealth] (7.5,4.5) to (4.5,1.5);
}
\end{tikzpicture}
\end{center}
Et le dernier subira l'axiome 1~:
\begin{center}
\begin{tikzpicture}
\draw[black,thick] (0,0) grid (8,8); 
\foreach \a in {0,...,7}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (7-\a,-0.5) {\aa};
\node[rotate=45] at (8.5,1+\a) {\aa};
\draw (7-\a,-1) -- (8-\a,0);
\draw (8,\a) -- (9,1+\a);
\node at (0.5+\a,7.5-\a) {.};
\node[pion] at (3.5,-1) {};
\node[pion] at (4.5,-1) {};
\node[pion] at (6.5,-1) {};
\node[pion] at (9,2.5) {};
\node[pion] at (9,4.5) {};
\node[jeton] at (8.7,4.2) {};
\node[jeton] at (8.3,3.8) {};
\node[jeton] at (7.5,4.5) {};
\node[jeton] at (4.5,1.5) {};
\node[jeton] at (5.5,4.5) {};
\node[jeton] at (4.5,4.5) {};
%\draw[->,>=stealth] (7.5,4.5) to (4.5,1.5);
}
\end{tikzpicture}
\end{center}
Un des 3 jetons de valeur 8 va en bas selon le corollaire 1~:
\begin{center}
\begin{tikzpicture}
\draw[black,thick] (0,0) grid (8,8); 
\foreach \a in {0,...,7}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (7-\a,-0.5) {\aa};
\node[rotate=45] at (8.5,1+\a) {\aa};
\draw (7-\a,-1) -- (8-\a,0);
\draw (8,\a) -- (9,1+\a);
\node at (0.5+\a,7.5-\a) {.};
\node[pion] at (3.5,-1) {};
\node[pion] at (4.5,-1) {};
\node[pion] at (6.5,-1) {};
\node[pion] at (9,2.5) {};
\node[pion] at (9,4.5) {};
\node[jeton] at (7.5,3.5) {};
\node[jeton] at (8.3,3.8) {};
\node[jeton] at (7.5,4.5) {};
\node[jeton] at (4.5,1.5) {};
\node[jeton] at (5.5,4.5) {};
\node[jeton] at (4.5,4.5) {};
\draw[->,>=stealth] (7.5,3.5) to (4.5,0.5);
}
\end{tikzpicture}
\end{center}
Le second continue l'écriture du 13 de la ligne 2~:
\begin{center}
\begin{tikzpicture}
\draw[black,thick] (0,0) grid (8,8); 
\foreach \a in {0,...,7}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (7-\a,-0.5) {\aa};
\node[rotate=45] at (8.5,1+\a) {\aa};
\draw (7-\a,-1) -- (8-\a,0);
\draw (8,\a) -- (9,1+\a);
\node at (0.5+\a,7.5-\a) {.};
\node[pion] at (3.5,-1) {};
\node[pion] at (4.5,-1) {};
\node[pion] at (6.5,-1) {};
\node[pion] at (9,2.5) {};
\node[pion] at (9,4.5) {};
\node[jeton] at (7.5,3.5) {};
\node[jeton] at (4.5,0.5) {};
\node[jeton] at (7.5,4.5) {};
\node[jeton] at (4.5,1.5) {};
\node[jeton] at (5.5,4.5) {};
\node[jeton] at (4.5,4.5) {};
\draw[->,>=stealth] (7.5,3.5) to (5.5,1.5);
}
\end{tikzpicture}
\end{center}
et le dernier subit l'axiome 1~:
\begin{center}
\begin{tikzpicture}
\draw[black,thick] (0,0) grid (8,8); 
\foreach \a in {0,...,7}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (7-\a,-0.5) {\aa};
\node[rotate=45] at (8.5,1+\a) {\aa};
\draw (7-\a,-1) -- (8-\a,0);
\draw (8,\a) -- (9,1+\a);
\node at (0.5+\a,7.5-\a) {.};
\node[pion] at (3.5,-1) {};
\node[pion] at (4.5,-1) {};
\node[pion] at (6.5,-1) {};
\node[pion] at (9,2.5) {};
\node[jeton] at (9,3.5) {};
\node[jeton] at (8.5,3) {};
\node[jeton] at (5.5,1.5) {};
\node[jeton] at (4.5,0.5) {};
\node[jeton] at (7.5,4.5) {};
\node[jeton] at (4.5,1.5) {};
\node[jeton] at (5.5,4.5) {};
\node[jeton] at (4.5,4.5) {};
%\draw[->,>=stealth] (7.5,3.5) to (5.5,1.5);
}
\end{tikzpicture}
\end{center}
Un des deux jetons de valeur 4 va bouger pour continuer à former un 13 (en bas)~:
\begin{center}
\begin{tikzpicture}
\draw[black,thick] (0,0) grid (8,8); 
\foreach \a in {0,...,7}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (7-\a,-0.5) {\aa};
\node[rotate=45] at (8.5,1+\a) {\aa};
\draw (7-\a,-1) -- (8-\a,0);
\draw (8,\a) -- (9,1+\a);
\node at (0.5+\a,7.5-\a) {.};
\node[pion] at (3.5,-1) {};
\node[pion] at (4.5,-1) {};
\node[pion] at (6.5,-1) {};
\node[pion] at (9,2.5) {};
\node[jeton] at (9,3.5) {};
\node[jeton] at (7.5,2.5) {};
\node[jeton] at (5.5,1.5) {};
\node[jeton] at (4.5,0.5) {};
\node[jeton] at (7.5,4.5) {};
\node[jeton] at (4.5,1.5) {};
\node[jeton] at (5.5,4.5) {};
\node[jeton] at (4.5,4.5) {};
\draw[->,>=stealth] (7.5,2.5) to (5.5,0.5);
}
\end{tikzpicture}
\end{center}
Et l'autre, n'ayant nulle part où aller, subit l'axiome 1~:
\begin{center}
\begin{tikzpicture}
\draw[black,thick] (0,0) grid (8,8); 
\foreach \a in {0,...,7}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (7-\a,-0.5) {\aa};
\node[rotate=45] at (8.5,1+\a) {\aa};
\draw (7-\a,-1) -- (8-\a,0);
\draw (8,\a) -- (9,1+\a);
\node at (0.5+\a,7.5-\a) {.};
\node[pion] at (3.5,-1) {};
\node[pion] at (4.5,-1) {};
\node[pion] at (6.5,-1) {};
\node[pion] at (9,2.5) {};
\node[jeton] at (8.7,2.2) {};
\node[jeton] at (8.3,1.8) {};
\node[jeton] at (5.5,0.5) {};
\node[jeton] at (5.5,1.5) {};
\node[jeton] at (4.5,0.5) {};
\node[jeton] at (7.5,4.5) {};
\node[jeton] at (4.5,1.5) {};
\node[jeton] at (5.5,4.5) {};
\node[jeton] at (4.5,4.5) {};
}
\end{tikzpicture}
\end{center}
Parmi les 3 jetons de valeur 2, il y en a un qui complète l'écriture binaire du 13 à la deuxième ligne~:
\begin{center}
\begin{tikzpicture}
\draw[black,thick] (0,0) grid (8,8); 
\foreach \a in {0,...,7}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (7-\a,-0.5) {\aa};
\node[rotate=45] at (8.5,1+\a) {\aa};
\draw (7-\a,-1) -- (8-\a,0);
\draw (8,\a) -- (9,1+\a);
\node at (0.5+\a,7.5-\a) {.};
\node[pion] at (3.5,-1) {};
\node[pion] at (4.5,-1) {};
\node[pion] at (6.5,-1) {};
\node[pion] at (9,2.5) {};
\node[jeton] at (7.5,1.5) {};
\node[jeton] at (8.3,1.8) {};
\node[jeton] at (5.5,0.5) {};
\node[jeton] at (5.5,1.5) {};
\node[jeton] at (4.5,0.5) {};
\node[jeton] at (7.5,4.5) {};
\node[jeton] at (4.5,1.5) {};
\node[jeton] at (5.5,4.5) {};
\node[jeton] at (4.5,4.5) {};
}
\end{tikzpicture}
\end{center}
un qui subit l'axiome 1, et un qui reste (où irait-il~?)~:
\begin{center}
\begin{tikzpicture}
\draw[black,thick] (0,0) grid (8,8); 
\foreach \a in {0,...,7}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (7-\a,-0.5) {\aa};
\node[rotate=45] at (8.5,1+\a) {\aa};
\draw (7-\a,-1) -- (8-\a,0);
\draw (8,\a) -- (9,1+\a);
\node at (0.5+\a,7.5-\a) {.};
\node[pion] at (3.5,-1) {};
\node[pion] at (4.5,-1) {};
\node[pion] at (6.5,-1) {};
\node[pion] at (9,2.5) {};
\node[jeton] at (7.5,1.5) {};
\node[pion] at (9,1.5) {};
\node[jeton] at (9.5,2) {};
\node[jeton] at (5.5,0.5) {};
\node[jeton] at (5.5,1.5) {};
\node[jeton] at (4.5,0.5) {};
\node[jeton] at (7.5,4.5) {};
\node[jeton] at (4.5,1.5) {};
\node[jeton] at (5.5,4.5) {};
\node[jeton] at (4.5,4.5) {};
}
\end{tikzpicture}
\end{center}
Enfin, l'un des jetons de valeur 1 sert à compléter le 13 d'en bas~:
\begin{center}
\begin{tikzpicture}
\draw[black,thick] (0,0) grid (8,8); 
\foreach \a in {0,...,7}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (7-\a,-0.5) {\aa};
\node[rotate=45] at (8.5,1+\a) {\aa};
\draw (7-\a,-1) -- (8-\a,0);
\draw (8,\a) -- (9,1+\a);
\node at (0.5+\a,7.5-\a) {.};
\node[pion] at (3.5,-1) {};
\node[pion] at (4.5,-1) {};
\node[pion] at (6.5,-1) {};
\node[pion] at (9,2.5) {};
\node[jeton] at (7.5,1.5) {};
\node[pion] at (9,1.5) {};
\node[jeton] at (7.5,0.5) {};
\node[jeton] at (5.5,0.5) {};
\node[jeton] at (5.5,1.5) {};
\node[jeton] at (4.5,0.5) {};
\node[jeton] at (7.5,4.5) {};
\node[jeton] at (4.5,1.5) {};
\node[jeton] at (5.5,4.5) {};
\node[jeton] at (4.5,4.5) {};
}
\end{tikzpicture}
\end{center}
La division est terminée. On lit le quotient sur n'importe quelle colonne (\texttt{10011} soit $16+2+1 = 19$) et dans la marge de droite, les jetons qui restent représentent le reste $2+1 = 3$. En bref, $250 = 19 \times 13 + 3$

\section{Chapitre XI : Des extractions de racines carrées}
\subsection{Les gnomons}
Pour extraire une racine carrée (entière), on essaye de reproduire un produit cartésien comme ceux du chapitre IX, mais qui soit symétrique par rapport à la diagonale principale (celle avec les points). L'algorithme de Neper pour extraire une racine carrée consiste à construire une telle figure, qui approche le plus possible, par en-dessous, le nombre dont on veut extraire la racine. Neper construit ce carré cartésien par des couches qu'il appelle \textit{gnomons}  Voici quelques gnomons~:

\bigskip

\begin{center}
\begin{tabular}{ccc}
\begin{tikzpicture}
\draw[gray,thin] (0,0) grid (2,2); 
\draw[black, ultra thick] (0,1) -- (1,1) -- (1,2) --  (0,2) -- cycle; 
\draw[black, ultra thick] (0,0) -- (2,0) -- (2,2) --  (1,2) -- (1,1) -- (0,1) -- cycle; 
\foreach \a in {0,1}{
\node at (0.5+\a,1.5-\a) {.};
\node[jeton] at (0.5,0.5) {};
\node[jeton] at (1.5,0.5) {};
\node[jeton] at (0.5,1.5) {};
\node[jeton] at (1.5,1.5) {};
}
\end{tikzpicture}

&
\begin{tikzpicture}
\draw[gray,thin] (0,0) grid (3,3); 
\draw[black, ultra thick] (0,2) -- (1,2) -- (1,3) --  (0,3) -- cycle; 
\draw[black, ultra thick] (0,1) -- (2,1) -- (2,3) --  (1,3) -- (1,2) -- (0,2) -- cycle; 
\draw[black, ultra thick] (0,0) -- (3,0) -- (3,3) --  (2,3) -- (2,1) -- (0,1) -- cycle; 
\foreach \a in {0,1,2}{
\node at (0.5+\a,2.5-\a) {.};
\node[jeton] at (0.5,0.5) {};
\node[jeton] at (2.5,0.5) {};
\node[jeton] at (0.5,2.5) {};
\node[jeton] at (2.5,2.5) {};
}
\end{tikzpicture}
&
\begin{tikzpicture}
\draw[gray,thin] (0,0) grid (3,3); 
\draw[black, ultra thick] (0,2) -- (1,2) -- (1,3) --  (0,3) -- cycle; 
\draw[black, ultra thick] (0,1) -- (2,1) -- (2,3) --  (1,3) -- (1,2) -- (0,2) -- cycle; 
\draw[black, ultra thick] (0,0) -- (3,0) -- (3,3) --  (2,3) -- (2,1) -- (0,1) -- cycle; 
\foreach \a in {0,1,2}{
\node at (0.5+\a,2.5-\a) {.};
\node[jeton] at (0.5,0.5) {};
\node[jeton] at (1.5,0.5) {};
\node[jeton] at (2.5,0.5) {};
\node[jeton] at (0.5,1.5) {};
\node[jeton] at (1.5,1.5) {};
\node[jeton] at (2.5,1.5) {};
\node[jeton] at (0.5,2.5) {};
\node[jeton] at (1.5,2.5) {};
\node[jeton] at (2.5,2.5) {};
}
\end{tikzpicture}
\\
$3^2=9=4+5$&$5^2=25=16+0+9$&$7^2=49=16+20+13$\\
2 gnomons & 3 gnomons & 3 gnomons\\
\end{tabular}
\end{center}


\begin{center}
\begin{tabular}{ccc}

\begin{tikzpicture}
\draw[gray,thin] (0,0) grid (4,4); 
\draw[black, ultra thick] (0,3) -- (1,3) -- (1,4) --  (0,4) -- cycle; 
\draw[black, ultra thick] (0,2) -- (2,2) -- (2,4) --  (1,4) -- (1,3) -- (0,3) -- cycle; 
\draw[black, ultra thick] (0,1) -- (3,1) -- (3,4) --  (2,4) -- (2,2) -- (0,2) -- cycle; 
\draw[black, ultra thick] (0,0) -- (4,0) -- (4,4) --  (3,4) -- (3,1) -- (0,1) -- cycle; 
\foreach \a in {0,...,3}{
\node at (0.5+\a,3.5-\a) {.};
\node[jeton] at (0.5,0.5) {};
\node[jeton] at (2.5,0.5) {};
\node[jeton] at (3.5,0.5) {};
\node[jeton] at (0.5,1.5) {};
\node[jeton] at (2.5,1.5) {};
\node[jeton] at (3.5,1.5) {};
\node[jeton] at (0.5,3.5) {};
\node[jeton] at (2.5,3.5) {};
\node[jeton] at (3.5,3.5) {};
}
\end{tikzpicture}

&&
\begin{tikzpicture}
\draw[gray,thin] (0,0) grid (5,5); 
\draw[black, ultra thick] (0,4) -- (1,4) -- (1,5) --  (0,5) -- cycle; 
\draw[black, ultra thick] (0,3) -- (2,3) -- (2,5) --  (1,5) -- (1,4) -- (0,4) -- cycle; 
\draw[black, ultra thick] (0,2) -- (3,2) -- (3,5) --  (2,5) -- (2,3) -- (0,3) -- cycle; 
\draw[black, ultra thick] (0,1) -- (4,1) -- (4,5) --  (3,5) -- (3,2) -- (0,2) -- cycle; 
\draw[black, ultra thick] (0,0) -- (5,0) -- (5,5) --  (4,5) -- (4,1) -- (0,1) -- cycle; 
\foreach \a in {0,...,4}{
\node at (0.5+\a,4.5-\a) {.};
\node[jeton] at (0.5,0.5) {};
\node[jeton] at (2.5,0.5) {};
\node[jeton] at (4.5,0.5) {};
\node[jeton] at (0.5,2.5) {};
\node[jeton] at (2.5,2.5) {};
\node[jeton] at (4.5,2.5) {};
\node[jeton] at (0.5,4.5) {};
\node[jeton] at (2.5,4.5) {};
\node[jeton] at (4.5,4.5) {};
}
\end{tikzpicture}
\\
$11^2=121=64+0+36+21$&&$21^2=441=256+0+144+0+41$\\
4 gnomons && 5 gnomons

\end{tabular}
\end{center}

\subsection{Exemple}

On veut extraire la racine carrée de 1238. On représente donc 1238 en binaire dans la marge basse~:

\begin{center}
\begin{tikzpicture}[scale=0.9]
\draw[black,thick] (0,0) grid (11,6); 
\foreach \a in {0,...,10}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (10-\a,-0.5) {\aa};
\draw (10-\a,-1) -- (11-\a,0);
}
\foreach \a in {0,...,5}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (11.5,1+\a) {\aa};
\draw (11,\a) -- (12,1+\a);
\node at (10.5-\a,0.5+\a) {.};
}
\node[pion] at (-0.5,-1) {};
\node[pion] at (2.5,-1) {};
\node[pion] at (3.5,-1) {};
\node[pion] at (5.5,-1) {};
\node[pion] at (7.5,-1) {};
\node[pion] at (8.5,-1) {};
\end{tikzpicture}
\end{center}

Le jeton de valeur 1024 peut aller jusqu'à la diagonale ($1024=32^2$)~:

\begin{center}
\begin{tikzpicture}[scale=0.9]
\draw[black,thick] (0,0) grid (11,6); 
\foreach \a in {0,...,10}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (10-\a,-0.5) {\aa};
\draw (10-\a,-1) -- (11-\a,0);
}
\foreach \a in {0,...,5}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (11.5,1+\a) {\aa};
\draw (11,\a) -- (12,1+\a);
\node at (10.5-\a,0.5+\a) {.};
}
\node[jeton] at (0.5,0.5) {};
\node[pion] at (2.5,-1) {};
\node[pion] at (3.5,-1) {};
\node[pion] at (5.5,-1) {};
\node[pion] at (7.5,-1) {};
\node[pion] at (8.5,-1) {};
\draw[->,>=stealth] (0.5,0.5) to (5.5,5.5);
\end{tikzpicture}
\end{center}

Alors il y va, constituant ainsi le \textit{gnomon de tête}. Mais pour continuer la construction, il faudrait 
\begin{itemize}
\item ou bien placer le jeton valant 128 sur la diagonale (ce qui est impossible car 128 n'est pas un carré)
\item ou alors placer deux jetons valant 128 de part et d'autre de la diagonale, l'un tout en haut, l'autre tout à gauche. Or il n'y a qu'un seul jeton valant 128, cela est donc également impossible.
\end{itemize}

\begin{comment}

Alors il y va. Mais le jeton de valeur 128 ne peut pas aller à la diagonale (elle n'est pas sur son chemin)~:

\begin{center}
\begin{tikzpicture}
\draw[black,thick] (0,0) grid (11,6); 
\foreach \a in {0,...,10}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (10-\a,-0.5) {\aa};
\draw (10-\a,-1) -- (11-\a,0);
}
\foreach \a in {0,...,5}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (11.5,1+\a) {\aa};
\draw (11,\a) -- (12,1+\a);
\node at (10.5-\a,0.5+\a) {.};
}
\node[jeton] at (5.5,5.5) {};
\node[jeton] at (3.5,0.5) {};
\node[pion] at (3.5,-1) {};
\node[pion] at (5.5,-1) {};
\node[pion] at (7.5,-1) {};
\node[pion] at (8.5,-1) {};
\draw[->,>=stealth] (3.5,0.5) to (6.5,3.5);
\end{tikzpicture}
\end{center}

\end{comment}



Alors on lui applique l'axiome 1~:
\begin{center}
\begin{tikzpicture}[scale=0.9]
\draw[black,thick] (0,0) grid (11,6); 
\foreach \a in {0,...,10}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (10-\a,-0.5) {\aa};
\draw (10-\a,-1) -- (11-\a,0);
}
\foreach \a in {0,...,5}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (11.5,1+\a) {\aa};
\draw (11,\a) -- (12,1+\a);
\node at (10.5-\a,0.5+\a) {.};
}
\node[jeton] at (5.5,5.5) {};
\node[jeton] at (3.9,-0.6) {};
\node[jeton] at (4.3,-0.3) {};
\node[pion] at (3.5,-1) {};
\node[pion] at (5.5,-1) {};
\node[pion] at (7.5,-1) {};
\node[pion] at (8.5,-1) {};
%\draw[->,>=stealth] (3.5,0.5) to (6.5,3.5);
\end{tikzpicture}
\end{center}
L'un des jetons de valeur 64 va sur la ligne supérieure du carré (les trois gnomons suivant le gnomon de tête de valeur $32^2 = 1024$ sont donc vides)~:
\begin{center}
\begin{tikzpicture}[scale=0.9]
\draw[black,thick] (0,0) grid (11,6); 
\foreach \a in {0,...,10}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (10-\a,-0.5) {\aa};
\draw (10-\a,-1) -- (11-\a,0);
}
\foreach \a in {0,...,5}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (11.5,1+\a) {\aa};
\draw (11,\a) -- (12,1+\a);
\node at (10.5-\a,0.5+\a) {.};
}
\node[jeton] at (5.5,5.5) {};
\node[jeton] at (4.5,0.5) {};
\node[jeton] at (4.3,-0.3) {};
\node[pion] at (3.5,-1) {};
\node[pion] at (5.5,-1) {};
\node[pion] at (7.5,-1) {};
\node[pion] at (8.5,-1) {};
\draw[->,>=stealth] (4.5,0.5) to (9.5,5.5);
\end{tikzpicture}
\end{center}
Un autre va sur la colonne le plus à gauche, continuant à constituer l'avant-dernier gnomon~:
\begin{center}
\begin{tikzpicture}[scale=0.9]
\draw[black,thick] (0,0) grid (11,6); 
\foreach \a in {0,...,10}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (10-\a,-0.5) {\aa};
\draw (10-\a,-1) -- (11-\a,0);
}
\foreach \a in {0,...,5}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (11.5,1+\a) {\aa};
\draw (11,\a) -- (12,1+\a);
\node at (10.5-\a,0.5+\a) {.};
}
\node[jeton] at (5.5,5.5) {};
\node[jeton] at (4.5,0.5) {};
\node[jeton] at (9.5,5.5) {};
\node[pion] at (3.5,-1) {};
\node[pion] at (5.5,-1) {};
\node[pion] at (7.5,-1) {};
\node[pion] at (8.5,-1) {};
\draw[->,>=stealth] (4.5,0.5) to (5.5,1.5);
\end{tikzpicture}
\end{center}

Ce gnomon (de valeur 129) peut être complété avec le jeton de valeur 4~:
\begin{comment}
\begin{center}
\begin{tikzpicture}[scale=0.9]
\draw[black,thick] (0,0) grid (11,6); 
\foreach \a in {0,...,10}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (10-\a,-0.5) {\aa};
\draw (10-\a,-1) -- (11-\a,0);
}
\foreach \a in {0,...,5}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (11.5,1+\a) {\aa};
\draw (11,\a) -- (12,1+\a);
\node at (10.5-\a,0.5+\a) {.};
}
\node[jeton] at (5.5,5.5) {};
\node[jeton] at (4.5,0.5) {};
\node[jeton] at (4.3,-0.3) {};
\node[pion] at (3.5,-1) {};
\node[pion] at (5.5,-1) {};
\node[pion] at (7.5,-1) {};
\node[pion] at (8.5,-1) {};
\draw[->,>=stealth] (4.5,0.5) to (9.5,5.5);
\end{tikzpicture}
\end{center}
Un autre va sur la colonne le plus à gauche, continuant à constituer l'avant-dernier gnomon~:
\end{comment}
\begin{center}
\begin{tikzpicture}[scale=0.9]
\draw[black,thick] (0,0) grid (11,6); 
\foreach \a in {0,...,10}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (10-\a,-0.5) {\aa};
\draw (10-\a,-1) -- (11-\a,0);
}
\foreach \a in {0,...,5}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (11.5,1+\a) {\aa};
\draw (11,\a) -- (12,1+\a);
\node at (10.5-\a,0.5+\a) {.};
}
\node[jeton] at (5.5,5.5) {};
\node[jeton] at (5.5,1.5) {};
\node[jeton] at (9.5,5.5) {};
\node[pion] at (3.5,-1) {};
\node[pion] at (5.5,-1) {};
\node[jeton] at (8.5,0.5) {};
\node[pion] at (8.5,-1) {};
\draw[->,>=stealth] (8.5,0.5) to (9.5,1.5);
\end{tikzpicture}
\end{center}





Le troisième jeton de valeur 64 subit l'axiome 1~:
\begin{center}
\begin{tikzpicture}[scale=0.9]
\draw[black,thick] (0,0) grid (11,6); 
\foreach \a in {0,...,10}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (10-\a,-0.5) {\aa};
\draw (10-\a,-1) -- (11-\a,0);
}
\foreach \a in {0,...,5}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (11.5,1+\a) {\aa};
\draw (11,\a) -- (12,1+\a);
\node at (10.5-\a,0.5+\a) {.};
}
\node[jeton] at (5.5,5.5) {};
\node[jeton] at (5.5,1.5) {};
\node[jeton] at (9.5,5.5) {};
\node[jeton] at (4.5,-1) {};
\node[jeton] at (5.2,-0.3) {};
\node[pion] at (5.5,-1) {};
\node[jeton] at (9.5,1.5) {};
\node[pion] at (8.5,-1) {};
%\draw[->,>=stealth] (4.5,0.5) to (5.5,1.5);
\end{tikzpicture}
\end{center}
Un des jetons de valeur 32 peut aller sur le dernier gnomon~:
\begin{center}
\begin{tikzpicture}[scale=0.9]
\draw[black,thick] (0,0) grid (11,6); 
\foreach \a in {0,...,10}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (10-\a,-0.5) {\aa};
\draw (10-\a,-1) -- (11-\a,0);
}
\foreach \a in {0,...,5}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (11.5,1+\a) {\aa};
\draw (11,\a) -- (12,1+\a);
\node at (10.5-\a,0.5+\a) {.};
}
\node[jeton] at (5.5,5.5) {};
\node[jeton] at (5.5,1.5) {};
\node[jeton] at (9.5,5.5) {};
\node[jeton] at (4.5,-1) {};
\node[jeton] at (5.5,0.5) {};
\node[pion] at (5.5,-1) {};
\node[jeton] at (9.5,1.5) {};
\node[pion] at (8.5,-1) {};
\draw[->,>=stealth] (5.5,0.5) to (10.5,5.5);
\end{tikzpicture}
\end{center}

L'autre aussi~:

\begin{center}
\begin{tikzpicture}[scale=0.9]
\draw[black,thick] (0,0) grid (11,6); 
\foreach \a in {0,...,10}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (10-\a,-0.5) {\aa};
\draw (10-\a,-1) -- (11-\a,0);
}
\foreach \a in {0,...,5}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (11.5,1+\a) {\aa};
\draw (11,\a) -- (12,1+\a);
\node at (10.5-\a,0.5+\a) {.};
}
\node[jeton] at (5.5,5.5) {};
\node[jeton] at (5.5,1.5) {};
\node[jeton] at (9.5,5.5) {};
\node[jeton] at (10.5,5.5) {};
\node[jeton] at (5.5,0.5) {};
\node[pion] at (5.5,-1) {};
\node[jeton] at (9.5,1.5) {};
\node[pion] at (8.5,-1) {};
%\draw[->,>=stealth] (5.5,0.5) to (10.5,5.5);
\end{tikzpicture}
\end{center}

Pour compléter le dernier gnomon, il faut deux jetons de valeur 2 et un jeton de valeur 1. On n'en dispose pas, donc on applique l'axiome 1 au jeton de valeur 16~:

\begin{center}
\begin{tikzpicture}[scale=0.9]
\draw[black,thick] (0,0) grid (11,6); 
\foreach \a in {0,...,10}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (10-\a,-0.5) {\aa};
\draw (10-\a,-1) -- (11-\a,0);
}
\foreach \a in {0,...,5}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (11.5,1+\a) {\aa};
\draw (11,\a) -- (12,1+\a);
\node at (10.5-\a,0.5+\a) {.};
}
\node[jeton] at (5.5,5.5) {};
\node[jeton] at (5.5,1.5) {};
\node[jeton] at (9.5,5.5) {};
\node[jeton] at (10.5,5.5) {};
\node[jeton] at (5.5,0.5) {};
\node[pion] at (6.5,-1) {};
\node[jeton] at (7.2,-0.3) {};
\node[jeton] at (9.5,1.5) {};
\node[pion] at (8.5,-1) {};
%\draw[->,>=stealth] (6.5,0.5) to (8.5,2.5);
\end{tikzpicture}
\end{center}

puis à un des deux jetons de valeur 8~:

\begin{center}
\begin{tikzpicture}[scale=0.9]
\draw[black,thick] (0,0) grid (11,6); 
\foreach \a in {0,...,10}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (10-\a,-0.5) {\aa};
\draw (10-\a,-1) -- (11-\a,0);
}
\foreach \a in {0,...,5}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (11.5,1+\a) {\aa};
\draw (11,\a) -- (12,1+\a);
\node at (10.5-\a,0.5+\a) {.};
}
\node[jeton] at (5.5,5.5) {};
\node[jeton] at (5.5,1.5) {};
\node[jeton] at (9.5,5.5) {};
\node[jeton] at (10.5,5.5) {};
\node[jeton] at (5.5,0.5) {};
\node[jeton] at (9.5,1.5) {};
\node[pion] at (6.5,-1) {};
%\node[jeton] at (8.2,-0.3) {};
\node[jeton] at (7.9,-0.7) {};
\node[pion] at (7.5,-1) {};
\node[pion] at (8.5,-1) {};
%\draw[->,>=stealth] (6.5,0.5) to (8.5,2.5);
\end{tikzpicture}
\end{center}
\begin{comment}
L'un des jetons de valeur 4 peut maintenant aller sur le gnomon (la diagonale)~:

\begin{center}
\begin{tikzpicture}[scale=0.9]
\draw[black,thick] (0,0) grid (11,6); 
\foreach \a in {0,...,10}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (10-\a,-0.5) {\aa};
\draw (10-\a,-1) -- (11-\a,0);
}
\foreach \a in {0,...,5}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (11.5,1+\a) {\aa};
\draw (11,\a) -- (12,1+\a);
\node at (10.5-\a,0.5+\a) {.};
}
\node[jeton] at (5.5,5.5) {};
\node[jeton] at (5.5,1.5) {};
\node[jeton] at (9.5,5.5) {};
\node[jeton] at (10.5,5.5) {};
\node[jeton] at (5.5,0.5) {};
\node[pion] at (6.5,-1) {};
\node[jeton] at (8.2,-0.3) {};
\node[jeton] at (8.5,0.5) {};
\node[pion] at (7.5,-1) {};
\node[pion] at (8.5,-1) {};
\draw[->,>=stealth] (8.5,0.5) to (9.5,1.5);
\end{tikzpicture}
\end{center}
\end{comment}
puis à un des deux jetons de valeur 4~:

\begin{center}
\begin{tikzpicture}[scale=0.9]
\draw[black,thick] (0,0) grid (11,6); 
\foreach \a in {0,...,10}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (10-\a,-0.5) {\aa};
\draw (10-\a,-1) -- (11-\a,0);
}
\foreach \a in {0,...,5}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (11.5,1+\a) {\aa};
\draw (11,\a) -- (12,1+\a);
\node at (10.5-\a,0.5+\a) {.};
}
\node[jeton] at (5.5,5.5) {};
\node[jeton] at (5.5,1.5) {};
\node[jeton] at (9.5,5.5) {};
\node[jeton] at (10.5,5.5) {};
\node[jeton] at (5.5,0.5) {};
\node[pion] at (6.5,-1) {};
\node[jeton] at (9.2,-0.3) {};
\node[jeton] at (8.9,-0.7) {};
\node[jeton] at (9.5,1.5) {};
\node[pion] at (7.5,-1) {};
\node[pion] at (8.5,-1) {};
\end{tikzpicture}
\end{center}

Un des jetons de valeur 2 peut aller sur le dernier gnomon à droite~:

\begin{center}
\begin{tikzpicture}[scale=0.9]
\draw[black,thick] (0,0) grid (11,6); 
\foreach \a in {0,...,10}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (10-\a,-0.5) {\aa};
\draw (10-\a,-1) -- (11-\a,0);
}
\foreach \a in {0,...,5}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (11.5,1+\a) {\aa};
\draw (11,\a) -- (12,1+\a);
\node at (10.5-\a,0.5+\a) {.};
}
\node[jeton] at (5.5,5.5) {};
\node[jeton] at (5.5,1.5) {};
\node[jeton] at (9.5,5.5) {};
\node[jeton] at (10.5,5.5) {};
\node[jeton] at (5.5,0.5) {};
\node[pion] at (6.5,-1) {};
\node[jeton] at (9.2,-0.3) {};
\node[jeton] at (9.5,0.5) {};
\node[jeton] at (9.5,1.5) {};
\node[pion] at (7.5,-1) {};
\node[pion] at (8.5,-1) {};
\draw[->,>=stealth] (9.5,0.5) to (10.5,1.5);
\end{tikzpicture}
\end{center}

Alors il y va~:

\begin{center}
\begin{tikzpicture}[scale=0.9]
\draw[black,thick] (0,0) grid (11,6); 
\foreach \a in {0,...,10}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (10-\a,-0.5) {\aa};
\draw (10-\a,-1) -- (11-\a,0);
}
\foreach \a in {0,...,5}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (11.5,1+\a) {\aa};
\draw (11,\a) -- (12,1+\a);
\node at (10.5-\a,0.5+\a) {.};
}
\node[jeton] at (5.5,5.5) {};
\node[jeton] at (5.5,1.5) {};
\node[jeton] at (9.5,5.5) {};
\node[jeton] at (10.5,5.5) {};
\node[jeton] at (5.5,0.5) {};
\node[pion] at (6.5,-1) {};
\node[jeton] at (9.2,-0.3) {};
\node[jeton] at (10.5,1.5) {};
\node[jeton] at (9.5,1.5) {};
\node[pion] at (7.5,-1) {};
\node[pion] at (8.5,-1) {};
\end{tikzpicture}
\end{center}
Un autre jeton de valeur 2 peut aller sur le gnomon~:
\begin{center}
\begin{tikzpicture}[scale=0.9]
\draw[black,thick] (0,0) grid (11,6); 
\foreach \a in {0,...,10}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (10-\a,-0.5) {\aa};
\draw (10-\a,-1) -- (11-\a,0);
}
\foreach \a in {0,...,5}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (11.5,1+\a) {\aa};
\draw (11,\a) -- (12,1+\a);
\node at (10.5-\a,0.5+\a) {.};
}
\node[jeton] at (5.5,5.5) {};
\node[jeton] at (5.5,1.5) {};
\node[jeton] at (9.5,5.5) {};
\node[jeton] at (10.5,5.5) {};
\node[jeton] at (5.5,0.5) {};
\node[pion] at (6.5,-1) {};
\node[jeton] at (9.5,0.5) {};
\node[jeton] at (10.5,1.5) {};
\node[jeton] at (9.5,1.5) {};
\node[pion] at (7.5,-1) {};
\node[pion] at (8.5,-1) {};
\end{tikzpicture}
\end{center}

Pour compléter le gnomon, il faudrait un jeton de valeur 1. Pour en obtenir un, il faut donc appliquer l'axiome 1 au jeton de valeur 2~: 
\begin{center}
\begin{tikzpicture}[scale=0.9]
\draw[black,thick] (0,0) grid (11,6); 
\foreach \a in {0,...,10}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (10-\a,-0.5) {\aa};
\draw (10-\a,-1) -- (11-\a,0);
}
\foreach \a in {0,...,5}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (11.5,1+\a) {\aa};
\draw (11,\a) -- (12,1+\a);
\node at (10.5-\a,0.5+\a) {.};
}
\node[jeton] at (5.5,5.5) {};
\node[jeton] at (5.5,1.5) {};
\node[jeton] at (9.5,5.5) {};
\node[jeton] at (10.5,5.5) {};
\node[jeton] at (5.5,0.5) {};
\node[pion] at (6.5,-1) {};
\node[jeton] at (9.5,0.5) {};
\node[jeton] at (10.5,1.5) {};
\node[jeton] at (9.5,1.5) {};
\node[pion] at (7.5,-1) {};
\node[pion] at (9.5,-1) {};
\node[jeton] at (10.2,-0.3) {};
\end{tikzpicture}
\end{center}



Un jeton de valeur 1 peut aller sur le gnomon~:
\begin{center}
\begin{tikzpicture}[scale=0.9]
\draw[black,thick] (0,0) grid (11,6); 
\foreach \a in {0,...,10}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (10-\a,-0.5) {\aa};
\draw (10-\a,-1) -- (11-\a,0);
}
\foreach \a in {0,...,5}{
\pgfmathtruncatemacro\aa{2^\a}
\node[rotate=45] at (11.5,1+\a) {\aa};
\draw (11,\a) -- (12,1+\a);
\node at (10.5-\a,0.5+\a) {.};
}
\node[jeton] at (5.5,5.5) {};
\node[jeton] at (5.5,1.5) {};
\node[jeton] at (9.5,5.5) {};
\node[jeton] at (10.5,5.5) {};
\node[jeton] at (5.5,0.5) {};
\node[pion] at (6.5,-1) {};
\node[jeton] at (9.5,0.5) {};
\node[jeton] at (10.5,1.5) {};
\node[jeton] at (9.5,1.5) {};
\node[pion] at (7.5,-1) {};
\node[pion] at (9.5,-1) {};
\node[jeton] at (10.5,0.5) {};
\end{tikzpicture}
\end{center}

Le calcul de la racine carrée est terminé, et on voit en bas un reste égal à $8+4+1=13$. La racine carrée entière valant $32+2+1=35$, on en déduit l'égalité $1238 = 35^2 + 13$.

Le gnomon de tête a pour valeur intrinsèque 1 (il n'y a qu'un seul jeton le constituant). Mais pour le placer on le fait glisser de 5 cases sur la diagonale, et sa valeur sur l'abaque est donc $4^5 = 1024$. Les deux autres gnomons non vides sont

\begin{center}
\begin{tabular}{cc}
\begin{tikzpicture}
\draw[black,ultra thick] (0,0) -- (5,0) -- (5,5) -- (4,5) -- (4,1) -- (0,1) -- cycle ;
\foreach \x in {1,...,4} {
\draw[gray,thin] (\x,0) -- (\x,1);
\draw[gray,thin] (4,\x) -- (5,\x);
}
\node[jeton] at (0.5,0.5) {};
\node[jeton] at (4.5,0.5) {};
\node[jeton] at (4.5,4.5) {};
\end{tikzpicture}
&
\begin{tikzpicture}
\draw[black,ultra thick] (0,0) -- (6,0) -- (6,6) -- (5,6) -- (5,1) -- (0,1) -- cycle ;
\foreach \x in {1,...,5} {
\draw[gray,thin] (\x,0) -- (\x,1);
\draw[gray,thin] (5,\x) -- (6,\x);
}
\node[jeton] at (0.5,0.5) {};
\node[jeton] at (4.5,0.5) {};
\node[jeton] at (5.5,0.5) {};
\node[jeton] at (5.5,1.5) {};
\node[jeton] at (5.5,5.5) {};
\end{tikzpicture}
\\
valeur $16\times 2 + 1 = 33$&valeur $32\times 2 + 2\times 2 + 1=69$
\end{tabular}
\end{center} 

Mais le gnomon de gauche, dont la valeur intrinsèque est 33, a été glissé d'une case sur la diagonale. Sa valeur sur l'abaque est donc $33 \times 4 = 132$. La valeur totale du carré est donc $1024+132+69 = 1225 = 35^2$.

\end{document}
