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\begin{document}
\mbox{}\hfill{\VeryHuge
\"Oklid algoritmas\i\ ve ortak \"ol\c c\"us\"u olmama}\hfill\mbox{}
%\begin{multicols}{2}
\centering
\begin{equation*}
  2x^2=y^2
\end{equation*}
Diofant denkleminin tamsay\i\ \c c\"oz\"um\"u yoktur.
Ayn\i\ anlamda, bir karenin kenar ve k\"o\c segeninin ortak \"ol\c
c\"us\"u yoktur.\\
Nitekim varsa, kenar, \"ol\c c\"un\"un $s$ kat\i, ve k\"o\c segen,
\"ol\c c\"un\"un $d$ kat\i, olsun.  O zaman
\begin{align*}
  2s^2&=d^2,&
s^2&=d^2-s^2=(d+s)(d-s),&
\frac s{d+s}&=\frac{d-s}s,&
0&<d-s<s.
\end{align*}
Ayr\i ca $\gcd(d,s)=\gcd(d+s,s)$, ve \"Oklid Algoritmas\i yla
bulunabilir.
\.Ilk ad\i m olarak
\begin{gather*}
  (d+s)=s\cdot2+(d-s)
\end{gather*}
e\c sitli\u gimiz vard\i r.  Ama
\begin{math}
%  \displaystyle
\frac{d+s}s=\frac s{d-s}
\end{math}
e\c sitli\u gi y\"uz\"unden Algoritman\i n her ad\i m\i\ ayn\i\ \c
sekilde olacak:
\begin{gather*}
  s=(d-s)\cdot2+(3s-2d),\\
(d-s)=(3s-2d)\cdot2+(5d-7s),\\
(3s-2d)=(5d-7s)\cdot2+(17s-12d),\\
\makebox[15cm]{\dotfill}
\end{gather*}
$s$ ile $d$ say\i lar\i n\i n bu kombinezonlar\i, fig\"urdeki
uzunlukturlar.  Algoritma ve fig\"ur bitmez; dolay\i s\i yla $s$ ile
$d$, tamsay\i lar olamaz.\\ 
Ancak Algoritma, $\surd2$ limitine yak\i nsayan
\begin{align*}
  &1,&
&\frac32,&
&\frac75,&
&\frac{17}{12},&
&\frac{41}{29},&
&\frac{99}{70},&
&\frac{239}{169},&
&\frac{577}{408},&
&\frac{1393}{985},&
&\frac{3363}{2378},&
&\frac{8119}{5741},&
&\dots
\end{align*}
 bir dizi olu\c sturur.


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%\scriptsize
\footnotesize
\centering
    \setlength\abovedisplayskip{2ex}
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Genel bir y\"ontem vard\i r.\\
  $x$, pozitif bir ger\c cel
say\i\ olsun.  Bir $n$ i\c cin $x_n$ tan\i mlanm\i\c s olursa
\begin{equation*}
a_n=[x_n]
\end{equation*}
olsun.  (Yani $a_n$, $x_n$ say\i s\i n\i n en b\"uy\"uk tamsay\i s\i
d\i r.)  E\u ger $a_n<x_n$ ise
\begin{equation*}
  x_{n+1}=\frac1{x_n-a_n}
\end{equation*}
olsun.  O zaman
\begin{equation*}
  x_n=a_n+\frac1{x_{n+1}}
\end{equation*}
olur.
\"Ozel olarak
\begin{equation*}
  x_0
=a_0+\cfrac1{x_1}
=a_0+\cfrac1{a_1+\cfrac1{x_2}}
=a_0+\cfrac1{a_1+\cfrac1{a_2+\cfrac1{x_3}}}
=a_0+\cfrac1{a_1+\cfrac1{a_2+\cfrac1{a_3+\cfrac1{x_4}}}}
=\dots
\end{equation*}
Ayr\i ca $p_n$ ve $q_n$ tamsay\i lar\i\ i\c cin
\begin{align*}
  x_1&=\frac1{q_0x_0-p_0},&
x_{n+2}&=\frac{p_n-q_nx_0}{q_{n+1}x_0-p_{n+1}}
\end{align*}
olur.  \c Simdi
\begin{align*}
  [a_0;]&=a_0,&
[a_0;a_1]&=a_0+\frac1{a_1},&
[a_0;a_1,\dots,a_{n+1}]&=\bigl[a_0;a_1,\dots,a_{n-1},a_n+\frac1{a_{n+1}}\bigr]
\end{align*}
olsun.  O zaman
\begin{equation*}
  x_0=[x_0;]=[a_0;x_1]=[a_0;a_1,x_2]=[a_0;a_1,a_2,x_3]=\dots
=[a_0;a_1,\dots,a_n,x_{n+1}]=\cdots
\end{equation*}
Ayr\i ca
\begin{gather*}
  [a_0;a_1,\dots,a_n]=\frac{p_n}{q_n},\\
\frac{p_0}{q_0}<\frac{p_2}{q_2}
<\frac{p_4}{q_4}<\dots<x<\dots<\frac{p_5}{q_5}<\frac{p_3}{q_3}<\frac{p_1}{q_1},\\
\frac{p_{n+1}}{q_{n+1}}-\frac{p_n}{q_n}=\frac{(-1)^n}{q_{n+1}q_n}.
\end{gather*}
\"Oyleyse $p_n/q_n$ kesirleri, $x$ limitine yak\i nsar.  Bunu
ispatlamak i\c cin, ilk olarak, 
\begin{align*}
&\begin{gathered}
    p_0=a_0,\\
q_0=1,
  \end{gathered}&
&
  \begin{gathered}
    p_1=a_0a_1+1,\\
q_1=a_1,
  \end{gathered}&
&
  \begin{gathered}
    p_{n+2}=a_{n+2}p_{n+1}+p_n,\\
q_{n+2}=a_{n+2}q_{n+1}+q_n
  \end{gathered}
\end{align*}
e\c sitliklerini g\"osterin.  Ondan sonra,
\begin{equation*}
  [a_0;a_1,\dots,a_{n+2}]=\frac{p_{n+2}}{q_{n+2}}  
\end{equation*}
ise, o zaman
\begin{gather*}
  [a_0;a_1,\dots,a_{n+2}]=\frac{a_{n+2}p_{n+1}+p_n}{a_{n+2}q_{n+1}+q_n},\\
  [a_0;a_1,\dots,a_{n+3}]
=\bigl[a_0;a_1,\dots,a_{n+1},a_{n+2}+\frac1{a_{n+3}}\bigr]
=\frac{\bigl(a_{n+2}+\frac1{a_{n+3}}\bigr)p_{n+1}+p_n}
{\bigl(a_{n+2}+\frac1{a_{n+3}}\bigr)q_{n+1}+q_n}, 
\end{gather*}
\c c\"unk\"u $p_0$, \dots, $p_{n+1}$, $q_0$, \dots, $q_{n+1}$ say\i
lar\i, $a_{n+2}$ say\i s\i na ba\u gl\i\ de\u gildir.
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\tiny \hfill 
David Pierce\hfill
\url{dpierce@msgsu.edu.tr}\hfill
\url{http://mat.msgsu.edu.tr/~dpierce/}\hfill
Matematik B\"ol\"um\"u, Mimar Sinan G\"uzel Sanatlar \"Universitesi,
\.Istanbul\hfill\mbox{} 
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