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 \begin{document}
%\frontmatter
 \title{Cantor normal bi\c cimleri}
 \author{David Pierce}
%\date{\today, \printtime}
\date{\today}
 \publishers{Matematik B\"ol\"um\"u\\
Mimar Sinan G\"uzel Sanatlar \"Universitesi\
\.Istanbul\\
\url{dpierce@msgsu.edu.tr}\\
\url{http://mat.msgsu.edu.tr/~dpierce/}}

\maketitle

\tableofcontents

Bu metinde
Cantor normal bi\c cimi hesaplama kurallar\i\ elde edilir.
Cantor normal bi\c ciminde
\begin{gather*}
  \alpha=\upomega^{\alpha_0}\cdot a_0+\dots+\upomega^{\alpha_m}\cdot a_m,\\
\beta=\upomega^{\beta_0}\cdot b_0+\dots+\upomega^{\beta_n}\cdot b_n
\end{gather*}
olsun.
O zaman
\begin{gather*}
\{m,n\}\included\upomega,\\
  \alpha_0>\dots>\alpha_m,\qquad
\beta_0>\dots>\beta_n,\\
\{a_0,\dots,a_m,b_0,\dots,b_n\}\included\upomega\setminus\{0\},\\
\deg(\alpha)=\alpha_0,\qquad\deg(\beta)=\beta_0.
\end{gather*}

\section{Toplama}

\.Ilk olarak
$\alpha+\beta$ toplam\i n\i n Cantor normal bi\c cimini bulmak isteriz.
Ordinaller toplamas\i\ birle\c smeli oldu\u gundan
\begin{equation*}
\text{\fbox{$\alpha+\beta=
\upomega^{\alpha_0}\cdot a_0+\dots+\upomega^{\alpha_m}\cdot a_m
+\upomega^{\beta_0}\cdot b_0+\dots+\upomega^{\beta_n}\cdot b_n$,}}
\end{equation*}
ama $\alpha_m\leq\beta_0$ ise bu ifade Cantor normal bi\c ciminde de\u gildir.
\begin{enumerate}
\item
\fbox{$\alpha_m=\beta_0$} ise
ordinaller \c carpmas\i\ soldan da\u g\i ld\i\u g\i ndan
\begin{multline*}
  \alpha+\beta=
\upomega^{\alpha_0}\cdot a_0+\dots+\upomega^{\alpha_{m-1}}\cdot a_{m-1}\\
{}+\upomega^{\alpha_m}\cdot(a_m+b_0)
+\upomega^{\beta_1}\cdot b_1+\dots+\upomega^{\beta_n}\cdot b_n,
\end{multline*}
ve bu ifade $\alpha+\beta$ toplam\i n\i n Cantor normal bi\c cimidir.
\item
\fbox{$\alpha_m<\beta_0$} ise,
a\c sa\u g\i daki teoreme g\"ore,
$\upomega^{\alpha_m}\cdot a_m$ terimini \c c\i kar\-arak
yuka\-r\i daki ifadeyi sadele\c stirebiliriz.
\end{enumerate}


\begin{theorem}%\label{thm:deg-a<deg-b}
$\deg(\gamma)<\deg(\delta)$ ise
$\gamma+\delta=\delta$.
\end{theorem}

\begin{proof}
  $\gamma<\delta$ ise 
$\upomega^{\gamma}+\upomega^{\delta}=\upomega^{\delta}$ e\c sitli\u gini 
kan\i tlamak yeter.
Bu durumda $0$ olmayan bir $\theta$ i\c cin $\gamma+\theta=\delta$,
dolay\i s\i yla
\begin{equation*}
  \upomega^{\gamma}+\upomega^{\delta}=\upomega^{\gamma}\cdot(1+\upomega^{\theta}).
\end{equation*}
T\"umevar\i mdan $1+\upomega^{\theta}=\upomega^{\theta}$:
\begin{asparaenum}
  \item
$1+\upomega=\sup_{\xi<\upomega}(1+\xi)=\upomega$.
\item
$1+\upomega^{\iota}=\upomega^{\iota}$ ise,
o zaman (t\"umevar\i mdan)
$0$ olmayan t\"um $k$ do\u gal say\i lar\i\ i\c cin
\begin{equation*}
  1+\upomega^{\iota}\cdot k=\upomega^{\iota}\cdot k,
\end{equation*}
dolay\i s\i yla
\begin{multline*}
    1+\upomega^{\iota+1}
=1+\upomega^{\iota}\cdot\upomega
=\sup_{\xi<\upomega}(1+\upomega^{\iota}\cdot\xi)\\
=\sup_{0<\xi<\upomega}(1+\upomega^{\iota}\cdot\xi)
=\sup_{0<\xi<\upomega}(\upomega^{\iota}\cdot\xi)
=\upomega^{\iota+1}.
\end{multline*}
\item
$\iota$ bir limit ve
$\theta<\iota$ oldu\u gu zaman $1+\upomega^{\theta}=\upomega^{\theta}$ ise
  \begin{equation*}
    1+\upomega^{\iota}
=\sup_{\xi<\iota}(1+\upomega^{\xi})
=\sup_{\xi<\iota}(\upomega^{\xi})
=\upomega^{\iota}.\qedhere
  \end{equation*}
\end{asparaenum}
\end{proof}

\"Orne\u gin
\begin{gather*}
  \begin{aligned}
    (\upomega^{\upomega}\cdot2+\upomega^{17}\cdot34)&+(\upomega^{10}\cdot3+63)\\
&=\upomega^{\upomega}\cdot2+\upomega^{17}\cdot34+\upomega^{10}\cdot3+63,
  \end{aligned}\\
  \begin{aligned}
(\upomega^{\upomega}\cdot2+\upomega^{17}\cdot3)&+(\upomega^{17}\cdot34+63)
=\upomega^{\upomega}\cdot2+\upomega^{17}\cdot37+63,\\
(\upomega^{\upomega}\cdot2+\upomega^{10}\cdot3)&+(\upomega^{17}\cdot34+63)
=\upomega^{\upomega}\cdot2+\upomega^{17}\cdot34+63.
  \end{aligned}
 \end{gather*}

\section{\c Carpma}

\c Simdi $\alpha\cdot\beta$ \c carp\i m\i n\i n Cantor normal bi\c cimini 
bulmak isteriz.
Ordinaller \c carpmas\i\ soldan da\u g\i lmal\i\ oldu\u gundan
\begin{equation*}
  \alpha\cdot\beta
=\alpha\cdot\upomega^{\beta_0}\cdot b_0+\dots+\alpha\cdot\upomega^{\beta_n}\cdot b_n.
\end{equation*}
Bu toplam\i n $\alpha\cdot\upomega^{\beta_k}\cdot b_k$ terimlerinin 
Cantor normal bi\c cimini bulmak i\c cin
\begin{equation*}
  \gamma=\upomega^{\alpha_1}\cdot a_1+\dots+\upomega^{\alpha_m}\cdot a_m
\end{equation*}
olsun ($\gamma=0$ olabilir), dolay\i s\i yla
\begin{equation*}
  \alpha=\upomega^{\alpha_0}\cdot a_0+\gamma,
\end{equation*}
ve yukar\i daki Teoreme %{thm:deg-a<deg-b} 
g\"ore 
\begin{equation*}
  \gamma+\upomega^{\alpha_0}=\upomega^{\alpha_0}.
\end{equation*}
\c Simdi $\beta_k$ \"uslerine g\"ore 
iki durum vard\i r: $\beta_k=0$ ve $\beta_k>0$.
\begin{enumerate}
  \item
\fbox{$0<c<\upomega$} ise
\begin{align*}
  \alpha\cdot c
&=(\upomega^{\alpha_0}\cdot a_0+\gamma)\cdot c\\
&=\underbrace{(\upomega^{\alpha_0}\cdot a_0+\gamma)+\dots
+(\upomega^{\alpha_0}\cdot a_0+\gamma)}_c\\
&=\upomega^{\alpha_0}\cdot a_0
+\underbrace{(\gamma+\upomega^{\alpha_0}\cdot a_0)+\cdots
+(\gamma+\upomega^{\alpha_0}\cdot a_0)}_{c-1}+\gamma\\
&=\upomega^{\alpha_0}\cdot a_0
+\underbrace{\upomega^{\alpha_0}\cdot a_0+\cdots+\upomega^{\alpha_0}\cdot a_0}_{c-1}
+\gamma\\
&=\upomega^{\alpha_0}\cdot a_0\cdot c+\gamma\\
&=\upomega^{\alpha_0}\cdot a_0\cdot c
+\upomega^{\alpha_1}\cdot a_1+\dots +\upomega^{\alpha_m}\cdot a_m,
\end{align*}
ve bu son ifade, Cantor normal bi\c cimindedir.
\item
\fbox{$\delta>0$} olsun.
O zaman bir $\theta$ i\c cin $\delta=1+\theta$,
dolay\i s\i yla
\begin{equation*}
  \alpha\cdot\upomega^{\delta}
=\alpha\cdot\upomega\cdot\upomega^{\theta}.
\end{equation*}
Ayr\i ca
\begin{align*}
  \alpha\cdot\upomega
&=\sup_{x<\upomega}(\alpha\cdot x)\\
&=\sup_{x<\upomega}\bigl((\upomega^{\alpha_0}\cdot a_0+\gamma)\cdot x\bigr)\\
%&=\sup_{x<\upomega}(\alpha\cdot x)\\
&=\sup_{x<\upomega}(\upomega^{\alpha_0}\cdot a_0\cdot x+\gamma)\\
&\leq\sup_{x<\upomega}\bigl(\upomega^{\alpha_0}\cdot a_0\cdot(x+1)\bigr)\\
&=\upomega^{\alpha_0}\cdot\sup_{x<\upomega}\bigl(a_0\cdot(x+1)\bigr)\\
&=\upomega^{\alpha_0}\cdot\upomega,
\end{align*}
ve benzer \c sekilde
\begin{equation*}
\upomega^{\alpha_0}\cdot\upomega
=\sup_{x<\upomega}(\upomega^{\alpha_0}\cdot x)
\leq\sup_{x<\upomega}(\alpha\cdot x)
=\alpha\cdot\upomega,
\end{equation*}
dolay\i s\i yla
\begin{gather*}
  \alpha\cdot\upomega=\upomega^{\alpha_0}\cdot\upomega,\\
  \alpha\cdot\upomega^{\delta}
=\upomega^{\alpha_0}\cdot\upomega\cdot\upomega^{\theta}
=\upomega^{\alpha_0+\delta}.
\end{gather*}
\end{enumerate}
\c Su andan itibaren
\begin{align*}
  \beta_n&>0,&
0&<c<\upomega
\end{align*}
olsun.  Hem $\alpha\cdot\beta$ hem $\alpha\cdot(\beta+c)$ 
\c carp\i m\i n\i n
Cantor normal bi\c cimini bulmu\c stuk.
Asl\i nda
\begin{equation*}
  \alpha\cdot\beta
=\upomega^{\alpha_0+\beta_0}\cdot b_0+\dots+\upomega^{\alpha_0+\beta_n}\cdot b_n,
\end{equation*}
ve k\i saca
\begin{equation*}
  \text{\fbox{$\alpha\cdot\beta
=\upomega^{\alpha_0}\cdot\beta;$}}
\end{equation*}
ayr\i ca
\begin{equation*}
  \text{\fbox{$\alpha\cdot(\beta+c)
=\upomega^{\alpha_0}\cdot\beta+\upomega^{\alpha_0}\cdot a_0\cdot c+\gamma,$}}
\end{equation*}
ve buradan
\begin{multline*}
%  \alpha\cdot(\beta+c)=
\upomega^{\alpha_0+\beta_0}\cdot b_0+\dots+\upomega^{\alpha_0+\beta_n}\cdot b_n\\
{}+\upomega^{\alpha_0}\cdot a_0\cdot c
+\upomega^{\alpha_1}\cdot a_1+\dots+\upomega^{\alpha_m}\cdot a_m
\end{multline*}
Cantor normal bi\c cimi \c c\i kar.
\"Orne\u gin
\begin{gather*}
5\cdot(\upomega^2\cdot 3+\upomega\cdot16+7)=\upomega^2\cdot 3+\upomega\cdot16+35,\\
(\upomega^{\upomega}\cdot2+\upomega+5)
\cdot(\upomega^2\cdot 3+\upomega\cdot16)
=\upomega^{\upomega+2}\cdot 3+\upomega^{\upomega+1}\cdot16,
\end{gather*}
ve
\begin{multline*}
(\upomega^{\upomega}\cdot2+\upomega+5)
\cdot(\upomega^2\cdot 3+\upomega\cdot16+7)\\
=\upomega^{\upomega+2}\cdot 3+\upomega^{\upomega+1}\cdot16+\upomega^{\upomega}\cdot14+\upomega+5.
\end{multline*}

\section{Kuvvet alma}

Son olarak
$\alpha^{\beta}$ kuvvetinin Cantor normal bi\c cimini isteriz.
Kuvvet alma kurallar\i na g\"ore
\begin{equation*}
  \alpha^{\beta}
=\alpha^{\upomega^{\beta_0}\cdot b_0}
\cdot\alpha^{\upomega^{\beta_1}\cdot b_1}
\cdots\alpha^{\upomega^{\beta_n}\cdot b_n}.
\end{equation*}
\.Iki durum vard\i r.
\begin{enumerate}
  \item
\fbox{$1<\alpha<\upomega$} olsun.  O zaman
\begin{equation*}
  \alpha^{\upomega}=\upomega,
\end{equation*}
dolay\i s\i yla $k<\upomega\leq\delta$ ise
\begin{align*}
  \alpha^{\upomega^{k+1}}&=\upomega^{\upomega^k},&
\alpha^{\upomega^{\delta}}&=\upomega^{\upomega^{\delta}}
\end{align*}
(\c c\"unk\"u $1+\delta=\delta$).
Sonu\c c olarak $\beta_{\ell-1}\geq\upomega>\beta_{\ell}$ ise
\begin{equation*}
  \beta^*=\upomega^{\beta_0}\cdot b_0+\dots
+\upomega^{\beta_{\ell-1}}\cdot b_{\ell-1}
+\upomega^{\beta_{\ell}-1}\cdot b_{\ell}+\dots
+\upomega^{\beta_n-1}\cdot b_n
\end{equation*}
olsun; o zaman
\begin{align*}
  \alpha^{\beta}&=\upomega^{\beta^*},&
  \alpha^{\beta+c}&=\upomega^{\beta^*}\cdot a^c,
\end{align*}
ve bunlar, Cantor normal bi\c cimindedir.
\"Orne\u gin
\begin{equation*}
2^{\upomega^{\upomega}\cdot3+\upomega^5\cdot4+\upomega\cdot7+5}
=\upomega^{\upomega^{\upomega}\cdot3+\upomega^4\cdot4+7}\cdot32.
\end{equation*}
\item
\fbox{$\alpha\geq\upomega$} olsun.  O zaman
  \begin{align*}
\alpha^2
%&=\alpha\cdot\alpha\\
&=\alpha\cdot(\upomega^{\alpha_0}\cdot a_0+\gamma)\\
&=\alpha\cdot\upomega^{\alpha_0}\cdot a_0+\alpha\cdot\gamma\\
&=(\upomega^{\alpha_0}\cdot a_0+\gamma)\cdot\upomega^{\alpha_0}\cdot a_0
+\alpha\cdot\gamma\\
&=\upomega^{\alpha_0+\alpha_0}\cdot a_0+\alpha\cdot\gamma\\
&=\upomega^{\alpha_0\cdot2}\cdot a_0+\alpha\cdot\gamma,
  \end{align*}
ve genelde
\begin{equation*}
  \alpha^c=\upomega^{\alpha_0\cdot c}\cdot a_0+\alpha^{c-1}\cdot\gamma.
\end{equation*}
\"Ozel olarak
\begin{equation*}
  \left(\upomega^{\alpha_0}\right)^c\leq\alpha^c
<\left(\upomega^{\alpha_0}\right)^{c+1},
\end{equation*}
dolay\i s\i yla
\begin{gather*}
  \alpha^{\upomega}
%=\upomega^{\alpha_0\cdot\upomega}
=\left(\upomega^{\alpha_0}\right)^{\upomega},\\
\alpha^{\upomega^{\beta_k}\cdot b_k}=\left(\upomega^{\alpha_0}\right)^{\upomega^{\beta_k}\cdot b_k}
\end{gather*}
(\c c\"unk\"u $\beta_k\geq\beta_n>0$),
ve sonu\c c olarak
\begin{equation*}
  \text{\fbox{$
    \begin{gathered}
  \alpha^{\beta}=\upomega^{\alpha_0\cdot\beta},\\
    \alpha^{\beta+c}
=\upomega^{\alpha_0\cdot\beta}\cdot\alpha^c.
    \end{gathered}$}}
\end{equation*}
O zaman
  \begin{align*}
    \alpha^{\beta+c}
%&=\upomega^{\alpha_0\cdot\beta}\cdot\alpha^c\\
&=\upomega^{\alpha_0\cdot\beta}\cdot(\upomega^{\alpha_0\cdot c}\cdot a_0
+\alpha^{c-1}\cdot\gamma)\\
&=\upomega^{\alpha_0\cdot(\beta+c)}\cdot a_0
+\upomega^{\alpha_0\cdot\beta}\cdot\alpha^{c-1}\cdot\gamma.
  \end{align*}
\end{enumerate}


\"Orne\u gin
\begin{align*}
  (\upomega^{\upomega+1}+\upomega^2+1)^{\upomega^2+\upomega\cdot3+2}
&=\upomega^{(\upomega+1)\cdot(\upomega^2+\upomega\cdot3)}
\cdot(\upomega^{\upomega+1}+\upomega^2+1)^2\\
&=\upomega^{\upomega^3+\upomega^2\cdot3}
\cdot(\upomega^{\upomega+1}+\upomega^2+1)^2,
\end{align*}
ve
\begin{align*}
  (\upomega^{\upomega+1}+\upomega^2+1)^2
&=\upomega^{(\upomega+1)\cdot2}
+(\upomega^{\upomega+1}+\upomega^2+1)\cdot(\upomega^2+1)\\
&=\upomega^{(\upomega+1)\cdot2}
+\upomega^{\upomega+1+2}+\upomega^{\upomega+1}+\upomega^2+1\\
&=\upomega^{\upomega\cdot2+1}
+\upomega^{\upomega+3}+\upomega^{\upomega+1}+\upomega^2+1,
\end{align*}
dolay\i s\i yla
\begin{multline*}
(\upomega^{\upomega+1}+\upomega^2+1)^{\upomega^2+\upomega\cdot3+2}\\
=\upomega^{\upomega^3+\upomega^2\cdot3+\upomega\cdot2+1}
+\upomega^{\upomega^3+\upomega^2\cdot3+\upomega+3}\\
+\upomega^{\upomega^3+\upomega^2\cdot3+\upomega+1}
+\upomega^{\upomega^3+\upomega^2\cdot3+2}
+\upomega^{\upomega^3+\upomega^2\cdot3}.
\end{multline*}
 \end{document}