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\begin{document}

\title{Say\i lar Kuram\i na Giri\c s}
\subtitle{Al\i\c st\i rmalar II}
\author{David Pierce}
\date{13 Ekim 2017}
\publishers{Matematik B\"ol\"um\"u, MSGS\"U\\
\url{dpierce@msgsu.edu.tr}\\
\url{http://mat.msgsu.edu.tr/~dpierce/}}

\maketitle
\thispagestyle{empty}
\addsec{Tamsay\i lar}

Tan\i m\i m\i za g\"ore
\begin{align*}
\N&=\{\text{sayma say\i lar\i}\}=\{1,2,3,\dots\},\\
\upomega&=\{\text{do\u gal say\i lar}\}=\{0\}\cup\N,\\
\Z&=\{\text{tamsay\i lar}\}=\{-x\colon x\in\N\}\cup\upomega.
\end{align*}
Burada $\upomega$ harf\/i $w$ (``\c cifte \emph u'') de\u gil, 
omegad\i r.

\begin{exercise}
$\Z$'de
\begin{align*}
&\begin{gathered}
	a+(b+c)=(a+b)+c,\\
	a+b=b+a,\\
	a+0=a,\\
	a+(-a)=0,
\end{gathered}&
&\begin{gathered}
	a\cdot(b\cdot c)=(a\cdot b)\cdot c,\\
	a\cdot b=b\cdot a,\\
	a\cdot1=a,\\
	a\cdot(b+c)=a\cdot b+a\cdot c.
\end{gathered}
\end{align*}
Bunlar\i\ kullanarak
\begin{equation*}
a\cdot0=0
\end{equation*}
e\c sitli\u gini kan\i tlay\i n.
\end{exercise}

\begin{exercise}
$\Z$'de
\begin{equation*}
a+x=b
\end{equation*}
denkleminin \emph{sayma say\i s\i} \c c\"oz\"um\"u varsa,
\begin{equation*}
a<b
\end{equation*}
ifadesini yazar\i z.
Ayr\i ca
\begin{equation*}
a\in\N\iff-a\notin\upomega.
\end{equation*}
Tabii ki iki sayma say\i s\i n\i n toplam\i\ ve \c carp\i m\i\ da
sayma say\i s\i d\i r.
Bunlardan
\begin{gather*}
a<b\And b<c\implies a<c,\\
a<b\implies	a+c<b+c,\\
a<b\And	c\in\N\implies a\cdot c<b\cdot c
\end{gather*}
gerektirmelerini kan\i tlay\i n.
\end{exercise}

\addsec{\"Oklid Algoritmas\i, B\'ezout Lemmas\i}

E\u ger
\begin{equation*}
a_1>a_2>0
\end{equation*}
ko\c sulunu sa\u glayan 
$a_1$ ve $a_2$ tamsay\i lar\i\ verilirse,
o zaman \"Oklid Algoritmas\i\ ile,
bir $n$ sayma say\i s\i\ i\c cin,
\"oyle sonlu bir $(a_k\colon 1\leq k\leq n)$ dizisini buluruz ki
\begin{equation*}
a_1>a_2>\dots>a_n>0,
\end{equation*}
ve $1<k<n$ olmak \"uzere bir $b_k$ i\c cin
\begin{equation*}
a_{k-1}=a_k\cdot b_k+a_{k+1},
\end{equation*}
ama
\begin{equation*}
a_n\divides a_{n-1}.
\end{equation*}
Bu durumda
\begin{equation*}
a_n=\gcd(a_1,a_2).
\end{equation*}
Ayr\i ca
\begin{equation}\label{eqn:B}
\gcd(a_1,a_2)=a_1\cdot x+a_2\cdot y
\end{equation}
denklemi $\Z$'de \c c\"oz\"ulebilir:
bu sonu\c c, B\'ezout Lemmas\i d\i r.

\begin{exercise}
$(a_1,a_2)$ a\c sa\u g\i daki s\i ral\i\ ikilisi olmak \"uzere
\eqref{eqn:B} denkleminin herhangi 
\emph{iki} \c c\"oz\"um\"un\"u bulun.
Sonu\c clar\i n\i z\i\ kontrol edin.
\begin{enumerate}[(a)]
\item
$(610,377)$
\item
$(5101,5099)$
\item
$(1690,455)$
\end{enumerate}
\end{exercise}
\pagebreak
\begin{exercise}
$\Z$'de,
  buradaki her \"onerme i\c cin,
  \"onerme her zaman do\u gru ise kan\i tlay\i n,
  de\u gilse kar\c s\i t \"ornek verin.
  \begin{enumerate}[(a)]
  \item
	$\gcd(a,b)=\gcd(a^2,b)$
	\item
	$\gcd(a,b)=\gcd(a,a+b)$
	\item
	$\gcd(ac,bc)=c\cdot\gcd(a,b)$
	\item
	$\gcd(a,b)=\gcd(2a+b,a+3b)$
	\item
	$\gcd(a,b)=\gcd(2a+3b,5a+8b)$
	\item
	$\gcd(a,c)=\gcd(b,c)=1\implies\gcd(a+b,c)=1$
	\item
	$\gcd(a,c)=\gcd(b,c)=1\implies\gcd(ab,c)=1$
\item
	$\gcd(a,b)=1\And d\divides ac\And d\divides bc\implies d\divides c$
  \end{enumerate}
\end{exercise}
\vfill
\begin{exercise}
E\u ger $\gcd(a,b)=1$ ise,
a\c sa\u g\i daki ifadelerin alabildi\u gi de\u gerler nedir?
  \begin{enumerate}[(a)]
  \item
$\gcd(a+b,a^2+b^2)$
  \item
$\gcd(a+b,a^2-b^2)$
  \end{enumerate}
\end{exercise}
\vfill
\end{document}
