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\begin{document}

\title{Say\i lar Kuram\i na Giri\c s}
\subtitle{Al\i\c st\i rmalar (taslak)}
\author{David Pierce}
\date{3 Ekim 2017}
\publishers{Matematik B\"ol\"um\"u, MSGS\"U\\
\url{dpierce@msgsu.edu.tr}\\
\url{http://mat.msgsu.edu.tr/~dpierce/}}

\maketitle

\addsec{\"Ozyineleme ve T\"umevar\i m}

\begin{definition}[\"u\c cgen say\i lar]
$t_1=1$, ve bir $n$ i\c cin $t_n$ tan\i mlanm\i\c s olursa
\begin{equation*}
t_{n+1}=t_n+n+1.
\end{equation*}
K\i saca her $n$ sayma say\i s\i\ i\c cin
$t_n=\sum_{k=1}^nk=1+\dots+n$.
\end{definition}

\begin{theorem}
Her $n$ sayma say\i s\i\ i\c cin
\begin{equation}\label{eqn:t_n}
t_n=\frac12\cdot n\cdot(n+1).
\end{equation}
\end{theorem}

\begin{proof}[Kan\i t 1.]
\begin{compactitem}
\item
$\frac12\cdot1\cdot(1+1)=1=t_1$.
\item
Iddia $n=m$ durumunda do\u gruysa
\begin{align*}
t_{m+1}
&=t_m+m+1&&\text{[tan\i m]}\\
&=\frac12\cdot m\cdot(m+1)+m+1&&\text{[hipotez]}\\
&=\left(\frac12\cdot m+1\right)\cdot(m+1)&&\\
&=\frac12(m+2)(m+1)&&\\
&=\frac12(m+1)(m+2),&&
\end{align*}
ve b\"oylece iddia $n=m+1$ durumunda do\u grudur.
\end{compactitem}
T\"umevar\i mdan her $n$ sayma say\i s\i\ i\c cin
\eqref{eqn:t_n} do\u grudur.
\end{proof}

\begin{proof}[Kan\i t 2.]
$(k+1)^2-k^2=2k+1$ oldu\u gundan
\begin{equation*}
\sum_{k=1}^n\bigl((k+1)^2-k^2\bigr)
=\sum_{k=1}^n(2k+1).
\end{equation*}
Ayr\i ca
\begin{gather*}
\sum_{k=1}^n\bigl((k+1)^2-k^2\bigr)=(n+1)^2-1^2=n^2+2n,\\
\sum_{k=1}^n(2k+1)=2\sum_{k=1}^nk+n,
\end{gather*}
dolay\i s\i yla
\begin{gather*}
	n^2+2n=2\sum_{k=1}^nk+n,\\
	2\sum_{k=1}^nk=n^2+n.\qedhere
\end{gather*}
\end{proof}

\begin{exercise}
A\c sa\u g\i daki e\c sitlikleri
her $n$ sayma say\i s\i\ i\c cin kan\i tlay\i n.
\begin{enumerate}[a)]
\item 
$\sum_{k=1}^n(2k-1)=n^2$.
\item
$t_n+t_{n+1}=(n+1)^2$.
\item
$\sum_{k=1}^nk^2=\frac13n^3+\frac12n^2+\frac16n$.
\end{enumerate}
\end{exercise}

\begin{exercise}
\"Oyle $a$, $b$, ve $c$ bulun ki
her $n$ sayma say\i s\i\ i\c cin
\begin{equation*}
\sum_{k=1}^nt_k=an^3+bn^2+cn.
\end{equation*}
\end{exercise}

\begin{exercise}
\"Oyle $a$, $b$, ve $c$ bulun ki
her $n$ sayma say\i s\i\ i\c cin
\begin{equation*}
\sum_{k=1}^nk^3=an^4+bn^3+cn^2.
\end{equation*}
\end{exercise}

\begin{exercise}
Her $n$ sayma say\i s\i\ i\c cin
\begin{equation*}
n^3=\sum_{k=t_{n-1}+1}^{t_n}(2k-1)
\end{equation*}
kan\i tlay\i n.  (Burada $t_0=0$.)
\end{exercise}

\addsec{Hesaplama kurallar\i}

Yukar\i da okulda \"o\u gretilmi\c s hesaplama kurallar\i n\i\ varsayd\i k,
ama bu kurallar\i\ kan\i tlayabiliriz.
Toplaman\i n birle\c smeli ve de\u gi\c smeli olmas\i,
geometriden gelir.

\begin{definition}[\c co\u galtma]
Her $a$ ger\c cel say\i s\i\ i\c cin
$a\cdot1=a$, ve bir $n$ i\c cin $a\cdot n$ tan\i mlan\i rsa
\begin{equation*}
a\cdot(n+1)=a\cdot n+a.
\end{equation*}
B\"oylece sayma say\i lar\i nda \c carpma i\c slemi tan\i mlan\i r.
\end{definition}

\begin{exercise}
A\c sa\u g\i daki e\c sitlikleri,
her $n$ sayma say\i s\i\ i\c cin kan\i tlay\i n.
Burada $a$ ve $b$, ger\c cel say\i dir ve $m$, sayma say\i s\i d\i r.
\begin{enumerate}[a)]
\item
$1\cdot n=n$.
\item
$(a+b)\cdot n=a\cdot n+b\cdot n$.
\item
$m\cdot n=n\cdot m$.
\item
$a\cdot(m+n)=a\cdot m+a\cdot n$.
\item
$a\cdot(m\cdot n)=(a\cdot m)\cdot n$.
\end{enumerate}
\end{exercise}

\begin{definition}[kuvvetler]
Her $a$ ger\c cel say\i s\i\ i\c cin
$a^1=a$, ve bir $n$ i\c cin $a^n$ tan\i mlan\i rsa
\begin{equation*}
a^{n+1}=a^n\cdot a.
\end{equation*}
\end{definition}

\begin{exercise}
A\c sa\u g\i daki e\c sitlikleri,
her $n$ sayma say\i s\i\ i\c cin kan\i tlay\i n.
Burada $a$, ger\c cel say\i dir ve $m$, sayma say\i s\i d\i r.
\begin{enumerate}[a)]
\item
$a^{m+n}=a^m\cdot a^n$.
\item
$a^{m\cdot n}=(a^m)^n$.
\end{enumerate}
\end{exercise}

\begin{exercise}
Baz\i\ $a$ ve $b$ ger\c cel say\i lar\i\ i\c cin $a<b$ olsun.
Geometriden her $c$ i\c cin $a+c<b+c$.
Her $n$ sayma say\i s\i\ i\c cin
\begin{equation*}
a\cdot n<b\cdot n
\end{equation*}
kan\i tlay\i n.
\end{exercise}

\addsec{\.Iyis\i ralama}

\begin{exercise}
  Sayma say\i lar\i nda
  $x^2=3y^2$ denkleminin
  hi\c c \c c\"oz\"um\"u olmad\i\u g\i n\i\ kan\i tlay\i n.
\end{exercise}

\begin{theorem}
  $1$'den b\"uy\"uk her sayma say\i s\i,
  $p_k$ \c carpanlar\i\ asal olmak \"uzere
  \begin{equation*}
    p_1\cdot p_2\cdots p_n
  \end{equation*}
  \c carp\i m\i\ olarak yaz\i labilir.
\end{theorem}

\begin{exercise}
  Teoremi kan\i tlay\i n.
\end{exercise}

\addsec{B\"olme}

Bu b\"ol\"umde her say\i\ bir sayma say\i s\i d\i r.

\begin{definition}
$ax=b$ denklemi \c c\"oz\"ulebilirse $a\divides b$ yaz\i l\i r.
\end{definition}

\begin{exercise}
A\c sa\u g\i daki \"onermeleri kan\i tlay\i n.
\begin{enumerate}[a)]
\item
$1\divides a$.
\item
$a\divides a$.
\item
$a\divides b$ ve $b\divides c$ ise $a\divides c$.
\end{enumerate}
\end{exercise}

\begin{exercise}
$a\divides b$ ise $a\leq b$ kan\i tlay\i n.
\end{exercise}

\begin{exercise}
$a\divides b$ ve $b\divides a$ ise $a=b$ kan\i tlay\i n.
\end{exercise}

\begin{exercise}
  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}
  \item
    $3\divides a^2$ ve $3\divides a^2+1$.
  \item
    $3\divides a^2$ veya $3\divides a^2+1$.
  \item
    $ab\divides c$ ise $a\divides c$ ve $b\divides c$.
  \item
    $ab\divides c$ ise $a\divides c$ veya $b\divides c$.
  \item
    $a\divides b$ ise $a\ndivides b+1$.
  \item
    $a+1\divides b$ ise $a\ndivides b$.
  \item
    $a\divides bc$ ise $a\divides b$.
  \item
    $a\divides bc$ ise $a\divides b$ veya $a\divides c$.
  \item
    $a\divides b$ ve $c\divides d$ ise $ac\divides bd$.
  \item
    $ac\divides bd$ ise $a\divides b$ ve $c\divides d$.
  \end{enumerate}
\end{exercise}

\end{document}
