\input cyracc.def
\font\tencyr=wncyr10
\font\tencyrb=wncyb10
\font\tencyri=wncyi10
\font\eightcyr=wncyr8
\input amstex
\def\cyr#1{{\tencyr\cyracc#1}}
\def\cyrb#1{{\tencyrb\cyracc#1}}
\def\cyri#1{{\tencyri\cyracc#1}}
\def\scyr#1{{\eightcyr\cyracc#1}}
\def\N{\bold N}
\def\Q{\bold Q}
\def\R{\bold R}
\def\sledi{\;\Longrightarrow\;}
\def\akko{\;\Longleftrightarrow\;}
\def\liml{\lim\limits}
\def\tezi{\rightarrow}

{\cyrb{Stav:}}
\cyr{Neka su} $r_1, r_2\in\Q$. \cyr{Tada vazhi}:
\smallskip

(1$^\circ$) $a^{r_1}a^{r_2}=a^{r_1+r_2}$,
$(a^{r_1})^{r_2}=a^{r_1r_2}$;
\smallskip

(2$^\circ$) $r_1<r_2\sledi a^{r_1}<a^{r_2}$;
\smallskip

(3$^\circ$) $\liml_{\Q\ni r\tezi r_0\in\Q}a^r=a^{r_0}$.
\bigskip

\cyri{Dokaz\/.} (1$^\circ$)

\cyr{Neka su} $n_1,
n_2\in\N$. \cyr{Tada je}:
$$a^{n_1}a^{n_2}=(\underbrace{a\cdot a\dotsb a}_{n_1\text{
\scyr{puta}}})\cdot(\underbrace{a\cdot a\dotsb a}_{n_2\text{
\scyr{puta}}})=(\underbrace{a\cdot a\dotsc a}_{n_1+n_2\text{
\scyr{puta}}})=a^{n_1+n_2}.$$
\cyr{Ovo tvrdjenje se mozhe dokazati indukcijom.}

\cyr{Neka su} $n_1, n_2\in\N$. \cyr{Tada je}:
$$(a^{n_1})^{n_2}=(\underbrace{a\cdot a\dotsb a}_{n_1\text{
\scyr{puta}}})^{n_2}=\bigl(\underbrace{(\underbrace{a\cdot a\dotsb
a}_{n_1\text{ \scyr{puta}}})\dotsb(\underbrace{a\cdot a\dotsb a}_{n_1\text{
\scyr{puta}}})}_{n_2\text{ \scyr{puta}}}\bigr)=(\underbrace{a\cdot a\dotsb
a}_{n_1\cdot n_2\text{ \scyr{puta}}})=a^{n_1n_2}.$$
\cyr{I ovo tvrdjenje se mozhe dokazati indukcijom.}

\cyr{Neka je sad} $\frac{1}{n_1}, \frac{1}{n_2}\in\Q$. \cyr{Tada je,
prema teoremi o postojanju korena}, (\cyr{T}2.4.3 \cyr{iz
Kadelburga}), $(\exists b_1\in\R)\ b_1=a^{\frac{1}{n_1}}\akko b_1^{n_1}=a$
\cyr{i} $(\exists b_2\in\R)\ b_2=a^{\frac{1}{n_2}}\akko b_2^{n_2}=a$,
\cyr{pa je}
$$a^{\frac{1}{n_1}}a^{\frac{1}{n_2}}=b_1b_2.$$

\cyrb{Shta dalje??} \cyr{Kako dokazati} $a^{\frac{1}%
{n_1}}a^{\frac{1}{n_2}}=a^{\frac{1}{n_1}+\frac{1}{n_2}}$ \cyr{i}
$(a^{\frac{1}{n_1}})^{\frac{1}{n_2}}=a^{\frac{1}{n_1}\cdot\frac{1}
{n_2}}$??

\bye