master > master: ÜB5

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RD 2 years ago
parent b1869a1cb0
commit a5e83b3f00
  1. BIN
      docs/loesungen.pdf
  2. 405
      docs/loesungen.tex

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@ -53,6 +53,14 @@
%% |
%% — body/uebung/ueb4.tex;
%% |
%% — body/uebung/ueb5.tex;
%% |
%% — body/ska/ska1.tex;
%% |
%% — body/ska/ska2.tex;
%% |
%% — body/ska/ska3.tex;
%% |
%% — body/ska/ska4.tex;
%% |
%% — body/ska/ska5.tex;
@ -64,6 +72,8 @@
%% — body/quizzes/quiz3.tex;
%% |
%% — body/quizzes/quiz4.tex;
%% |
%% — body/quizzes/quiz5.tex;
%% |
%% — back/index.tex;
%% |
@ -1301,6 +1311,10 @@
\def\ntrlzero{\mathbb{N}_{0}}
\def\reellNonNeg{\reell_{+}}
\def\imageinh{\imath}
\def\ReTeil{\mathop{\mathfrak{R}\text{\upshape e}}}
\def\ImTeil{\mathop{\mathfrak{I}\text{\upshape m}}}
\def\leer{\emptyset}
\def\restr#1{\vert_{#1}}
\def\ohne{\setminus}
@ -3535,11 +3549,249 @@ für $a,b\in\intgr$.
Darum gilt $\Phi(n)$ per Induktion für alle $n\in\ntrl$.
\end{proof}
%% ********************************************************************************
%% FILE: body/uebung/ueb5.tex
%% ********************************************************************************
\setcounternach{chapter}{5}
\chapter[Woche 5]{Woche 5}
\label{ueb:5}
\textbf{ACHTUNG.}
Diese Lösungen dienen \emph{nicht} als Musterlösungen sondern eher als Referenz.
Hier wird eingehender gearbeitet, als generell verlangt wird.
Das Hauptziel hier ist, eine Variant anzubieten, gegen die man seine Versuche vergleichen kann.
%% AUFGABE 5-1
\let\altsectionname\sectionname
\def\sectionname{Aufgabe}
\section[Aufgabe 1]{}
\label{ueb:5:ex:1}
\let\sectionname\altsectionname
Fixiere eine natürliche Zahl $n\in\ntrlzero$.
Sei $a_{i}\in\{0,1,\ldots,10-1\}$ die eindeutige Zahlen,
so dass
\begin{mathe}[mc]{rcl}
n &= &\sum_{i\in\ntrlzero}a_{i}10^{i}\\
\end{mathe}
gilt.
\begin{enumerate}{\bfseries (a)}
%% AUFGABE 5-1a
\item
\begin{claim*}
$3\divides n$ $\Leftrightarrow$ $3\mid\sum_{i\in\ntrlzero}a_{i}$.
\end{claim*}
\begin{proof}
Beachte zunächst, dass $10\equiv 1\mod 3$.
Also gilt modulo $3$
\begin{mathe}[mc]{rcccl}
n &\equiv &\sum_{i\in\ntrlzero}a_{i}1^{i} &\equiv &\sum_{i\in\ntrlzero}a_{i}.\\
\end{mathe}
Folglich gilt
\begin{mathe}[mc]{rcccccl}
3\mid n
&\Longleftrightarrow
&n\equiv 0\mod 3
&\Longleftrightarrow
&\sum_{i\in\ntrlzero}a_{i}\equiv 0\mod 3
&\Longleftrightarrow
&3\mid\sum_{i\in\ntrlzero}a_{i}
\end{mathe}
wie behauptet.
\end{proof}
%% AUFGABE 5-1b
\item
\begin{claim*}
$11\divides n$ $\Leftrightarrow$ $1\mid\sum_{i\in\ntrlzero}(-1)^{i}a_{i}$.
\end{claim*}
\begin{proof}
Beachte zunächst, dass $10=-1\mod 11$.
Also gilt modulo $11$
\begin{mathe}[mc]{rcccl}
n &\equiv &\sum_{i\in\ntrlzero}a_{i}(-1)^{i}.\\
\end{mathe}
Folglich gilt
\begin{mathe}[mc]{rcccccl}
11\mid n
&\Longleftrightarrow
&n\equiv 0\mod 11
&\Longleftrightarrow
&\sum_{i\in\ntrlzero}a_{i}\equiv 0\mod 11
&\Longleftrightarrow
&11\mid\sum_{i\in\ntrlzero}(-1)^{i}a_{i}
\end{mathe}
wie behauptet.
\end{proof}
\end{enumerate}
%% AUFGABE 5-2
\let\altsectionname\sectionname
\def\sectionname{Aufgabe}
\section[Aufgabe 2]{}
\label{ueb:5:ex:2}
\let\sectionname\altsectionname
\begin{enumerate}{\bfseries (a)}
%% AUFGABE 5-2a
\item
Seien $a=142$ und $b=84$.
Wir berechnen $\ggT(a,b)$ mittels des Euklidischen Algorithmus
(siehe \cite[Satz 3.4.7]{sinn2020}).
\begin{longtable}[mc]{|c|c|}
\hline
\hline
Restberechnung (symbolisch) &Restberechnung (Werte)\\
\hline
\endhead
$a = b\cdot q_{1} + r_{1}$ &$142 = 84\cdot 1 + 58$\\
$b = r_{1}\cdot q_{2} + r_{2}$ &$84 = 58\cdot 1 + 26$\\
$r_{1} = r_{2}\cdot q_{3} + r_{3}$ &$58 = 26\cdot 2 + 6$\\
$r_{2} = r_{3}\cdot q_{4} + r_{4}$ &$26 = 6\cdot 4 + \boxed{\mathbf{2}}$\\
$r_{3} = r_{4}\cdot q_{5} + r_{5}$ &$6 = 2\cdot 3 + 0$\\
\hline
\hline
\end{longtable}
Darum gilt $\ggT(a,b)=r_{2}=2$.
%% AUFGABE 5-2b
\item
\begin{claim}
\makelabel{claim:main:ueb:5:ex:2b}
Seien $a,b,c\in\intgr$ mit $a,b\neq 0$.
Die folgenden Aussagen sind äquivalent:
\begin{kompaktenum}{\bfseries (i)}[\rtab][\rtab]
\item\punktlabel{1}
$\exists{x,y\in\intgr:~}ax+by=c$
\item\punktlabel{2}
$\ggT(a,b)\divides c$
\end{kompaktenum}
\end{claim}
\begin{proof}
Fixiere zunächst $d:=\ggT(a,b)$.
Da $a,b\in\intgr\ohne\{0\}$, ist $d\in\ntrl$ eine wohldefinierte positive Zahl.
\hinRichtung{1}{2}
Angenommen, $ax+by=c$ für ein $x,y\in\intgr$.\\
Da $x,y\in\intgr$,
erhalten wir $c=ax+by\equiv 0x+0z\equiv 0$ modulo $d$.\\
Also $\ggT(a,b)=d\divides c$.
\hinRichtung{2}{1}
Angenommen, $\ggT(a,b)\divides c$.\\
Dann existiert ein $k\in\intgr$, so dass $c=k\cdot\ggT(a,b)$.\\
Laut des Lemmas von B\'ezout (siehe \cite[Lemma 3.4.8]{sinn2020})
existiere nun $u,v\in\intgr$, so dass $\ggT(a,b)=au+bv$.\\
Daraus folgt ${c=k\cdot\ggT(a,b)=aku+bkv}$.\\
Da $ku,kv\in\intgr$, haben wir \punktcref{1} bewiesen.
\end{proof}
\end{enumerate}
%% AUFGABE 5-3
\let\altsectionname\sectionname
\def\sectionname{Aufgabe}
\section[Aufgabe 3]{}
\label{ueb:5:ex:3}
\let\sectionname\altsectionname
\begin{enumerate}{\bfseries (a)}
%% AUFGABE 5-3a
\item
Sei $H:=\intgr/2\intgr$ die (abelsche) Gruppe von Restklassen modulo $2$ unter Addition.\\
Sei $G:=H\times H$ mit Neutralelement $e=([0],[0])$ und versehen mit der Produktstruktur.\\
Als Produkt von (abelschen) Gruppen ist $G$ automatisch eine (abelsche) Gruppe.
Und offensichtlich hat $G$ genau $|G|=|H\times H|=|H|\cdot|H|=2\cdot 2=4$ Elemente.\\
Es bleibt \textbf{zu zeigen}, dass $\forall{a\in G:~}a\ast a=e$.\\
Sei also $a=([k],[j])\in H\times H=G$ ein beliebiges Element.
Es gilt
\begin{mathe}[mc]{rcl}
a\ast a
&= &([k],[j])\ast([k],[j])\\
&= &([k]+[k],[j]+[j])\\
&= &([k+k],[j+j])\\
&= &([2k],[2j])
=([0],[0])
=e,
\end{mathe}
da $2\equiv 0\mod 2$.\\
Also ist unsere Konstruktion von $G$ ein passendes Beispiel.
%% AUFGABE 5-3b
\item
\begin{claim*}
Sei $(G,\ast,e)$ eine Gruppe.
Angenommen, $\forall{a\in G:~}a\ast a=e$.
Dann ist $G$ kommutativ.
\end{claim*}
\begin{proof}
Beachte zunächst, dass wegen Eindeutigkeit des Inverses
die Annahme zu
\begin{mathe}[mc]{c}
\eqtag[eq:1:ueb:5:ex:3]
\forall{a\in G:~}a^{-1}=a
\end{mathe}
äquivalent ist.\\
\textbf{Zu zeigen:} Für alle $a,b\in G$ gilt $a\ast b=b\ast a$.\\
Seien also $a,b\in G$ beliebige Elemente.
Es gilt
\begin{mathe}[mc]{rcccccl}
a\ast b
&\eqcrefoverset{eq:1:ueb:5:ex:3}{=}
&a^{-1}\ast b^{-1}
&= &(b\ast a)^{-1}
&\eqcrefoverset{eq:1:ueb:5:ex:3}{=}
&b\ast a.\\
\end{mathe}
Also ist $G$ eine kommutative Gruppe.
\end{proof}
\end{enumerate}
\setcounternach{part}{2}
\part{Selbstkontrollenaufgaben}
\def\chaptername{SKA Blatt}
%% ********************************************************************************
%% FILE: body/ska/ska1.tex
%% ********************************************************************************
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%% FILE: body/ska/ska2.tex
%% ********************************************************************************
%% ********************************************************************************
%% FILE: body/ska/ska3.tex
%% ********************************************************************************
%% ********************************************************************************
%% FILE: body/ska/ska4.tex
%% ********************************************************************************
@ -5040,14 +5292,14 @@ Um diese Urteil also leichter treffen zu können ersetzen wir die Elemente durch
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\hraum
@ -5057,54 +5309,54 @@ Um diese Urteil also leichter treffen zu können ersetzen wir die Elemente durch
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\node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,200;green,200;blue,200}, draw] (gh_1_0) at (1*\habst, -2*\vabst) {};
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\node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,64;blue,64}, draw] (gh_2_0) at (1*\habst, -3*\vabst) {};
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\node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,200;blue,64}, draw] (gh_4_1) at (2*\habst, -5*\vabst) {};
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\node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,200;green,64;blue,200}, draw] (gh_4_3) at (4*\habst, -5*\vabst) {};
\node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,200;blue,200}, draw] (gh_4_4) at (5*\habst, -5*\vabst) {};
\node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,64;blue,64}, draw] (gh_4_5) at (6*\habst, -5*\vabst) {};
\node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,200;blue,64}, draw] (gh_5_0) at (1*\habst, -6*\vabst) {};
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\node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,200;blue,200}, draw] (gh_5_2) at (3*\habst, -6*\vabst) {};
\node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,64;green,64;blue,64}, draw] (gh_5_3) at (4*\habst, -6*\vabst) {};
\node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,200;green,200;blue,200}, draw] (gh_5_4) at (5*\habst, -6*\vabst) {};
\node[rectangle, line width=0.5pt, minimum size=0.9*\rad, fill={rgb,255:red,200;green,64;blue,200}, draw] (gh_5_5) at (6*\habst, -6*\vabst) {};
\end{tikzpicture}
}
\hraum
@ -5386,6 +5638,47 @@ Wir betrachten die Komposition ${g\circ f:X\to Z}$
\end{proof}
\end{enumerate}
%% ********************************************************************************
%% FILE: body/quizzes/quiz5.tex
%% ********************************************************************************
\setcounternach{chapter}{5}
\chapter[Woche 5]{Woche 5}
\label{quiz:5}
\begin{claim*}
Seien $n\in\ntrlpos$ und $p\in\mathbb{P}$ mit $n<p\leq 2n$.
Dann gilt $p\divides\begin{svector}2n\\n\\\end{svector}$.
\end{claim*}
\begin{proof}
Aus $n<p\leq 2n$, d.\,h. $p\in\{n+1,n+2,\ldots,2n\}$, folgt (i)~$p\divides\prod_{i=n+1}^{2n}i$.\\
Es gilt nun
\begin{mathe}[mc]{rcccccl}
\eqtag[eq:1:quiz:5:ex:1]
\prod_{i=n+1}^{2n}i
&= &\dfrac{\prod_{i=1}^{2n}i}{n!}
&= &n!\dfrac{(2n)!}{n!(2n-n)!}
&= &n!\begin{vector}2n\\n\\\end{vector}.\\
\end{mathe}
Aus (i) und \eqcref{eq:1:quiz:5:ex:1} folgt also
(ii)~$p\divides \begin{svector}2n\\n\\\end{svector}\cdot n!$.\\
Beachte, dass $p$ eine Primzahl ist und ${n!,\begin{svector}2n\\n\\\end{svector}\in\intgr}$.\\
Aus (ii) und \cite[Satz 3.4.14]{sinn2020} folgt also
$p\divides\begin{svector}2n\\n\\\end{svector}$ oder $p\divides n!$.\\
\fbox{Angenommen, $p\ndivides\begin{svector}2n\\n\\\end{svector}$.}\\
Dann muss laut des o.\,s. Arguments ${p\divides n!(=\prod_{i=1}^{n}i)}$ gelten.\\
Eine weitere Anwendung von \cite[Satz 3.4.14]{sinn2020} liefert,
dass ${p\divides i_{0}}$ für ein $i_{0}\in\{1,2,\ldots,n\}$.\\
Aber dann gilt $1\leq p\leq i_{0}\leq n$.
Das widerspricht der Voraussetzung, dass $n<p$.\\
Darum stimmt die Annahme nicht.
Das heißt, $p\divides\begin{svector}2n\\n\\\end{svector}$.
\end{proof}
%% ********************************************************************************
%% FILE: back/index.tex
%% ********************************************************************************

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