// Numbas version: exam_results_page_options {"name": "Weitere \u00dcbungen (Lektion 2)", "metadata": {"description": "

Übungen zu Mengenlehre und Beweisprinzipien

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Introductory exercise about set equality

", "licence": "Creative Commons Attribution-ShareAlike 4.0 International"}, "statement": "

Nehmen wir an, wir haben die drei Elemente $1, 1$ und $2$. Wenn wir diese als ungeordnete Zusammenstellung einzelner unterscheidbarer Objekte auffassen dann nennen wir dies die Menge $\\left\\{1,1,2\\right\\}$. Weil Mengen ungeordnet sind, ist das dieselbe Menge wie $\\left\\{2,1,1\\right\\}$ und weil wir nur unterscheidbare Objekte zusammenstellen auch dieselbe Menge wie $\\left\\{1,2\\right\\}$.

\n

Nehmen wir als Beispiel die folgenden Mengen her: $A=\\left\\{\\var{A[0]},\\var{A[1]},\\var{A[2]},\\var{A[3]}\\right\\}, B=\\left\\{\\var{B[0]},\\var{B[1]},\\var{B[2]},\\var{B[3]},\\var{B[4]}\\right\\}$ und $C=\\left\\{\\var{C[0]},\\var{C[1]},\\var{C[2]},\\var{C[3]},\\var{C[4]},\\var{C[5]}\\right\\}$.

", "advice": "

Sie können überprüfen, ob $M \\subset N$ ist, in dem Sie nach und nach für jedes Element von $M$ prüfen, ob es auch in $N$ enthalten ist. Es sind 6 Aussagen zu prüfen, also müssen Sie das Ganze sechs mal so machen.

\n
\n

Der zweite Teil baut auf dem ersten auf. Sie müssen sich Ihre Antworten ansehen und nachschauen, welche Mengen einander als Teilmengen enthalten. Das ist sehr ähnlich wie bei der 'kleiner oder gleich'-Relation insofern Sie, wenn Sie zwei Zahlen $x$ und $y$ haben bei denen $x \\leq y$ und $y\\leq x$ gilt, dann eben auch $x=y$ gelten muss.

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Wenn jedes Elememt der Menge $M$ auch ein Element der Menge $N$ ist, dann nennen wir $M$ eine Teilmenge von $N$: $M \\subseteq N$. Welche der oben angegebenen Mengen sind Teilmengen voneinander?

\n

Kreuzen Sie alle korrekten Antworten an!

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$A \\subseteq B$

", "

$B \\subseteq A$

", "

$C \\subseteq B$

", "

$B \\subseteq C$

", "

$C \\subseteq A$

", "

$A \\subseteq C$

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Mengen sind gleich, wenn Sie gegenseitige Teilmengen voneinander sind. Welche der Mengen sind gleich?

\n

Kreuzen Sie die korrekte Antwort an!

\n

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$A=B$

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$B=C$

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$C=A$

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Simple exercises introducing the fundamental set operations, and NUMBAS syntax for sets.

", "licence": "Creative Commons Attribution-ShareAlike 4.0 International"}, "statement": "

Betrachten Sie die Mengen $A = \\var{A}$ und $B = \\var{B}$. Bestimmen Sie die Vereinigungsmenge, die Schnittmenge und die Differenzmengen. 

\n

Benutzen Sie dabei bitte die Schreibweise z.B. set(1,2,3) für die Menge $\\left\\{1,2,3\\right\\}$.

", "advice": "

Die Vereinigungsmenge ist die Menge aller Elemente aus einer der beiden Mengen.

\n

$ A \\cup B = \\left\\{ x\\, |\\, x \\in A \\text{ or } x \\in B\\right\\}$

\n
\n

Die Schnittmenge ist die Menge aller Elemente, die in beiden Mangen enthalten sind.

\n

$ A \\cap B = \\left\\{ x\\, |\\, x \\in A \\text{ and } x \\in B\\right\\}$

\n
\n

Die Differenzmenge $A\\setminus B$ ist die Menge der Elemente aus $A$, welche nicht in$B$ enthalten sind:

\n

$ A - B = \\left\\{ x \\in A |\\, x \\notin B\\right\\}$

\n
\n

Analog, ist $B\\setminus A$ is die Menge der Elemente aus $B$, die nicht in $A$ enthalten sind:

\n

$ B - A = \\left\\{ x \\in B |\\, x \\notin A\\right\\}$

\n
\n

Die symmetrische Differenz $A \\Delta B$ is die Vereinigung der beiden Differenzmengen, kann aber auch als Vereinigungsmenge minus Schnittmenge dargestellt werden:

\n

$ A \\Delta B = (A \\cup B) - (A \\cap B)$.

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Was ist die Vereinigungsmenge $A \\cup B$?

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Was ist die Schnittmenge $A \\cap B$?

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Was ist $A\\setminus B$, die Menge der Elemente von $A$ die nicht in $B$ enthalten ist?

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Was ist $B\\setminus A$, die Menge der Elemente von $B$ die nicht in $A$ enthalten ist?

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What ist die Vereinigung der Differenzmengen $(A\\setminus B)\\cup(B\\setminus A)$? Diese Menge heißt auch symmetrische Differenz $A\\Delta B$.

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Ordnen Sie zu: Welche Menge gehört zu welchem Venn-Diagramm, wenn der linke Kreis die Menge $A$ und der rechte Kreis die Menge $B$ darstellt.

\n

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$A\\cup B$

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$A \\cap B$

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$A \\Delta B$

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$A \\setminus B$

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", "

", "

", "

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Einfache Wahrheitstabelle mit zwei Aussagen

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Für die Aussagen $A$ und $B$ sei die folgende kombinierte Aussage gegeben: {full_tex}.

\n

Vervollständigen Sie die zugehörige Wahrheitstabelle!

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Sie können oben die korrekten Eintragungen sehen.

\n

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$A$$B${column1}{column3}{full_tex}
ww[[0]][[4]][[8]]
wf[[1]][[5]][[9]]
fw[[2]][[6]][[10]]
ff[[3]][[7]][[11]]
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Ein einfacher, geführter Induktionsbeweis

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Zu zeigen: Für alle $n\\in\\mathbb{N}$ gilt: $\\,\\var{a}^{n+1} + \\var{a + 1}^{2n−1}$ ist durch $\\var{a^2 + a + 1}$ teilbar.

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Induktionsanfang: Wir setzen 1 für $n$ ein: $A(1):\\ \\,\\var{a}^{1+1} + \\var{a + 1}^{2\\cdot 1−1}=\\var{a}^{2} + \\var{a + 1}=\\var{a1}$, was offensichtlich durch $\\var{a1}$ teilbar ist.

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Induktionsbehauptung: Wir setzen zunächst $(n+1)$ für $n$ ein: $A(n+1):\\ \\,\\var{a}^{(n+1)+1} + \\var{a + 1}^{2\\cdot(n+1)−1}=\\var{a}^{n+2} + \\var{a + 1}^{2n+1}$

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Die Umformungsschritte nutzen dann die Potenzrechenregel $a^{(b+c)}=a^b+a^c$ und das Distibutivgesetz $(a+b)\\cdot c=a\\cdot c+b\\cdot c$ 

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Womit sich ergibt:

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$A(n+1):\\ \\,\\var{a}^{n+2} +\\var{a + 1}^{2n+1}=\\var{k1}\\cdot\\,\\var{a}^{n+1} +\\var{k2}\\cdot\\,\\var{a + 1}^{2n−1}$

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$=\\var{k1}\\cdot\\,\\var{a}^{n+1} +\\var{k1}\\cdot\\,\\var{a + 1}^{2n−1}+\\var{k2-k1}\\cdot\\,\\var{a + 1}^{2n−1}=\\var{k3}\\cdot\\left(\\var{a}^{n+1} + \\var{a + 1}^{2n−1}\\right)+\\var{k4}\\cdot\\,\\var{a + 1}^{2n−1}$

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Der erste Summand ist gemäß Induktionsvoraussetzung durch $\\var{a1}$ teilbar, der zweite Summand, weil er ein Vielfaches von $\\var{a1}$ ist. Die Summe ist dann auch durch $\\var{a1}$ teilbar.

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Induktionsanfang $A(1)$:

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Durch Einsetzen ergibt sich: [[0]], was ein ganzzahliges Vielfaches von $\\var{a^2 + a + 1}$ ist.

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Induktionsschluss $A(n)\\Rightarrow A(n+1)$:

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Es ist zu zeigen, dass unter der Induktionsvoraussetzung

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$A(n):\\ \\,\\var{a}^{n+1} + \\var{a + 1}^{2n−1}$ ist durch $\\var{a^2 + a + 1}$ teilbar (für beliebiges aber festes $n$)

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die Induktionsbehauptung

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$A(n+1):\\,$ [[0]] ist durch $\\var{a^2 + a + 1}$ teilbar 

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richtig ist.

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Wir formen jetzt den Term in der Aussage $A(n+1)$ so um, dass wir den Term aus der Aussage $A(n)$ wieder erkennen, dazu heben wir wie folgt heraus:

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$A(n+1):\\,$ [[0]]$=$[[1]]$\\cdot\\,\\var{a}^{n+1} +$[[2]]$\\cdot\\,\\var{a + 1}^{2n−1}$

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als nächstes müssen wir den gesamten Term $\\var{a}^{n+1} + \\var{a + 1}^{2n−1}$ herausheben:

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$\\qquad\\qquad=$[[3]]$\\cdot\\left(\\var{a}^{n+1} + \\var{a + 1}^{2n−1}\\right)+$[[4]]$\\cdot\\,\\var{a + 1}^{2n−1}$

\n

Dieser Term ist durch $\\var{a^2 + a + 1}$ teilbar, weil es der erste Summand wegen der Induktionsvoraussetzung ist, warum es auch der zweite ist, erkennen Sie, wenn Sie korrekt herausgehoben haben. Sind zwei Summanden durch eine Zahl teilbar, dann ist es immer auch deren Summe.

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