// Numbas version: finer_feedback_settings
{"name": "Dezimalzahlen in Bruchzahlen umwandeln", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Dezimalzahlen in Bruchzahlen umwandeln", "tags": [], "metadata": {"description": "
Identify well-known fractional equivalents of decimals. Convert obscure decimals and recurring decimals into fractions.
\nhttps://numbas.mathcentre.ac.uk/question/22784/decimals-to-fractions/ by Lauren Richards
\nTranslated to German.
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Brüche lassen sich als Dezimalzahlen schreiben (und periodische Dezimalzahlen können als Bruchzahlen geschrieben werden). In den folgenden Fragen können üben, Dezimalzahlen in Brüche umzuwandeln.
\nDer Überstrich zeigt an, dass sich die Ziffernfolge darunter periodisch wiederholt: $0,\\overline{3} = 0,3333\\dots$.
", "advice": "a)
\nUm eine endliche Dezimalzahl in einen Bruch umzuwandeln, \"erweitern\" wir mit einer geeigneten Zehnerpotenz und kürzen dann den resultierenden Bruch.
\ni)
\n$\\var{a}$
\n\\[
\\frac{\\var{a}}{1}\\times\\frac{10}{10}=\\frac{\\var{10a}}{10}\\text{.}
\\]
\nii)
\n$\\var{b}$
\n\\[
\\frac{\\var{b}}{1}\\times\\frac{100}{100}=\\frac{\\var{100b}}{100}=\\simplify{{100b}/{100}}\\text{.}
\\]
\niii)
\n\n$\\var{d}$
\n\\[
\\frac{\\var{d}}{1}\\times\\frac{10}{10}=\\frac{\\var{10d}}{10}=\\simplify{{10d}/{10}}\\text{.}
\\]
\niv)
\n\n$0.\\bar{\\var{c}}$
\nUm eine periodische Dezimalzahl in eine Bruchzahl umzuschreiben, machen wir den folgenden Ansatz:
\n\\[
x=0.\\bar{\\var{c}}\\text{.}
\\]
\nWir multiplizieren beide Seiten mit $10$ und erhalten
\n\\[
10x=\\var{c}.\\bar{\\var{c}}\\text{.}
\\]
\nJetzt ziehen wir die erste Gleichung von der zweiten Gleichung ab:
\n\\[
\\begin{align}
&&\\var{c}.\\bar{\\var{c}}&={10}x\\\\
-&&{0.\\bar{\\var{c}}}&=x\\\\
&&\\overline{\\qquad} & \\overline{\\qquad}\\\\
&&{\\var{c}}&=9x\\\\
\\\\
&&\\frac{\\var{c}}{9}&=x
\\end{align}
\\]
\nZum Schluss kürzen wir $\\displaystyle\\frac{\\var{c}}{9}$ mit $3$ und bekommen $\\simplify{{c}/{9}}$. Also gilt $0.\\bar{\\var{c}}=\\simplify{{c}/{9}}$.
\n\nb)
\n$\\displaystyle\\var{f}$
\n\\[
\\var{f}\\times\\frac{\\var{f1000}}{\\var{f1000}}=\\frac{\\var{f2}}{\\var{f1000}}\\text{.}
\\]
\nWir berechnen den größten gemeinsamen Teiler von Zähler und Nenner; dies ist $\\var{mygcd}$.
\nFolglich kann der Bruch nicht weiter gekürzt werden, und die Antwort ist
\nIndem wir damit kürzen, erhalten wir
\n\\[\\frac{\\var{f3}}{\\var{f4}}.\\]
\nc)
\n$\\var{h}.\\overline{\\var{j}\\var{k}}.$
\nWir gehen ähnlich vor wie beim letzten Teil von a):
\n$x=\\var{h}.\\overline{\\var{j}\\var{k}}.$
\n
Um in diesem Fall zum Ziel zu kommen, multiplizieren wir mit $100 = 10^2$, weil der periodische Teil die Länge $2$ hat.
\n$100x=\\var{h}\\var{j}\\var{k}.\\overline{\\var{j}\\var{k}}.$
\n\nWir ziehen die erste Gleichung von der zweiten ab und lösen nach $x$ auf.
\n\\[
\\begin{align}
&&\\var{h}\\var{j}\\var{k}.\\overline{\\var{j}\\var{k}}&=100x\\\\
-&&\\var{h}.\\overline{\\var{j}\\var{k}}&=x\\\\
&&\\overline{\\qquad} & \\overline{\\qquad}
\\\\
&&{{\\var{h}}\\var{j}\\var{k-h}}&=99x\\\\
\\\\
&&\\frac{\\var{numerator}}{\\var{g}}&=x\\text{.}\\\\
\\end{align}
\\]
\nZum Schluss prüfen wir noch, ob der Bruch gekürzt werden kann. Der ggT von Zähler und Nenner ist $\\var{gcd1 }$.
\nAlso lässt sich der Bruch nicht weiter kürzen und wir erhalten
\nIndem wir damit kürzen, erhalten wir $\\displaystyle\\simplify{{{numerator}}/{g}}$ und somit
\n\\[
\\begin{align}
\\var{h}.\\overline{\\var{j}\\var{k}}=\\simplify{{{numerator}}/{g}}\\text{ als Bruchzahl.}\\\\
\\end{align}
\\]
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\ni)
\n$\\var{a}=$ [[0]] [[1]]
\nii)
\n$\\var{b}=$ [[2]] [[3]]
\niii)
\n$\\var{d}=$ [[6]] [[7]]
\niv)
\n$0.\\bar{\\var{c}}=$ [[4]] [[5]]
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\n$\\displaystyle\\var{f} = $ [[0]] [[1]]
\n
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\n$\\var{h},\\overline{\\var{j}\\var{k}} = $ [[0]] [[1]]
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