Lösung 1.2:2c

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We divide up the two numerators into the smallest possible integer factors,
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<center> [[Image:1_2_2c.gif]] </center>
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<math>\begin{align}
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& 12=2\centerdot 6=2\centerdot 2\centerdot 3 \\
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& 14=2\centerdot 7 \\
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\end{align}</math>
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The expression can thus be written as
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<math>\frac{1}{2\centerdot 2\centerdot 3}-\frac{1}{2\centerdot 7}</math>
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Here, we see that the denominators have a factor
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<math>2</math>
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in common. We multiply the top and bottom of the first fraction by
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<math>7</math>
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and the second by
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<math>2\centerdot 3</math>
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i.e. we leave out the common factor
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<math>2</math>, so that the fractions have the lowest common denominator
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<math>2\centerdot 2\centerdot 3\centerdot 7</math>,
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<math>\frac{1}{12}-\frac{1}{14}=\frac{1}{2\centerdot 2\centerdot 3}-\frac{1}{2\centerdot 7}=\frac{1}{2\centerdot 2\centerdot 3}\centerdot \frac{7}{7}-\frac{1}{2\centerdot 7}\centerdot \frac{2\centerdot 3}{2\centerdot 3}</math>
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The lowest common denominator is
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<math>84</math>.

Version vom 09:39, 22. Sep. 2008

We divide up the two numerators into the smallest possible integer factors,


\displaystyle \begin{align} & 12=2\centerdot 6=2\centerdot 2\centerdot 3 \\ & 14=2\centerdot 7 \\ \end{align}

The expression can thus be written as


\displaystyle \frac{1}{2\centerdot 2\centerdot 3}-\frac{1}{2\centerdot 7}

Here, we see that the denominators have a factor \displaystyle 2 in common. We multiply the top and bottom of the first fraction by \displaystyle 7 and the second by \displaystyle 2\centerdot 3 i.e. we leave out the common factor \displaystyle 2, so that the fractions have the lowest common denominator \displaystyle 2\centerdot 2\centerdot 3\centerdot 7,


\displaystyle \frac{1}{12}-\frac{1}{14}=\frac{1}{2\centerdot 2\centerdot 3}-\frac{1}{2\centerdot 7}=\frac{1}{2\centerdot 2\centerdot 3}\centerdot \frac{7}{7}-\frac{1}{2\centerdot 7}\centerdot \frac{2\centerdot 3}{2\centerdot 3}

The lowest common denominator is \displaystyle 84.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_1.2:2c“