Lösung 1.2:2c
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Version vom 13:27, 22. Okt. 2008
We divide up the two numerators into the smallest possible integer factors,
| \displaystyle \begin{align}
12 &= 2\cdot 6 = 2\cdot 2\cdot 3\,,\\ 14 &= 2\cdot 7\,\textrm{.} \\ \end{align} |
The expression can thus be written as
|
\displaystyle \frac{1}{2\cdot 2\cdot 3}-\frac{1}{2\cdot 7}\,. |
Here, we see that the denominators have a factor 2 in common. We multiply the top and bottom of the first fraction by 7 and the second by \displaystyle 2\cdot 3 i.e. we leave out the common factor 2, so that the fractions have the lowest common denominator \displaystyle 2\cdot 2\cdot 3\cdot 7,
| \displaystyle \begin{align}
\frac{1}{12}-\frac{1}{14} &= \frac{1}{2\cdot 2\cdot 3}-\frac{1}{2\cdot 7}\\[5pt] &= \frac{1}{2\cdot 2\cdot 3}\cdot \frac{7}{7}-\frac{1}{2\cdot 7}\cdot \frac{2\cdot 3}{2\cdot 3}\\[5pt] &= \frac{7}{2\cdot 2\cdot 3\cdot 7} - \frac{2\cdot 3}{2\cdot 2\cdot 3\cdot 7}\\[5pt] &= \frac{7}{84} - \frac{6}{84}\,\textrm{.} \end{align} |
The lowest common denominator is 84.
