Lösung 1.2:4c
Aus Online Mathematik Brückenkurs 1
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Version vom 13:29, 22. Okt. 2008
Method 1
If we calculate the numerator in the main fraction first, we get
| \displaystyle \frac{\,\dfrac{1}{4}-\dfrac{1}{5}\vphantom{\Biggl(}\,}{\dfrac{3}{10}\vphantom{\Biggl(}} =\frac{\,\dfrac{1\cdot 5}{4\cdot 5}-\dfrac{1\cdot 4}{5\cdot 4}\vphantom{\Biggl(}\,}{\dfrac{3}{10}\vphantom{\Biggl(}} =\frac{\,\dfrac{5}{20}-\dfrac{4}{20}\vphantom{\Biggl(}\,}{\dfrac{3}{10}\vphantom{\Biggl(}} = \frac{\,\dfrac{1}{20}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,}\,. |
The double fraction on the right-hand side becomes, after multiplying top and bottom by \displaystyle {10}/{3}\,,
| \displaystyle \frac{\,\dfrac{1}{20}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{1}{20}\cdot \dfrac{10}{3}\vphantom{\Biggl(}\,}{\,\dfrac{\rlap{/}3}{\rlap{\,/}10}\cdot \dfrac{\rlap{\,/}10}{\rlap{/}3}\vphantom{\Biggl(}\,} = \dfrac{1}{20}\cdot \dfrac{10}{3}\,. |
Then, we remove the common factor 10,
| \displaystyle \dfrac{1}{20}\cdot \dfrac{10}{3}=\dfrac{1}{2\cdot{}\rlap{\,/}10}\cdot \dfrac{\rlap{\,/}10}{3}=\dfrac{1}{2\cdot 3}=\dfrac{1}{6}\,. |
Method 2
Another way to calculate the expression is to divide it up into two separate terms
| \displaystyle \frac{\,\dfrac{1}{4}-\dfrac{1}{5}\vphantom{\Biggl(}\,}{\dfrac{3}{10}\vphantom{\Biggl(}} = \frac{\,\dfrac{1}{4}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,}-\frac{\,\dfrac{1}{5}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,}\,. |
We simplify both double fractions on the right-hand side by multiplying top and bottom by 10/3
| \displaystyle \frac{\,\dfrac{1}{4}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,}-\frac{\,\dfrac{1}{5}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{1}{4}\cdot \dfrac{10}{3}\vphantom{\Biggl(}\,}{\,\dfrac{\rlap{/}3}{\rlap{\,/}10}\cdot \dfrac{\rlap{\,/}10}{\rlap{/}3}\vphantom{\Biggl(}\,}-\frac{\,\dfrac{1}{5}\cdot \dfrac{10}{3}\vphantom{\Biggl(}\,}{\,\dfrac{\rlap{/}3}{\rlap{\,/}10}\cdot \dfrac{\rlap{\,/}10}{\rlap{/}3}\vphantom{\Biggl(}\,} = \frac{1}{4}\cdot \frac{10}{3}-\frac{1}{5}\cdot \frac{10}{3}\,. |
Instead of multiplying, respectively, by \displaystyle 4\cdot 3 and \displaystyle 5\cdot 3, we keep the numerators factorized and observe that if we multiply the top and bottom of the first fraction by 5 and the second by 4, we obtain the common denominator
| \displaystyle \frac{10}{4\cdot 3}-\frac{10}{5\cdot 3}=\frac{10\cdot 5}{4\cdot 3\cdot 5}-\frac{10\cdot 4}{5\cdot 3\cdot 4}=\frac{50-40}{3\cdot 4\cdot 5}=\frac{10}{3\cdot 4\cdot 5}\,. |
Because \displaystyle 10=2\cdot 5 and \displaystyle 4=2\cdot 2, we can cancel out the common factors 2 and 5 and obtain the answer
| \displaystyle \frac{10}{3\cdot 4\cdot 5}=\frac{\rlap{/}2{}\cdot{}\rlap{/}5}{3\cdot 2\cdot{}\rlap{/}2\cdot{}\rlap{/}5}=\frac{1}{3\cdot 2}=\frac{1}{6}\,. |
