Lösung 1.2:2a

Aus Online Mathematik Brückenkurs 1

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A common way to calculate the expression in the exercise is to multiply top and bottom of each fraction by the other fraction's denominator, so as to obtain a common denominator,
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<center> [[Image:1_2_2a.gif]] </center>
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<math>\frac{1}{6}+\frac{1}{10}=\frac{1}{6}\centerdot \frac{10}{10}+\frac{1}{10}\centerdot \frac{6}{6}=\frac{10}{60}+\frac{6}{60}</math>
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However, this gives a common denominator, 60, which is larger than it really needs to be.
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If we instead divide up the fractions' denominators into their smallest possible integral factors,
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<math>\frac{1}{2\centerdot 3}+\frac{1}{2\centerdot 5}</math>
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we see that both denominators contain the factor 2 and it is then unnecessary to include that factor when we multiply the top and bottom of each fraction.
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<math>\frac{1}{6}+\frac{1}{10}=\frac{1}{6}\centerdot \frac{5}{5}+\frac{1}{10}\centerdot \frac{3}{3}=\frac{5}{30}+\frac{3}{30}</math>
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This gives the lowest common denominator (LCD), 30.

Version vom 12:22, 11. Sep. 2008

A common way to calculate the expression in the exercise is to multiply top and bottom of each fraction by the other fraction's denominator, so as to obtain a common denominator,


\displaystyle \frac{1}{6}+\frac{1}{10}=\frac{1}{6}\centerdot \frac{10}{10}+\frac{1}{10}\centerdot \frac{6}{6}=\frac{10}{60}+\frac{6}{60}

However, this gives a common denominator, 60, which is larger than it really needs to be.

If we instead divide up the fractions' denominators into their smallest possible integral factors,


\displaystyle \frac{1}{2\centerdot 3}+\frac{1}{2\centerdot 5}

we see that both denominators contain the factor 2 and it is then unnecessary to include that factor when we multiply the top and bottom of each fraction.


\displaystyle \frac{1}{6}+\frac{1}{10}=\frac{1}{6}\centerdot \frac{5}{5}+\frac{1}{10}\centerdot \frac{3}{3}=\frac{5}{30}+\frac{3}{30}


This gives the lowest common denominator (LCD), 30.

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