Lösung 1.2:2a
Aus Online Mathematik Brückenkurs 1
(Unterschied zwischen Versionen)
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)) |
K (hat „Solution 1.2:2a“ nach „Lösung 1.2:2a“ verschoben: Robot: moved page) |
Version vom 13:26, 22. Okt. 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}\cdot \frac{10}{10}+\frac{1}{10}\cdot \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\cdot 3}+\frac{1}{2\cdot 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}\cdot \frac{5}{5}+\frac{1}{10}\cdot \frac{3}{3}=\frac{5}{30}+\frac{3}{30}\,. |
This gives the lowest common denominator (LCD), 30.
