Lösung 1.2:2a

Aus Online Mathematik Brückenkurs 1

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 1: Zeile 1:
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,
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,
-
{{Displayed math||<math>\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}\,</math>.}}
+
{{Abgesetzte Formel||<math>\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}\,</math>.}}
However, this gives a common denominator, 60, which is larger than it really needs to be.
However, this gives a common denominator, 60, which is larger than it really needs to be.
Zeile 7: Zeile 7:
If we instead divide up the fractions' denominators into their smallest possible integral factors,
If we instead divide up the fractions' denominators into their smallest possible integral factors,
-
{{Displayed math||<math>\frac{1}{2\cdot 3}+\frac{1}{2\cdot 5}\,</math>,}}
+
{{Abgesetzte Formel||<math>\frac{1}{2\cdot 3}+\frac{1}{2\cdot 5}\,</math>,}}
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.
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.
-
{{Displayed math||<math>\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}\,</math>.}}
+
{{Abgesetzte Formel||<math>\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}\,</math>.}}
This gives the lowest common denominator (LCD), 30.
This gives the lowest common denominator (LCD), 30.

Version vom 08:14, 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.

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