Solution 3.1:5a
From Förberedande kurs i matematik 1
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| - | {{  | + | If we multiply the top and bottom of the fraction by   | 
| - | <  | + | <math>\sqrt{12}</math>, the new denominator will be   | 
| - | {{  | + | <math>\sqrt{12}\centerdot \sqrt{12}=12</math>  | 
| + | and we will get rid of the root sign in the denominator:  | ||
| + | |||
| + | |||
| + | <math>\frac{2}{\sqrt{12}}=\frac{2}{\sqrt{12}}\centerdot \frac{\sqrt{12}}{\sqrt{12}}=\frac{2\sqrt{12}}{12}=\frac{2\sqrt{12}}{2\centerdot 6}=\frac{\sqrt{12}}{6}</math>  | ||
| + | |||
| + | |||
| + | This expression can be simplified even further if we write   | ||
| + | <math>\text{12}=\text{2}\centerdot \text{6}=\text{2}\centerdot \text{2}\centerdot \text{3}=\text{2}^{\text{2}}\centerdot \text{3 }</math>  | ||
| + | and take   | ||
| + | <math>\text{2}^{\text{2}}</math>  | ||
| + | out from under the root, We get  | ||
| + | |||
| + | |||
| + | <math>\frac{\sqrt{12}}{6}=\frac{2\sqrt{3}}{6}=\frac{2\sqrt{3}}{2\centerdot 3}=\frac{\sqrt{3}}{3}.</math>  | ||
Revision as of 14:28, 22 September 2008
If we multiply the top and bottom of the fraction by \displaystyle \sqrt{12}, the new denominator will be \displaystyle \sqrt{12}\centerdot \sqrt{12}=12 and we will get rid of the root sign in the denominator:
\displaystyle \frac{2}{\sqrt{12}}=\frac{2}{\sqrt{12}}\centerdot \frac{\sqrt{12}}{\sqrt{12}}=\frac{2\sqrt{12}}{12}=\frac{2\sqrt{12}}{2\centerdot 6}=\frac{\sqrt{12}}{6}
This expression can be simplified even further if we write 
\displaystyle \text{12}=\text{2}\centerdot \text{6}=\text{2}\centerdot \text{2}\centerdot \text{3}=\text{2}^{\text{2}}\centerdot \text{3 }
and take 
\displaystyle \text{2}^{\text{2}}
out from under the root, We get
\displaystyle \frac{\sqrt{12}}{6}=\frac{2\sqrt{3}}{6}=\frac{2\sqrt{3}}{2\centerdot 3}=\frac{\sqrt{3}}{3}.
