Lösung 3.1:2b
Aus Online Mathematik Brückenkurs 1
(Unterschied zwischen Versionen)
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<math>9 = 3\cdot 3 = 3^{2}</math>, hence | <math>9 = 3\cdot 3 = 3^{2}</math>, hence | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\sqrt{(-3)^{2}} = \sqrt{9} = 9^{1/2} = \bigl(3^{2}\bigr)^{1/2} = 3^{2\cdot\frac{1}{2}} = 3^{1} = 3</math>.}} |
Note: | Note: | ||
The calculation <math>\sqrt{(-3)^{2}} = \bigl((-3)^{2}\bigr)^{1/2} = (-3)^{2\cdot \frac{1}{2}} = (-3)^1 = -3</math> is wrong at the second equals sign. Remember that the power rules apply when the base is positive. | The calculation <math>\sqrt{(-3)^{2}} = \bigl((-3)^{2}\bigr)^{1/2} = (-3)^{2\cdot \frac{1}{2}} = (-3)^1 = -3</math> is wrong at the second equals sign. Remember that the power rules apply when the base is positive. | ||
Version vom 08:35, 22. Okt. 2008
That which is under the root sign is the same as \displaystyle (-3)^{2} = 9 and because \displaystyle 9 = 3\cdot 3 = 3^{2}, hence
| \displaystyle \sqrt{(-3)^{2}} = \sqrt{9} = 9^{1/2} = \bigl(3^{2}\bigr)^{1/2} = 3^{2\cdot\frac{1}{2}} = 3^{1} = 3. |
Note:
The calculation \displaystyle \sqrt{(-3)^{2}} = \bigl((-3)^{2}\bigr)^{1/2} = (-3)^{2\cdot \frac{1}{2}} = (-3)^1 = -3 is wrong at the second equals sign. Remember that the power rules apply when the base is positive.
