Lösung 3.1:2b

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That which is under the root sign is the same as
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That which is under the root sign is the same as <math>(-3)^{2} = 9</math> and because
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<math>\left( -\text{3} \right)^{\text{2}}=\text{9 }</math>
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<math>9 = 3\cdot 3 = 3^{2}</math>, hence
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and because
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<math>\text{9}=\text{3}\centerdot \text{3}=\text{3}^{\text{2}}</math>, hence
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{{Displayed math||<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>.}}
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<math>\sqrt{\left( -3 \right)^{2}}=\sqrt{9}=9^{{1}/{2}\;}=\left( 3^{2} \right)^{{1}/{2}\;}=3^{2\centerdot \frac{1}{2}}=3^{1}=3</math>
 
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Note:
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NOTE:
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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.
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The calculation
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<math>\sqrt{\left( -3 \right)^{2}}=\left( \left( -3 \right)^{2} \right)^{{1}/{2}\;}=\left( -3 \right)^{2\centerdot \frac{1}{2}}=\left( -3 \right)^{1}=-3</math>
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is wrong at the second equals sign. Remember that the power rules apply when the base is positive.
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Version vom 07:52, 30. Sep. 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

Vorlage:Displayed math


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.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_3.1:2b“