Lösung 3.1:2b

Aus Online Mathematik Brückenkurs 1

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K (Lösning 3.1:2b moved to Solution 3.1:2b: Robot: moved page)
Zeile 1: Zeile 1:
-
{{NAVCONTENT_START}}
+
That which is under the root sign is the same as
-
<center> [[Image:3_1_2b.gif]] </center>
+
<math>\left( -\text{3} \right)^{\text{2}}=\text{9 }</math>
-
{{NAVCONTENT_STOP}}
+
and because
 +
<math>\text{9}=\text{3}\centerdot \text{3}=\text{3}^{\text{2}}</math>, hence
 +
 
 +
 
 +
<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>
 +
 
 +
 
 +
NOTE:
 +
The calculation
 +
<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>
 +
 
 +
is wrong at the second equals sign. Remember that the power rules apply when the base is positive.

Version vom 10:44, 22. Sep. 2008

That which is under the root sign is the same as \displaystyle \left( -\text{3} \right)^{\text{2}}=\text{9 } and because \displaystyle \text{9}=\text{3}\centerdot \text{3}=\text{3}^{\text{2}}, hence


\displaystyle \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


NOTE: The calculation \displaystyle \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

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“