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Solution 3.1:2g

From Förberedande kurs i matematik 1

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m (Lösning 3.1:2g moved to Solution 3.1:2g: Robot: moved page)
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Because
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<center> [[Image:3_1_2g.gif]] </center>
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<math>-\text{125 }</math>
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can be written as
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<math>-125=\left( -5 \right)\centerdot \left( -5 \right)\centerdot \left( -5 \right)=\left( -5 \right)^{3}</math>,
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<math>\sqrt[3]{-125}</math>
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is defined as
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<math>\sqrt[3]{-125}=-5</math>
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NOTE: As opposed to
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<math>\sqrt{-125}</math>
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(the square root of
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<math>-125</math>
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) which is not defined,
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<math>\sqrt[3]{-125}</math>
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is defined . In other words, there does not exist any number which satisfies
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<math>x^{\text{2}}=-\text{125}</math>, but there is a number
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<math>x</math>
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which satisfies
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<math>x^{\text{3}}=-\text{125}</math>.
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NOTE: It is possible to write the calculation in the solution as
 +
<math>\sqrt[3]{-125}=\sqrt[3]{\left( -5 \right)^{3}}=\left( -5 \right)^{1}=-5</math>, but one has to be careful when one calculates using negative numbers and fractional exponents. Sometimes, the expression is not defined and the usual power rules do not always hold. Look, for example, at the calculation
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<math>\begin{align}
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& -5=\left( -125 \right)^{{1}/{3}\;}=\left( -125 \right)^{{2}/{6}\;}=\left( \left( -125 \right)^{2} \right)^{{1}/{6}\;} \\
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& =15625^{{1}/{6}\;}=5 \\
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\end{align}</math>

Revision as of 09:00, 30 September 2008

Because −125 can be written as −125=−5−5−5=−53 , 3−125  is defined as


3−125=−5 


NOTE: As opposed to −125  (the square root of −125 ) which is not defined, 3−125  is defined . In other words, there does not exist any number which satisfies x2=−125, but there is a number x which satisfies x3=−125.

NOTE: It is possible to write the calculation in the solution as 3−125=3−53=−51=−5 , but one has to be careful when one calculates using negative numbers and fractional exponents. Sometimes, the expression is not defined and the usual power rules do not always hold. Look, for example, at the calculation


−5=−12513=−12526=−125216=1562516=5 

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