Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I was taught in school that

$$x^{a/b} = \sqrt[b]{x^a}$$

however, wolfram says this is not always true:

$\sqrt[3]{x^2} \ne x^{2/3}$
http://www.wolframalpha.com/input/?i=cbrt%28x%5E2%29+%3D%3D%3D+x%5E%282%2F3%29

but also says:

$\sqrt[3]{x} = x^{1/3}$
http://www.wolframalpha.com/input/?i=cbrt%28x%29+%3D%3D%3D+x%5E%281%2F3%29

Is he right? If so, can anyone explain why?

share|improve this question
    
On WolframAlpha, it is mentioned that this holds for $x \geq 0$. For negative $x$, $\sqrt[3]{x^2}$ is defined and is positive, while $x^{2/3}$ is taken negative as a principal value (at least there). These functions are best considered as functions of a complex variable though, with multiple branches. –  A Walker Oct 19 '11 at 7:03
    
For example, for $(-1)^{3/2}$ it returns $e^{\frac{2i\pi}{3}}=-\frac12+i\frac{\sqrt3}2$. –  Brian M. Scott Oct 19 '11 at 7:11
    
The problem is your cube root: there are always three (complex) numbers that when cubed yield your original number. Wolfram Alpha takes the "principal value" of the cube root of a negative number to be the cube root with positive imaginary part. –  J. M. Oct 19 '11 at 7:19
    
Thank you, that pretty much explained everything. –  rogi Oct 19 '11 at 7:28

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.