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If $x^{1/2}$ is the same as $\sqrt[2] x$ than why $x^{1/3}$ is not equal to $\sqrt[3] x$ and $x^{1/4}$ not equal to $\sqrt[4] x$ and so on...?

Thank you.

abel
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Muhammad Asif
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    I am sorry I do not understand what you mean. Also, do you mean $\sqrt[2]{x}$, and have difficulty typing it, or really $2 \sqrt{x}$. – quid Jan 30 '15 at 16:38
  • what do these symbols $\sqrt[3] x, x^{1/3}$ mean to you? can you compute some them, at least for some nice values of $x.$ – abel Jan 30 '15 at 17:17
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    $x^{1/3}=\sqrt[3]x$ and $x^{1/4}=\sqrt[4]x$ and also $x^{1/n}=\sqrt[n]x$. It would be great to think a little bit before to post as here:http://math.stackexchange.com/questions/1126743/why-271-3-sqrt327-3-as-271-3-9-and-sqrt327-3 – idm Jan 30 '15 at 17:18

2 Answers2

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$x^{1/2}$ is just $\sqrt[2]{x}$ since :

$$ x^{1/2}x^{1/2}=x^{1/2+1/2}=x=\sqrt[2]{x}\sqrt[2]{x}, \tag*{$x\geqslant0$} $$

hence $x^{1/2}=\sqrt[2]{x}$.

Using the same idea you can show that $x^{1/n}=\sqrt[n]{x}$, for $n=2,3,\ldots$

Workaholic
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    I think they meant $x^{1/2}$ is the same as $\sqrt[2]{x}$. And indeed, $x^{1/n}$ is the same as $\sqrt[n]{x}$ for positive integers $n$. – Ian Jan 30 '15 at 16:39
  • This http://web2.0calc.com/ and every other calculator tells me that 2√25=5 and √25=5. so there is not a difference. And if x^1/2 is just 2√x then why 25^1/2=12.5 and 2√25=5..? – Muhammad Asif Jan 30 '15 at 16:52
  • Please add that this only works (in a unique way) for non-negative real numbers. – k.stm Jan 30 '15 at 17:16
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    @Muhammad Asif: When you write 25^1/2 on your calculator it may actually computing, $(25^1)/2=25/2=12.5$. Try writing 25^(1/2) instead. – Peder Jan 30 '15 at 17:18
  • Thanks once again Peder. The way the calculator works was actually confusing me. – Muhammad Asif Jan 30 '15 at 17:33
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It mostly depends on conventions. According to the one I follow, the function $x\mapsto x^{1/3}$ is only defined for $x>0$, because I use the definition $$ a^b=e^{b\log a} $$ when $b$ is not integer. Since $\log a$ is defined only for $a>0$, the map $x\mapsto x^{1/3}$ is defined only for $x>0$.

In this way I can say without posing any condition that, for instance, $$ x^{1/3}=x^{2/6}=(x^2)^{1/6} $$ which would be utterly false when $x<0$ and I defined $x^{1/3}=\sqrt[3]{x}$.

Other people accept $x^{1/3}$ also for negative $x$, but, in this case, algebraic manipulations like the one above require care.

egreg
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