I know that $\sqrt[2k]{a}$ is defined only when $a \geq 0$.

but I don't understand why this is true:

$(\sqrt{35}-6)^{\frac{1}{3}}$ Is a complex number, therefore is different of $ \sqrt[3]{\sqrt{35}-6}$ I thought the roots were defined when the radical index is odd independently if the base is negative, so what is wrong?

Basically why $a^\frac{1}{2k+1} \neq \sqrt[2k+1]{a}$ when $a \in \mathbb{Z^-}$ ?

  • $\begingroup$ Real numbers are a subset of complex numbers (all reals are complex). But the number you wrote is part of the reals. It doesn’t have to be imaginary. Why do you think it is imaginary? $\endgroup$ – Sina Babaei Zadeh Oct 8 '19 at 1:20
  • $\begingroup$ I don't think that is imaginary, is imaginary $\endgroup$ – Eduardo S. Oct 8 '19 at 1:22
  • $\begingroup$ Is $(-1)^{1/3}$ an imaginary number? I think no :) $\endgroup$ – Azif00 Oct 8 '19 at 1:30
  • 1
    $\begingroup$ wolframalpha.com/input/… Or it can be written as a real number. $\endgroup$ – Sina Babaei Zadeh Oct 8 '19 at 1:31
  • $\begingroup$ @Azif00 But the property $x^\frac{1}{3} = \sqrt[3]{x}$ holds for all $x$ ? or only for $x \geq 0$? $\endgroup$ – Eduardo S. Oct 8 '19 at 1:38

Every nonzero complex number (including real numbers) has $n$ $n$-th roots in $\Bbb C$, if $n$ is a positive integer. That is, if $z\in\Bbb C$ and $z\ne0$, then the polynomial $X^n-z$ has $n$ distinct roots in $\Bbb C$.

In case $z$ is a positive real number, then among those $n$, one root is distinguished, namely the one that’s positive, and we write this root $z^{1/n}=\sqrt[n]z$, indifferently.

In case $z$ is a negative real number and $n$ is odd, that is, if $n=2k+1$ for $k\ge0$, then there’s a distinguished $n$-th root, namely the negative one, and we write it $z^{1/n}=\sqrt[n]z$, indifferently.

If anyone tries to tell you in either of the two above cases that $z^{1/n}\ne\sqrt[n]z$, that person is wrong. I admit that this is a matter of notation rather than of the underlying mathematics, but in my oh so many years of mathematics and mathematics teaching, I have never seen it claimed that the two are different.

When $n$ is not a positive integer, and in the second case when $n$ is an even integer, then of course all bets are off, and the ambiguity of the situation overcomes all attempts at pedagogical and notational clarity.

  • $\begingroup$ My professor say that if $x<0$ then $x^{\frac{1}{3}}$ is not defined and that is different of $\sqrt[3]{x}$. He is wrong? $\endgroup$ – Eduardo S. Oct 8 '19 at 1:53
  • $\begingroup$ I say he’s wrong. There’s no question that each of these notations must denote some complex number. Perhaps he could say which two complex numbers are represented by $(-1)^{1/3}$ and $\sqrt[3]{-1}$ ? $\endgroup$ – Lubin Oct 8 '19 at 3:23

$x^{\frac{1}{2k+1}}=\sqrt[2k+1]{x}$ for all $x\in \mathbb{R}$ and $k\in \mathbb{Z}$.



Real numbers can be written as product of complex numbers such as:

$(1+i)(1-i)=2$ So is $2$ a complex number?

Sub fields/ sunsets are not the whole set. All reals are complex in the form of $a+0i$ it should not be surprising that they can be written as multiples or suns of imaginary numbers

  • $\begingroup$ But the property $x^\frac{1}{3} = \sqrt[3]{x}$ holds for all $x$ ? or only for $x \geq 0$ $\endgroup$ – Eduardo S. Oct 8 '19 at 1:36
  • $\begingroup$ Holds for all $x$. $\endgroup$ – Sina Babaei Zadeh Oct 8 '19 at 1:39

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