If $b$ is a nonzero real number and $p/q$ is a positive rational number, then, if $p/q$ is expressed in lowest terms, $b^{p/q}=\sqrt[q]{b^p}$.
Why if $p/q$ is not expressed in lowest terms, $b^{p/q}$ may or may not equal $\sqrt[q]{b^p}$?

  • $\begingroup$ First, I think you wanted to set $x=b.$ Second, who said the second equality does not hold when the fraction isn't in lowest terms? $\endgroup$
    – Allawonder
    May 5, 2019 at 14:47
  • 1
    $\begingroup$ Try $b=-1$, $p/q = 1/2 = 2/4$ $\endgroup$
    – Ned
    May 5, 2019 at 14:51
  • $\begingroup$ @Allawonder for the first, yes, thanks for that. For the second, it may or may not hold; I tried graphing $f(x)=x^{2÷6}$, but, I got a different graph than that of $g(x)=\sqrt[6]{x^2}$. $\endgroup$
    – Muhammad
    May 5, 2019 at 15:11
  • $\begingroup$ With $b=-1,p/q=1/2=2/4$ as Ned suggested, you have $(-1)^{1/2}=\{ i,-i \}$ (where here, unlike usual, we don't select a "privileged" root) but $((-1)^2)^{1/4}=\{ 1,i,-i,-1 \}$. The result after reduction is always a subset of the result before reduction, but as you can see there can be other stuff in there too. $\endgroup$
    – Ian
    May 5, 2019 at 15:12
  • $\begingroup$ I'd say $(-1)^{1/2} = \{i, -i\}$ is just wrong, since we do select the principal root so that $(-1)^{1/2} = i$. The formula works for unreduced fractions when $b>0$ but we have to be more careful when $b<0$, where it still works for reduced fractions. $\endgroup$
    – Christoph
    May 5, 2019 at 15:32

1 Answer 1


The reason is this: if we allow complex values (and there's no way to avoid that anyway since you say $b$ can be negative), then the property $$b^{xy}=(b^x)^y=(b^y)^x$$ fails, so that there's no way to define $b^{xy}$ unambiguously if $x$ and $1/y$ have common factors, where $1/y$ and $x$ are positive integers.

But why does this fail? The reason is simple -- each complex number (which include the reals) has exactly $n$ $n$-th roots, and since we allow these (and we have no choice when $b<0$ and $n$ is a positive even number), it follows that, for example, $$b^{2/6}$$ may be written as either $(b^2)^{1/6},$ or else $(b^{1/6})^2,$ each of which yields six different values; but if we invoke the property above and write $2\cdot\frac16=\frac13,$ to have $$b^{1/3},$$ then this gives us only three values. Thus, in general we have $$b^{xy}\ne (b^x)^y.$$

However, this problem vanishes if we restrict $b$ only to positive values. Then with the usual convention that for such $b,$ we mean by $b^{1/n}$ the positive $n$-th root when $n$ happens to be even, the property continues to hold. (Of course when $n$ is odd, there is always a unique real $n$-th root, even if $b$ happens to be negative, but that is irrelevant.)

PS. To see why $b^{1/n}$ has $n$ distinct values, note that we may write $b$ as $$r(\cos\phi+i\sin\phi)=r(\cos(\phi+2πk)+i\sin(\phi+2πk)),$$ where the last equation follows from the periodicity of the trigonometric functions. Also, $r\ge 0,\,\phi$ are real and $k\in\mathrm Z.$

Thus, we have $$b^{1/n}=r(\cos(\phi+2πk)+i\sin(\phi+2πk))^{1/n}=r^{1/n}\left(\cos\left(\frac{\phi+2πk}{n}\right)+i\sin\left(\frac{\phi+2πk}{n}\right)\right),$$ by De Moivre. It is then easy to see that these complex numbers are distinct for each $k\in[0,n-1],$ thus confirming the claim.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .