# Why rational exponents expressed in the simplest form are defined as roots but those not in the simplest form are not?

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 $$\sqrt[q]{b^p}$$?

• 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? May 5, 2019 at 14:47
• Try $b=-1$, $p/q = 1/2 = 2/4$
– Ned
May 5, 2019 at 14:51
• @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 $g(x)=\sqrt[6]{x^2}$. May 5, 2019 at 15:11
• 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.
– Ian
May 5, 2019 at 15:12
• 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. May 5, 2019 at 15:32

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.