# Raising to rational power - issues

Raising a real number to a rational power is very simple, right? Consider the following example:

$$−27 = (−27)^{\frac{2}{3}\frac{3}{2}} = ((−27)^{\frac{2}{3}})^\frac{3}{2} = 9^\frac{3}{2} = 27$$

The issue arose because of this part: $-27^{\frac{2}{3}}$. Instead of taking the negative cube root, we first raised it to second power to get $729$ and then calculated $\sqrt[3]729=9$.

Fine, but shouldn't it be like guaranteed by the definition of raising a real number to rational power, that no matter what order of operation I choose, I get the same (correct) answer? What if it was used in some proof? That would immediately invalidate it, but how certain can we be that proofreaders remembered about this issue?

Wikipedia describes it not as something that should never be done, instead it is claimed one needs extra care when raising negative numbers to rational powers. If you think it should never be done, then I guess the article needs to be corrected.

• I think that professional mathematicians know that write $(-1)^\frac{3}{2}$ doesn't have sense. Dec 19, 2015 at 23:45
• thats why the radicand must be non negative Dec 19, 2015 at 23:47
• The law $b^{pq}=(b^p)^q$ does not hold anymore with the usual ways of generalizing exponentiation to negative values of the "base" $b$. Also note that by some definitions $(-27)^\frac23$ is not the real number, but one of the complex (and irreal) solutions. Edit: See e.g. Wikipedia: Exponentiation § Failure of power and logarithm identities. Dec 19, 2015 at 23:55

$(-27)^{\frac 2 3} \ne [(-27)^2]^{\frac 1 3}$ and no mathematician ever claimed it did.

For $b > 0$ and $n > 0; n \in \mathbb Z$ it is provable that there is a unique real number $c > 0$ such that $c^n = b$. We use that to define $b^{\frac 1 n} = c$ and and $b^{\frac m n} = (b^{\frac 1 n})^m = ( b^m)^{\frac 1 n}$ but in doing so we were thoroughly and completely aware that the this definition was only valid for real $b > 0$.

THAT was our definition. $b > 0$ was a requirement.

We also noticed that for $b < 0; n > 0; n \text { odd }; n \in \mathbb Z$ that there $c = - |b|^{\frac 1 n}$ is a unique solution to $c^n = b$. Thus we can define, secondarily as a special except and NOT as part of the primary definition, for $b < 0$ $b^{\frac m n} = (b^{\frac 1 n})^m = -(|b|^m)^{\frac 1 n}$ but ONLY if $n$ is odd, and only if we are aware that $b^{\frac m n} \ne (b^m)^{\frac 1 n}$ in general.

Thus, yes indeed, care must be taken when raising negative bases to rational powers.

"Fine, but shouldn't it be like guaranteed by the definition of raising a real number to rational power, that no matter what order of operation I choose, I get the same (correct) answer?"

It is. The definition of raising to a rational number is very specific about the parity of the base.

" What if it was used in some proof?"

Nearly all proofs will specify that $b > 0$. If on a rare proof one is taking powers of negative values no faulty assumptions may be made.

By specifying directly in the proof the assumption is that $b > 0$. This is as basic and as fundamental as specifying $r$ is rational or $x$ is real or that the domain of log functions are positive. And by assuming proof readers have had this properly pounded into their heads.

"Wikipedia describes it not as something that should never be done, instead it is claimed one needs extra care when raising negative numbers to rational powers."

It isn't something the should never be done. It is something that one needs extra care.

"If you think it should never be done, then I guess the article needs to be corrected."

I don't think it should never be done. I don't think the article need to be corrected.

• 'The definition of raising to a rational number is very specific about the parity of the base' - what do you mean? Dec 20, 2015 at 9:16
• I mean the definition of raising a base to a rational power is very specific about whether the base is positive, negative, or zero. Dec 20, 2015 at 9:18

The problem is that the law $$b^{pq}=(b^p)^q$$ does not hold for $b<0$, so you are not allowed to use it.

Also note that an expression such as $$(-27)^\frac23$$ can be interpreted in two different ways. See the thread How do you compute negative numbers to fractional powers? Neither of these two conventions saves the law $b^{pq}=(b^p)^q$.

One convention is to take $$(-27)^\frac23 = \sqrt[3]{(-27)^2} = (-\sqrt[3]{27})^2$$ which gives $9$, and the other convention is to use principal value of the complex expression which (with usual choices of "principal") gives $$9\left( -\frac12 + i\frac{\sqrt3}{2} \right)$$ or approximately $-4.5+7.79i$.

Let me emphasize again that neither convention of raising negative real numbers to rational exponents preserves the law $b^{pq}=(b^p)^q$.

$$(-27)^{2/3} = (-1)^{2/3} (27)^{2/3}$$

so let's just consider $w = (-1)^{2/3}$. Note that $w^3 = 1$ so by definition, it is a third root of unity (https://en.wikipedia.org/wiki/Root_of_unity). Therefore, we should not blindly apply the exponent rules but instead complexify it first: $$w = (-1)^{2/3} = (e^{\pi i})^{2/3} = e^{\frac{2 \pi}{3} i}$$ Of course you don't need to explicitly find what it is, just leave it in exponential form, and say that now,

$$((-27)^{2/3})^{3/2} = (9w)^{3/2} = 27w^{3/2} = 27e^{\frac{2 \pi}{3} \frac{3}{2}i} = 27 e^{\pi i} = -27$$

• No, we were talking about rational POWERS. The result does not have to be a rational number. Dec 20, 2015 at 0:03
• This answer is misleading, in my opinion. You switch to another interpretation of what $(-27)^\frac23$ means. The asker uses another convention. Also, your answer could give the impression that these "paradoxes" do not arise when one uses the "principal value of the complex function" approach. But we know they do, for example: $$(-1)^{-\frac13} \ne ((-1)^{-1})^\frac13$$ (under your interpretation). Dec 20, 2015 at 0:55

No one should ever ever write $a^b$ when b is not an integer or $a$ is not a positive real number.

Actually there is no definition of a real number raised to any rational power. For example, $(-1)^{1/2}$ is undefined.

• Replace 'or' with 'and' x_x Dec 19, 2015 at 23:52
• Uh, No one should every right $a^x$ if x is not an integer? Um... no. Dec 20, 2015 at 9:10
• $b^x; b > 0$ is defined for all real x. $b^q; b < 0$ is defined for $q \in Q$ iff $q$ has an odd denominator when expressed in lowest terms. $b^x; b < 0$ is undefined if $q$ has an even denominator or if $q$ is irrational. (At least in real analysis. Complex analysis is another issue.) Dec 20, 2015 at 9:15