I've tried both calculations on Wolfram Alpha and it returns different results, but I can't get a grasp of why it is like that. From my point of view, both calculations should be the same, as $2.5=25/10,$ and $(-2)^{2.5}$ is equal to $(-2)^{25/10},$ relying on a general rule $(a^m)^n=a^{mn}$.

Links to sources:



  • 6
    $\begingroup$ Welcome to Math Stack Exchange. Fractional powers of negative numbers aren't uniquely defined $\endgroup$ Commented May 8, 2019 at 21:52
  • $\begingroup$ Thanks for welcoming. But, why it is supposed to mean 'uniquely defined'? $\endgroup$
    – Rizescu
    Commented May 8, 2019 at 21:53
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    $\begingroup$ For example, what is the square root of $-1$? Is it $i$? $-i$? $\endgroup$ Commented May 8, 2019 at 21:54
  • $\begingroup$ I guess it mathematically works for both i and -i, but why calculators like Google's or Wolfram Alpha's give i as result? $\endgroup$
    – Rizescu
    Commented May 8, 2019 at 21:58
  • 3
    $\begingroup$ @Alexandros Actually it might not be so much as "the second equality isn't true" as "the exponent rules don't work for negative numbers like you propose." Certainly $((-1)^2)^{1/2}=1^{1/2}$ but the user wants to say $((-1)^2)^{1/2}=(-1)^{2/2}$. $\endgroup$
    – rschwieb
    Commented Jun 4, 2019 at 16:26

6 Answers 6


J.W. Tanner has communicated the main point and provided some links to questions that provide more details. I'd like to try to tell the (mostly) whole story in one place.

Recall that the standard definition of $a^b$ for $a \in \mathbb{R}_{>0}$, $b\in \mathbb{R}$ is

$$a^b := e^{b\ln(a)}$$

Where the exponential function can be defined in several ways-- through its power series, as the solution to the differential equation $y'=y$, or the inverse to the natural logarithm (which is in turn defined as the integral $\ln(x)=\int_1^x\frac{1}{t}dt$). From this definition, it's clear that $b\ln(a)=\ln(a^b)$, so we have

$$a^{bc} = e^{bc\ln(a)}=e^{c\ln(a^b)}=(a^b)^c.$$

However, for $a \leq 0$, this definition requires us to make sense of $\ln(a)$, and the integral definition referenced above diverges. How might we do this? Since we're trying to understand exponentiation of negative numbers, we surely must include the case of $(-1)^{1/2} = \pm i \in \mathbb{C}$, so we can't get around working in the complex plane. If we want to try to extend our earlier definition of $a^b$, then, we're forced to confront the extension of the exponential function to the complex plane. Fortunately, the exponential function's power series definition extends naturally to the complex plane, and from it we can easily derive Euler's identity, which states

$$e^{i\theta} = \cos(\theta)+i\sin(\theta)$$

for $\theta \in \mathbb{R}$, so $e^{i\theta}$ is a point on the unit circle at angle $\theta$ from the positive real axis, measured counterclockwise. In particular, we see that any nonzero complex number $z$ can be written uniquely as $z=re^{i\theta}$ for some $r \in \mathbb{R}_{>0}$ and $-\pi < \theta \leq \pi$. If we want a defining property of our extension of the natural logarithm to be that the exponential function inverts it (which it had better, if the original formula is to always return $a^1=a$), then, one way to define the natural logarithm of $z$ is $\ln(z) := \ln(r)+i\theta$, as this gives $$e^{\ln(z)}=e^{\ln(r)+i\theta}=re^{i\theta}=z,$$ as desired. Note $z=r$ and $\theta=0$ if $z$ is real and positive, so this is indeed an extension of the usual natural logarithm.

However, this choice was not unique-- we had to restrict $-\pi < \theta \leq \pi$ to make this definition. If our defining property is just inversion by the exponential function, it's clear that $\ln(z)=\ln(r)+i(\theta+2\pi n)$ works just as well for any integer $n$, and in general one could define a natural logarithm by instead restricting $\theta$ to be in any interval of length $2\pi$ we want, even making the interval a function of $r$-- making this choice is called choosing a branch of the logarithm. The original definition I gave is called the principal branch, and this is what most calculators like Wolfram Alpha will use. Going back to our definition of $a^b$ and declaring it true for any $a,b \in \mathbb{C}$, we see the result depends on our choice of branch. This is what people mean when they say that exponentiation isn't uniquely defined in $\mathbb{C}$.

Now, let's finally see what goes wrong in your example using the principal branch of the logarithm to define $(-2)^{2.5}$ and $((-2)^{25})^{1/10}$. We have $$(-2)^{2.5}=e^{2.5\ln(-2)}=e^{2.5(\ln(2)+i\pi)}=e^{2.5\ln(2)+2.5\pi i}=e^{2.5\ln(2)}e^{i\frac{\pi}{2}} = 2^{2.5}i,$$ while $$((-2)^{25})^{1/10}=(-2^{25})^{1/10} = e^{\frac{1}{10}\ln(-2^{25})} = e^{\frac{1}{10}(\ln(2^{25})+i\pi)} = 2^{2.5}e^{i\pi/10}=2^{2.5}(\cos(\pi/10)+i\sin(\pi/10)),$$ and these are clearly different. This example demonstrates precisely that, in general, the identity $a^{bc}=(a^b)^c$ does not hold if $a$ is not a positive real number, and you can similarly see that this identity breaks down if $b$ is not real, even if $a \in \mathbb{R}_{>0}$.

  • 6
    $\begingroup$ +1, especially appreciate you working through the two OP examples showing why they're different. $\endgroup$
    – antkam
    Commented May 9, 2019 at 3:03
  • $\begingroup$ Useful to know, but I forgot to say that I am in 11th grade at secondary school and an answer regarding differential equation or integrals may not fit my knowlodge so far, but it's useful to be in touch with these things as I will deepen what I already know by now. $\endgroup$
    – Rizescu
    Commented May 9, 2019 at 8:32
  • 1
    $\begingroup$ @Rizescu Don't sell yourself short! Those concepts can be used in this construction, but they don't have to be, and you should have the tools to understand the rest. You can just as well define $e^x$ for $x \in \mathbb{R}$ via its power series $$e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$$ and $\ln(x)$ for $x \in \mathbb{R}_{>0}$ as the inverse function to $e^x$. In fact, this fits better with the parallel treatment in the complex case. $\endgroup$
    – jawheele
    Commented May 9, 2019 at 17:20

Fractional powers of negative numbers are not uniquely defined,

and the "general rule" $(a^m)^n=a^{m\times n}$ does not always work when $m$ and $n$ are not integers.

  • 2
    $\begingroup$ The answers in the sources you linked to give $(-2)^{2.5}=2^{2.5}e^{\frac{i\pi}2}$ and $(-2)^{25}$ $^{\frac1{10}}=2^{2.5}$ $e^{\frac{i\pi}{10}}$ but really multiples of $2\pi$ could be added in the exponents $\endgroup$ Commented May 8, 2019 at 23:03
  • $\begingroup$ Thank you so much for writing in English logarithm on my recent question. I had written it in the Italian form. I hope you can appreciate my Sicilian tradition. $\endgroup$
    – Sebastiano
    Commented Oct 20, 2021 at 20:18
  • 1
    $\begingroup$ @Sebastiano: you are welcome; yes, have a good day $\endgroup$ Commented Oct 20, 2021 at 20:44
  • 1
    $\begingroup$ Also for you and for the next days. $\ddot \smile$. $\endgroup$
    – Sebastiano
    Commented Oct 20, 2021 at 20:50

As @J. W. Tanner pointed out in his comment, $a^{m/n}$ is not defined in $\mathbb{R}$ when $a<0$ (and not uniquely defined in $\mathbb{C}$ for all $a$s). That's why softwares usually mess up with things like $(-2)^{2.5} = (-2)^{5/2}$.

Now, you should ask why $a^{m/n}$ is not defined in $\mathbb{R}$ when $a<0$. The motivation of this fact is not trivial, and depends on the importance Mathematicians have attributed to exponentiation identities as $(a^x)^y = a^{xy}$ or $a^{x+y}=a^x\cdot a^y$.

As you should know, the power $a^{1/n}$ (with $a\geq 0$ and $n \in \mathbb{N}$) is defined through the following theorem:

For each $a \geq 0$ and $n \in \mathbb{N}$, there exists a unique $\alpha \geq 0$ s.t. $\alpha^n = a$.

Such an $\alpha$ is called the arithmetic $n$-th root of $a$ and denoted with $\sqrt[n]{a}$ or $a^{1/n}$.

whose proof heavily relies on the completeness of $\mathbb{R}$. Theorem allows you to define the fractional power $a^{m/n}$ with $a\geq 0$ (or $a > 0$ when $m/n < 0$) by letting:

$$a^{m/n} := \sqrt[n]{a^m}\quad \text{(or equivalently } a^{m/n} := (\sqrt[n]{a})^m \text{)}$$

for each $m/n \in \mathbb{Q}$ (it is easy to prove that $\sqrt[n]{a^m} = (\sqrt[n]{a})^m$, hence definition does not depend on the order of application of $m$-th power and $n$-th root).

What happens if constraint $a\geq 0$ is dropped? The theorem cannot remain true for every value of the exponent $n \in \mathbb{N}$: in particular, if $n$ is even (i.e., $n=2,4,6,\ldots$) then $\alpha^n \geq 0$ for all $\alpha \in \mathbb{R}$, therefore equality $\alpha^n = a < 0$ is out of question for even $n$s. On the other hand, the situation for odd $n$s is straightforward:

When $n \in \mathbb{N}$ odd (i.e., $n=1,3,5,\ldots$), for each $a<0$ there exists only one $\alpha < 0$ s.t. $\alpha^n = a$, precisely:

$$\alpha = -\sqrt[n]{-a}\quad \text{(or equivalently } \alpha = -\sqrt[n]{|a|}\text{)}\; . $$

Previous statement allows you to define the arithmetic $n$-th root of $a$ also when $a<0$ and $n \in \mathbb{N}$ is odd by setting:

$$\tag{*} \sqrt[n]{a} := - \sqrt[n]{-a}\; ,$$

but doesn't allow you to define the fractional power $a^{1/n}$, nor $a^{m/n}$ when $a<0$!

In fact, it happens that the definition of rational power with base $a<0$ (by means of $a^{m/n} = \sqrt[n]{a^m}$) is incompatible with usual exponentiation identities, i.e. it causes failure of usual rules like $(a^x)^y = a^{xy}$. In order to see this, consider $a=-1$ and use (*) to get:

$$(-1)^{1/3} = \sqrt[3]{-1} \stackrel{\text{def.}}{=} - \sqrt[3]{-(-1)} = -\sqrt[3]{1} = -1\; ;$$

if usual exponentiation identities were in force then you would get:

$$-1 = (-1)^{1/3} = (-1)^{2/6} = \left[ (-1)^2 \right]^{1/6} = \left[ 1 \right]^{1/6} = 1$$

which is clearly wrong (for $-1 \neq 1$!), or oddities like:

$$-1 = (-1)^{1/3} = (-1)^{1/6 + 1/6} = (-1)^{1/6} \cdot (-1)^{1/6}$$

whose rightmost side has no meaning at all.

Therefore, there is a problem here: fractional powers with negative base and usual exponentiation identities do not fit together.

Mathematicians think it is way better to choose exponentiation identities to hold over the possibility of giving a definition to the symbol $a^{m/n}$ with $a<0$, because identities are of fundamental importance and almost ubiquitous in every possible kind of computation. ;-)

  • 1
    $\begingroup$ In your final equation line showing $-1 = ... = 1$, the use of $(-1)^{1/3}$ seems to me to be irrelevant. The whole thing can be done simply with $-1 = ((-1)^2)^{1/2} = 1$. So perhaps you are trying to make a different point and this is not the appropriate example? $\endgroup$
    – antkam
    Commented May 9, 2019 at 2:57
  • $\begingroup$ @antkam: Thanks for make me notice. I just edited my post. ;-) $\endgroup$
    – Pacciu
    Commented May 9, 2019 at 21:10

When $a$ is not a nonnegative real number and $n$ is not an integer, the number $a^{n}$ is not uniquely defined. This is because we might define the number $\sqrt{-1}$ to be a complex number $z$ so that $z^2=-1$, but the problem is that $z$ is not unique. In particular, we might have $z=i$ or $z=-i$. Similarly, numbers like $\sqrt[3]{-2}$ are also not unique, taking multiple possible values. That's why Wolfram gave you two different results for what looks like should be the same expression--because the complex values of the expressions are not uniquely determined.


You assume that exponentation of real numbers by real numbers satisfies $a^{p \cdot q}= (a^p)^q$. However, it is not that simple. It is true for any $a \in \mathbb R$ and any $p,q \in \mathbb N$. But what is $a^x$ for a non-integer $x$? For $a > 0$ there are various approaches to define it.

(a) $a^x = e^{x \ln a}$ for all $x \in \mathbb R$.

(b) $a^{r/s} = \sqrt[s]{a^r}$ for all $r/s \in \mathbb Q$ where we understand $s \in \mathbb N$.

The second approach can be used to define $a^x$ as $\lim_{r/s \to x} a^{r/s}$, but this requires some work.

For $a >0$ both approaches yield $a^{x \cdot y}= (a^x)^y$ for $x,y \in \mathbb R$ resp. $x,y\in \mathbb Q$.

For $a < 0$ we get troubles. The first approach fails beacuse $\ln a$ is not defined (as a real number). The second approach has serious problems:

(1) It can only work when $r$ is even or $s$ is odd, otherwise you get something undefined (at least if you want a real value for $a^{r/s}$).

(2) If both $r,s$ are even, then the $s$-th root has two possible values (a positive and a negative). You may think that we should always choose the positive value, but the consequences would be unpleasant as you will in the next point.

(3) We should expect that $a^{r/s} = a^{u/v}$ if $r/s = u/v$. But if both $r, s$ are odd, then $a^{r/s}$ is negative whereas $a^{2r/2s}$ is positive.

Choosing always the negative value for the $s$-th root, $s$ even, produces the same problem (consider $r$ even, $s$ odd). And choosing in an ad-hoc way cannot be a serious approach.

Thus, if $a < 0$, you cannot expect $a^{x \cdot y}= (a^x)^y$ to be true no matter how you define $a^{r/s}$. Here is an example, similar to your question:

$$((-1)^2)^{1/2} = 1^{1/2}= \sqrt{1} = 1 \ne (-1)^{2 \cdot 1/2} = (-1)^1 = -1$$ if we choose the positive root.

The lesson is: Be careful when using $a^{x \cdot y}= (a^x)^y$.

  • $\begingroup$ In general, sequential powering is not commutative for arbitrary real (let alone complex or quaternion) numbers. However, sequential powering is commutative in the following special cases: integer exponents, positive real bases, and rational exponents and real bases when the exponents' denominators are all odd. In such cases, $(a^b)^c=a^{bc}$ $\endgroup$ Commented Feb 18 at 19:00

In general, sequential powerings and root extractions are not commutative among the complex numbers. If $c_1$, $c_2$, and $c_3$ are complex numbers, $(c_1^{c_2})^{c_3}$ does not always equal $(c_1^{c_3})^{c_2}$.

For example, $\sqrt{(-1)^3}=\sqrt{-1}=i$, if $i^2=-1$, but $(\sqrt{-1})^3=i^3=-i\neq-1$.

It is always commutative in the special case of positive reals. For positive rational exponents, it is always commutative for all nonnegative real bases. It is also always commutative if the bases are nonzero real and all exponents are rational numbers whose lowest term denominator is an odd number. If the exponents are all positive, then it is commutative for all real bases. If the exponents are integers, then it is commutative for all complex bases.

A similar pattern can be seen regarding multiplication of quaternions. Let $i^2=J^2=k^2=ijk=-1$. Then $ij=k$, while $ji=-k$, so multiplication is not always commutative among quaternions. But multiplication is commutative among the complex numbers, which is a subset of the quaternions.

For this specific example, $2.5=\frac52$. the exponent's denominator $2$ is an even number, so there is no commutativity with sequential powerings/root extractions, as $-2$ is not a nonnegative real number.


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