On Math SE, I've seen several questions which relate to the following. By abusing the laws of exponents for rational exponents, one can come up with any number of apparent paradoxes, in which a number seems to be shown as equal to its opposite (negative). Possibly the most concise example:

$-1 = (-1)^1 = (-1)^\frac{2}{2} = (-1)^{2 \cdot \frac{1}{2}} = ((-1)^2)^\frac{1}{2} = (1)^\frac{1}{2} = \sqrt{1} = 1$

Of the seven equalities in this statement, I'm embarrassed to say that I'm not totally sure which one is incorrect. Restricting the discussion to real numbers and rational exponents, we can look at some college algebra/precalculus books and find definitions like the following (here, Ratti & McWaters, Precalculus: a right triangle approach, section P.6):

Ratti's definition of rational exponents Ratti's properties of rational exponents

The thing that looks the most suspect in my example above is the 4th equality, $(-1)^{2 \cdot \frac{1}{2}} = ((-1)^2)^\frac{1}{2}$, which seems to violate the spirit of Ratti's definition of rational exponents ("no common factors")... but technically, that translation from rational exponent to radical expression was not used at this point. Rather, we're still only manipulating rational exponents, which seems fully compliant with Ratti's 2nd property: $(a^r)^s = a^{rs}$, where indeed "all of the expressions used are defined". The rational-exponent-to-radical-expression switch (via the rational exponent definition) doesn't actually happen until the 6th equality, $(1)^\frac{1}{2} = \sqrt{1}$, and that seems to undeniably be a true statement. So I'm a bit stumped at exactly where the falsehood lies.

We can find effectively identical definitions in other books. For example, in Sullivan's College Algebra, his definition is (sec. R.8): "If $a$ is a real number and $m$ and $n$ are integers containing no common factors, with $n \ge 2$, then: $a^\frac{m}{n} = \sqrt[n]{a^m} = (\sqrt[n]{a})^m$, provided that $\sqrt[n]{a}$ exists"; and he briefly states that "the Laws of Exponents hold for rational exponents", but all examples are restricted to positive variables only. OpenStax College Algebra does the same (sec. 1.3): "In these cases, the exponent must be a fraction in lowest terms... All of the properties of exponents that we learned for integer exponents also hold for rational exponents."

So what exactly are the restrictions on the Laws of Exponents in the real-number context, with rational exponents? As one example, is there a reason missing from the texts above why $(-1)^{2 \cdot \frac{1}{2}} = ((-1)^2)^\frac{1}{2}$ is a false statement, or is it one of the other equalities that fails?

Edit: Some literature that discusses this issue:

  • Goel, Sudhir K., and Michael S. Robillard. "The Equation: $-2 = (-8)^\frac{1}{3} = (-8)^\frac{2}{6} = [(-8)^2]^\frac{1}{6} = 2$." Educational Studies in Mathematics 33.3 (1997): 319-320.

  • Tirosh, Dina, and Ruhama Even. "To define or not to define: The case of $(-8)^\frac{1}{3}$." Educational Studies in Mathematics 33.3 (1997): 321-330.

  • Choi, Younggi, and Jonghoon Do. "Equality Involved in 0.999... and $(-8)^\frac{1}{3}$" For the Learning of Mathematics 25.3 (2005): 13-36.

  • Woo, Jeongho, and Jaehoon Yim. "Revisiting 0.999... and $(-8)^\frac{1}{3}$ in School Mathematics from the Perspective of the Algebraic Permanence Principle." For the Learning of Mathematics 28.2 (2008): 11-16.

  • Gómez, Bernardo, and Carmen Buhlea. "The ambiguity of the sign √." Proceedings of the Sixth Congress of the European Society for Research in Mathematics Education. 2009.

  • Gómez, Bernardo. "Historical conflicts and subtleties with the √ sign in textbooks." 6th European Summer University on the History and Epistemology in Mathematics Education. HPM: Vienna University of Technology, Vienna, Austria (2010).

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    $\begingroup$ Rational exponents are not defined for negative real numbers. So, the paradoxa you describe do not exist. $\endgroup$ Jan 27, 2016 at 6:21
  • $\begingroup$ @BjörnFriedrich: No, the example that immediately follows Ratti's definition is: $(-8)^\frac{2}{6} = (-8)^\frac{1}{3} = \sqrt[3]{-8} = -2$. $\endgroup$ Jan 27, 2016 at 6:26
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    $\begingroup$ Your objection to his example is wrong, because he did not say what $a^\frac{m}{n}$ is when $\gcd(m,n) \ne 1$. He only specified what it is when $\gcd(m,n) = 1$, implying that to find the value in general you must first convert it to a form where the rational exponent is expressed in lowest terms. However, as my answer says, his example is wrong for a different reason, namely that his rules are inconsistent due to your example, and that the consistent versions are not enough to specify it but we further need additional definition for $n$-th roots where $n$ is odd. $\endgroup$
    – user21820
    Jan 30, 2016 at 3:13
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    $\begingroup$ An alternative definition that is equivalent to Ratti's is: "$\sqrt[n]{a}$, where $a\in\Bbb R$ and $n\in\Bbb Z-\{0\}$, is the largest real solution to $x^n=a$ if one exists, otherwise undefined", and "$a^{m/n}=(\sqrt[n]{a})^m$, provided that $\sqrt[n]{a}$ is defined". If you plug in $(-1)^{2/2}$ to this definition directly, it will say nothing about the value, because $\sqrt{-1}$ is not defined, but going at it indirectly you have $(-1)^{2/2}=(-1)^{1/1}=\sqrt[1]{-1}=-1$, so it is actually defined. $\endgroup$ Jan 31, 2016 at 20:32
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    $\begingroup$ I would like to note that this question has been asked before and I would like someone to please find where this question has been asked before. (I'm trying to find it too) $\endgroup$ Feb 2, 2016 at 21:57

12 Answers 12


You have put your finger precisely on the statement that is incorrect.

There are two competing conventions with regard to rational exponents.

The first convention is to define the symbol $a^x$ for $a > 0$ only. The symbol $\sqrt[n]{a}$ is defined for negative values of $a$ so long as $n$ is odd, but according to this convention, one wouldn't write $a^{1/n}$, for instance.

In defining $a^{p/q}$ to be $(\sqrt[q]{a})^p$, the author you quoted chose the fraction $p/q$ to be in lowest form so that the definition would be unambiguous. For example, $a^{10/15}$ is defined to be $(\sqrt[3]{a})^2$. However, it is preferable to define $a^{p/q}$ to be $(\sqrt[q]{a})^p$ in all cases and to prove that this definition is independent of the particular representation chosen for $p/q$; this is what more rigorous books tend to do. That is, you prove that if $p/q = r/s$, then $(\sqrt[q]{a})^p = (\sqrt[s]{a})^r$. There is no mention of lowest form.

The competing convention is to also allow $a^x$ to be defined for all $a \ne 0$ and all rational numbers $x = p/q$ that have at least one representation with an odd denominator. You then prove that $(\sqrt[q]{a})^p$ is independent of the particular representation $p/q$ chosen, so long as the denominator is odd. Thus you can write $a^{3/5} = (\sqrt[5]{a})^3 = (\sqrt[15]{a})^{9} = a^{9/15}$. All of that is fine. However, you cannot write $a^{6/10} = (\sqrt[10]{a})^6$, or even $a^{6/10} = \sqrt[10]{a^6}$. The number $a^{6/10}$ is well-defined, but to write down its definition, you must first select a fraction equivalent to $6/10$ that has an odd denominator, which could be $3/5$ or $9/15$ or something else. For $a^{1/2}$, this can't be done at all, so $a^{1/2}$ is undefined for $a < 0$.

The rules for exponents break down if you start allowing $a < 0$ and exponents that can't be written with an odd denominator. For example, the rule $a^{xy} = (a^x)^y$ is valid, but only so long as $x$ and $y$ are both rational numbers that can be written with an odd denominator. This is not the case if you write $a^1 = (a^2)^{1/2}$, despite the fact that both sides of the equation are defined since $a^2 > 0$.

Edit Reading the paper by Tirosh and Even, I was surprised to learn this matter has drawn serious attention from math educators.

A long time ago, I assumed that, apart from complex extensions, $a^x$ for non-integer $x$ should be defined only for $a > 0$. I reasoned that it made no sense to have a function $(-2)^x$ defined only for rational numbers $x$ with odd denominator. I objected strenuously to notations like $(-8)^{1/3}$.

But that was before I taught a calculus class, which is when I realized why some textbook authors are so happy to define $a^x$ for $a < 0$, following the second convention. The reason is that the formula $\frac{d}{dx}(x^r) = rx^{r-1}$ is perfectly valid for $x < 0$ and $r$ with odd denominator.

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    $\begingroup$ No, not at all. The fraction $1/1$ is equivalent to $2/2$, so (according to the second convention) there's nothing wrong with saying $(-1)^{2/2} = (\sqrt[1]{-1})^1$. (See the examples with $6/10$.) The error is exactly where you said it was, in $(-1)^{2 \cdot \frac{1}{2}} = [(-1)^2]^{1/2}$. When $a < 0$, the rule $a^{xy} = (a^x)^y$ is valid only when $x$ and $y$ can both be written with an odd denominator. There is no way to write $y = 1/2$ with an odd denominator. $\endgroup$
    – David
    Jan 27, 2016 at 6:38
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    $\begingroup$ I am probably not the best person to answer this question, so you might consider asking a new question like "Where can I find a reasonably rigorous exposition of the properties of rational exponents at the precalculus level?" (I say "reasonably rigorous" because the existence of $n$th roots is unlikely to be proved at that level.) But I will try to tell you about the books I know. Gelfand and Shen's Algebra (in English translation) discusses this question. The exposition is informal, and some proofs are left as exercises. However, the discussion is honest about the difficulties... $\endgroup$
    – David
    Jan 27, 2016 at 7:49
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    $\begingroup$ ...involved. Lang's Basic Mathematics is precise and honest, but skirts the difficulties by making strong, explicit assumptions at the outset. There is a rigorous discussion of this topic in Algèbre et Trigonométrie by Commeau (French) and in Algebra dlya 9 klassa by Vilenkin et al. (Russian). I'm sure I've seen precalculus books in English with fairly rigorous treatment of this subject, but I can't recall which ones. (I think Algebra and Trigonometry by Robison might work, but I don't have it to hand.) That's why I'm suggesting you ask a question here or on Math Educators. Books... $\endgroup$
    – David
    Jan 27, 2016 at 7:56
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    $\begingroup$ at the calculus level (such as Apostol's and Spivak's) are rigorous, but they avoid these difficulties by giving a different definition. ($a^x$ is defined as $e^{x \log a}$, where the functions $e^x$ and $\log a$ are given separate definitions using calculus.) $\endgroup$
    – David
    Jan 27, 2016 at 7:58
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    $\begingroup$ Actually, I have to retract my observation on Wolfram Alpha; if I write the exponent with parentheses instead of braces then it does say that the 2nd equation is true; and in fact the 4th equation, as you say, is the first one it identifies as being false. $\endgroup$ Jan 30, 2016 at 4:10

$-1 = (-1)^1 = (-1)^\frac{2}{2} = (-1)^{2 \cdot \frac{1}{2}} = ((-1)^2)^\frac{1}{2} = (1)^\frac{1}{2} = \sqrt{1} = 1$

The thing that looks the most suspect in my example above is the 4th equality, $(-1)^{2 \cdot \frac{1}{2}} = ((-1)^2)^\frac{1}{2}$, which seems to violate the spirit of Ratti's definition of rational exponents ("no common factors")... but technically, that translation from rational exponent to radical expression was not used as this point.

The 4th equality is indeed suspect, but not for the reason you suggest. It is an application the 2nd property of rational exponents that you list above:

If $r$ and $s$ are rational numbers and $a$ is a real number, then we have: $$(a^r)^s = a^{r\cdot s}$$

provided that all expressions used are defined.

More formally and less ambiguous would be:

$$\forall r,s \in \mathbb{Q}\colon \forall a \in \mathbb{R}\colon [a^r\in \mathbb{R} \land a^s\in \mathbb{R} \implies (a^r)^s=a^{r\cdot s}]$$

This statement makes it clear that we cannot infer $((-1)^2)^\frac{1}{2}=(-1)^{2 \times \frac{1}{2}}$ as in the "paradox" because $(-1)^\frac{1}{2} \notin \mathbb R$, i.e. because $(-1)^\frac{1}{2}$ is not defined.

That both restrictions are necessary can be seen from the fact that we must have $a^{r\cdot s}=a^{s\cdot r}=(a^s)^r=(a^r)^s$. If we had $a^s \notin \mathbb{R}$, we could not make this substitution.

With this in mind, we could restate the rule as follows:

$$\forall r,s \in \mathbb{Q}\colon \forall a \in \mathbb{R}\colon [a^r\in \mathbb{R} \land a^s\in \mathbb{R} \implies a^{r\cdot s}=(a^r)^s=(a^s)^r]$$

Though it has nothing to do with resolving the paradox, we may also need to define $x^\frac{1}{n}$ as follows:

$\forall x,y\in \mathbb{R}\colon\forall n\in \mathbb{N}\colon [Odd(n)\lor Even(n) \land n\neq 0 \land y\geq 0\implies [x^\frac{1}{n} =y\iff x=y^n ]]$

Using this rule, we could infer that $4^\frac{1}{2}=2$, but not $4^\frac{1}{2}=-2$.

BTW, as far as $\frac{m}{n}$ having to be in lowest terms, the definition given seems a bit sloppy. It cannot be, for example, that $4^\frac{2}{4}$ is undefined when $4^\frac{2}{4}= 4^\frac{1}{2}$ by substitution of $\frac{2}{4}=\frac{1}{2}$. I really don't think this notion can be source of the paradox.

  • $\begingroup$ I didn't include it in the question above, but in the lead-in to defining $a^\frac{m}{n}$ Ratti does define principal nth roots, and then $a^\frac{1}{n}$ in terms of those roots, in a way that's effectively equivalent to what suggest; and this is again standard in all of the college-algebra-level texts that I can put my hand on. So now I'm wondering why some texts bother to do it this way (as opposed to prohibiting negative bases which does seem cleaner). $\endgroup$ Jan 30, 2016 at 3:08
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    $\begingroup$ @DanielR.Collins That would be like throwing out the baby with the bathwater. We really need results like $(-8)^\frac{1}{3}=-2$. $\endgroup$ Jan 30, 2016 at 15:36
  • $\begingroup$ @DanielR.Collins See my edited version above. $\endgroup$ Jan 30, 2016 at 16:08
  • $\begingroup$ Yeah, I think it's convincing that the properties here need an additional restriction of the type that you specify (and I think that's effectively what @David is suggesting in the last paragraph of his answer). I already upvoted your answer previously. $\endgroup$ Jan 30, 2016 at 20:17
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    $\begingroup$ @MarioCarneiro Nevertheless, requiring both $a^r\in \mathbb{R}$ and $a^s \in \mathbb{R}$ avoids the paradox and allows the substitution $(a^r)^s = (a^s)^r$, as you would expect. Also, I wasn't suggesting that my formal statement of the rule was equivalent to the author's informal statement of the rule. I said, mine was less ambiguous. $\endgroup$ Jan 31, 2016 at 16:55

The issue is that $a^{\frac{1}{n}}$ is multivalued. You could arguably simplify the first calculation into $1 = \sqrt{1} = -1$. Taking different branch cuts is how the "paradox" arises.

Essentially, in the context of the reals (or even the complex numbers) $\sqrt{a}$ is one name for two functions, say $\sqrt[+]{a^2} = a$ and $\sqrt[-]{a^2} = -a$. All the laws are fine as long as you remain consistent with your choice. (Alternatively, by moving to a Riemann surface you don't have to make and track a choice... well, you have to decide when and how you are going to embed your reals into the Riemann surface, but once you do, no more choices.)

Whenever square roots entered the picture — you can say at $-1 = (-1)^{\frac{2}{2}}$ or at $((-1)^2)^{\frac{1}{2}}$ — it explicitly chose, going from left-to-right, the non-standard choice of $a^{\frac{1}{2}} = \sqrt[-]{a}$. If it chose the standard choice which it uses later on then, $-1 = -(-1)^{\frac{2}{2}} = -((-1)^2)^{\frac{1}{2}}$ and everything would work out. If it was consistent with the choice of $\sqrt[-]{}$ then $\sqrt{1} = \sqrt[-]{1} = -1$ would also have led to a correct result.

Moving my comment to the answer, a crucial source of confusion is that the definition of $a^{\frac{m}{n}}$ is not a well-defined function of the rationals in that it doesn't respect equality of the rationals. This is witnessed by the need for $\frac{m}{n}$ to be in lowest terms, and, relevant here, the fact that $1 = \frac{n}{n}$ does not imply $a^1 = a^{\frac{n}{n}}$. In fact, the ill-definedness of the provided definition of $a^\frac{m}{n}$ is entirely reduced to the question of what $a^\frac{n}{n}$ is.

So to put it in terms of rules: all the rules are valid, what's invalid is cancelling common factors in a "rational" exponent because the exponents aren't actually rational numbers.

  • $\begingroup$ Downvoted because this is inconsistent with the real definitions in use here. Ratti sec. P.6: "Definition of Principal Square Root: $\sqrt{a} = b$ means (1) $b^2 = a$ and (2) $b \ge 0$". Perhaps you could improve your answer this way: Granted that $a^\frac{1}{n}$ is definitely not multivalued under this real definition, what additional verbiage is required in the properties of exponents to prevent this falsehood? $\endgroup$ Jan 27, 2016 at 6:18
  • $\begingroup$ The key, based on the images you included, is that $a^{\frac{2}{2}}$ hasn't been defined since 2 definitely shares a common factor with 2. So the "failure" is that $a = a^{\frac{n}{n}}$ is not true in general. In particular, whenever you reduce a rational exponent to radicals, if the fraction isn't in lowest terms, you will end up with an expression like $a^{\frac{n}{n}}$ for some $a$. $\endgroup$ Jan 27, 2016 at 6:29
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    $\begingroup$ I would say both the current answers focus on how exactly are fractional exponents interpreted in to radicals. In particular, the definition from your first image should make you queasy. If the exponents were really rationals, then it shouldn't matter if they are in lowest terms or not. In technical terms, the provided definition of $a^{\frac{m}{n}}$ is not well-defined in the sense that it doesn't preserve the equivalence classes of pairs that form the rationals which is how $a^1 \neq a^{\frac{n}{n}}$ is possible even though $1 = \frac{n}{n}$. $\endgroup$ Jan 27, 2016 at 6:50
  • $\begingroup$ You are quite right to put rationals between quotes because if you are not allowed to cancel (or introduce) common factors, these beasts are not rationals but integer pairs. It is unreasonable to accept such a definition and notation, meaning that $a^{m/n}$ has nothing to do with $a^{(m/n)}$. Also unreasonable to speak of "rational powers". I can't adhere to such a convention and prefer to reject $a^{rs}=(a^r)^s$ for the negatives. $\endgroup$
    – user65203
    Feb 1, 2016 at 10:11

I can give a complete description of all the exceptions to the exponent laws. First, an auxiliary defintion.

For any integer $n\neq 0$, let $\nu(n)$ be the largest integer $k$ for which $2^k$ divides $n$. We extend this definition to rational $x=\frac{m}n$ by letting $\nu(x)=\nu(m)-\nu(n)$, which is independent of the choice of representation.

This means that $a^x$ is defined as long as $a\ge 0$, or if $x$ is rational with $\nu(x)\ge 0$.

Laws of exponents:

  1. $a^{x+y}=a^xa^y$ holds as long as all exponents are defined.

  2. $(ab)^x=a^xb^x$ holds as long as all exponents are defined.

  3. $(a^x)^y = a^{xy}$ holds as long as all exponents are defined, except when $a<0$ and $\nu(x)>0$ and $\nu(x)=-\nu(y)$.

    • If $\nu(x)>0$ and $\nu(x)=-\nu(y)$, then $(a^x)^y=|a|^{xy}$.

The exception in rule $3$ takes care of the fallacious proof $-1=(-1)^{\frac22}=((-1)^2)^{\frac12}=1^\frac12=1$. It also includes the identity $\sqrt{a^2}=|a|$, and more generally that $\sqrt[2n]{a^{2n}}=|a|$, as special cases.

  • $\begingroup$ Wow, this seems incredibly elegant and nifty. Thanks for sharing that! $\endgroup$ Sep 20, 2019 at 22:03
  • $\begingroup$ Related: While testing some stuff on Wolfram Alpha, I observed that it says "(-1)^(3/5) in reals" is -1, while "(-1)^(6/10) in reals" is undefined, which seems faulty. $\endgroup$ Sep 20, 2019 at 22:04

While I agree with everything in the answer by David, I'll give a different answer here just to put a different emphasis.

The fundamental error is to put the rule $(a^r)^s=a^{rs}$ in the box governed by the condition provided that all the expressions used are defined. That is not the right kind of condition for this rule, it requires specific limitations to the values of $a,r,s$. In this particular context ($a\neq0$ real and $\def\Q{\Bbb Q}r,s\in\Q$), the condition should be:

either $a>0$ or both $r$ and $s$ lie in the valuation ring $\def\Z{\Bbb Z}\Z_{(2)}$, the subring of $\Q$ of numbers that can be represented with an odd denominator.

Note that this condition ensures that both expressions are defined and that they are equal. Note also that these conditions are identical to those under which the powers $a^r$ and $a^s$ are both defined. However, neither of the expressions in the rule involves$~a^s$, so the conditions are not implied by "all expressions used in the rule are defined".

I am not a partisan of defining (certain) non-integer rational powers of negative numbers at all; it is of very little use, and if one wants to study the function $x\mapsto\sqrt[3]{x^2}$ on all of $\def\R{\Bbb R}\R$, there is not much against having to write just that, or $x\mapsto|x|^{2/3}$, rather than $x^{2/3}$. But if one does choose to go that way, I would suggest restating the definition as follows:

For $a\in\R_{\neq0}$ and $r\in\Q$, the power $a^r$ is defined provided that either $a>0$ or $r\in\Z_{(2)}$ (or both); in the former case one has $a^{m/n}=\sqrt[n]{a^m}=(\sqrt[n]a)^m$ for any fraction $m/n$ representing $r$, while in the latter case one has the same identities for any fraction $m/n$ representing $r$ in which $n$ is odd.

Given that the latter case has $r\in\Z_{(2)}$, restricting to odd $n$ there is quite natural (and it is necessary).

There exists some other contexts in which one might want to state the validity of $(a^r)^s=a^{rs}$ with the proviso that all occurring expressions are defined. I can think of the following two cases:

  • Exponents $r,s\in\Z$, and for instance $\def\C{\Bbb C}a\in\C$ unrestricted (it could even be something more general, like a square matrix). Here the rule basically derives from $a^{x+y}=a^xa^y$ (with the same proviso), and some considerations about how negative exponents combine. The proviso would serve to bar negative powers of $0$, and could be replaced by the explicit condition: $r,s\in\Bbb N$, or $a$ invertible.
  • Real $a\geq0$ and real exponents $r,s$. Here the proviso is needed for the same reason as in the previous point, to avoid negative powers of $0$; with $a>0$ the rule is valid unconditionally.

But the second point hints at a generalization where again the condition "all occurring expressions are defined" is insufficient. For real $a>0$, there is no difficulty in defining $a^r$ for all $r\in\C$. However (as I've mentioned in this answer to another question), the rule $(a^r)^s=a^{rs}$ is only valid with the restriction that $\def\R{\Bbb R}r\in\R$; this is strictly stronger than the condition $a^r\in\R$ ensuring that $(a^r)^s$ is defined, but which does not make the rule valid. The validity of the rule with the given restriction is easy to prove, see here.

  • $\begingroup$ I like this answer a lot. It nicely observes the issue of different possible definitions, and that the Ratti-style qualifier on the properties needs some improvement. Thanks for writing it, upvoted. $\endgroup$ Feb 4, 2016 at 16:16

No continuous definition of $a^r$ can be made for all real $a$ and $r$; and likewise, the familiar properties of exponents cannot be extended consistently to all real bases and powers. As a result, there are a number of competing definitions for $a^r$ for non-integer values $r$, depending on how much the author wishes to extend these properties, and in what direction. Here are some things that we can positively say for an identity like $(a^r)^s = a^{rs}$:

  • It is true for all natural numbers $r$ and $s$, and all real numbers $a$. [1]
  • It is true for all integers $r$ and $s$, and all nonzero reals $a$.
  • It is true for all real $r$ and $s$, and all positive reals $a$.

Note that the more permissive we are with $r$ and $s$, the more restrictions we must place on $a$. Some authors do further extend the real-valued definition of $a^r$ (and hence related properties) to negative real $a$'s and non-integer rationals $r$ (while others do not); but this is a fairly fragile definition, in that to be well-defined it requires that $r = m/n$ be written with an odd value for $n$ (books in this vein usually specify that it be in lowest terms). Among the greatest problems with such an approach is that a real-valued “principal $n$th root” will give contradictory results to the complex-valued “principal $n$th root” for negative bases. For example, if a real-valued definition is given, then $(-8)^{1/3} = -2$; but by the standard complex-valued definition, $(-8)^{1/3} = 1 + \sqrt{3}i$. This seems to create some confusion when discussing the issue across different contexts. Arguably it would be best to refrain from that very limited extension in reals, so as to not conflict with the more general complex-valued definition. (See the cited articles in the question above for some published debates on the wisdom of using such a real-valued definition for negative bases and non-integer exponents.)

Regarding the example in the question, most everyone agrees that $(-1)^{2 \cdot \frac{1}{2}} \ne ((-1)^2)^\frac{1}{2}$, if both sides are simplified in the standard order of operations; and this highlights the fact that the identity $(a^r)^s$ = $a^{rs}$ is not true unrestrictedly. Exactly what restrictions need to be honored depend on the definitions in use in a particular textbook. For Ratti, we might rescue the presentation by interpreting the clause “provided that all of the expressions used are defined” in the broad sense of every expression inside the box (not just the one identity being used), and since $a^s$ appears in other places in the box, and $(-1)^\frac{1}{2}$ is certainly undefined in real numbers, then the assertion $((-1)^2)^\frac{1}{2} = (-1)^{2 \cdot \frac{1}{2}}$ (the 4th equality) would thereby be proscribed.

[1]: And more generally for $a$ an element of any ring.

  • $\begingroup$ The definition $a^{m/n} = (\sqrt[n]{a})^m$ does not require that $m/n$ be written in lowest terms. It requires only that $m/n$ be written with an odd denominator (if it can be written that way). $\endgroup$
    – David
    Feb 4, 2016 at 7:05
  • $\begingroup$ My previous comment refers to the case $a < 0$. $\endgroup$
    – David
    Feb 4, 2016 at 7:25
  • $\begingroup$ @David: Right, I edited that part, thanks. $\endgroup$ Feb 4, 2016 at 7:29

$(a^r)^s=a^{rs}$ can indeed be false for $a<0$, as shown by your example.

You can "rescue" this rule by stating instead "$(a^r)^s=a^{rs}=(a^s)^r$, provided all three expressions are defined". (As the product is commutative, you cannot really distinguish $r$ and $s$.)

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    $\begingroup$ While the "rescue" would be technically correct, stating it thus would be a pedagogical disaster. Saying three expressions are all equal provided "all of the expressions used are defined" is just inviting the error of assuming any two of them are equal provided they are defined. If you need to compare those two expressions, you don't really need to use the third expression at all, so it is not even obvious that "all of the expressions used are defined" is being violated. $\endgroup$ Feb 4, 2016 at 5:23
  • $\begingroup$ @MarcvanLeeuwen: you are right, I shouldn't have used "used". I have rephrased to avoid the misinterpretation. $\endgroup$
    – user65203
    Feb 4, 2016 at 9:10

I wouldn't raise a negative number to a non-integer power without explicitly adopting a convention. What, for example, is $(-1)^{1/2}$? One could plausibly say it's $i$ or $-i$. There's no non-arbitrary way of singling one of those out.

Suppose we were to say as a matter of convention that $(-1)^x = \exp(i\pi x)$.

Could we then say that $(-1)^{xy} = ((-1)^x)^y$? The problem here is that our convention defines powers of $-1$ and not of any other number, such as $(-1)^x$. If $x=2$, then there's a problem.

I don't think one can define negative number raised to non-integer powers in such a way that they obey the usual laws of exponents. Integers powers, however, seem to pose no problem.

$$ -1 = (-1)^1 = (-1)^{2/2} = \overbrace{(-1)^{2 \cdot (1/2)} = ((-1)^2)^{1/2}}^\text{Therefore this step is not valid.} = (1)^{1/2} = \sqrt{1} = 1 $$


I think that the fourth equality introduces the error, since $a^{rs}=(a^r)^s$ may not hold if considering complex numbers. For example, \begin{align} -1=e^{i\pi}=e^{2i\pi \cdot \frac{1}{2}} \neq (e^{2i\pi})^{\frac{1}{2}}=1. \end{align}


Well, Algebra meets Calculus and they disagree on some points, literally (unless well guided).

Algebra says, "I have a polynomial $x^n=a$ with $n$ different complex roots. ($a\neq 0$, $n$ integer). And for positive real numbers I can have a function $\sqrt[n]a$ that is positive and solves the equation"

Convenience says: "Oh, so I can write $\sqrt[n]a=a^\frac 1 n$, for real positive $a$.

Student/teacher says: "Oh, it's true for some $n$ and negative $a$, too, so I'll write things like $\sqrt[3]{-8}$ because we all know what is meant. And this is the part where confusion leaks in.

On the other part, Calculus says, "I have a function $e^z$ that behaves like an algebraic polynomial for some $z$ ($z=n\ln x$). And I want it to be holomorphic (https://en.wikipedia.org/w/index.php?title=Holomorphic_function&oldid=699948452)".

The crucial point that no-one tells you is that, in calculus, $e^z:=\exp(z)$ is seen as the well-defined function $\sum_{k=0}^{\infty}\frac{z^n}{n!}$ rather than some dubious exponentiation.

For the function and any integer $k$, $1=e^{2i\pi k}$, and fixing any $k$, you have a branch for the exponentiation such that for any $a=e^\lambda= e^{\lambda+2i\pi k}$ you get a well-defined $a^z=\sum_{k=0}^{\infty}\frac{(z\cdot(\lambda +2i\pi k))^n}{n!}$.

Still, exponentiation is only unique for a fixed branch.


enter image description here

Graphically the results look very different. Reinforcing the differences.


Edit: I'm now disavowing this answer, but keeping it here to maintain the comments. Consider my newer answer elsewhere for, I think, a better treatment.

I would currently argue that it's the 2nd equality in the example, $(-1)^1 = (-1)^\frac{2}{2}$, that contains the initial error. It's illegitimate to write $(-1)^\frac{2}{2}$ because that's an undefined expression, although the exact reason depends on which of two common definitions of $a^\frac{m}{n}$ is in use in a given textbook:

  • Some books at the college-algebra level define rational exponents for any real number $a$, but only for rational exponents written in lowest terms (i.e., $m$ and $n$ with no common factors). In this case, $(-1)^\frac{2}{2}$ in invalid because of the common factor in the exponent.

  • Other books, including those at the real analysis level, define rational exponents only for nonnegative real $a$ (i.e., $a \ge 0$), but permit any $m$ and $n$ regardless of common factors. In this case, $(-1)^\frac{2}{2}$ is undefined because of the negative base.

In summary: for $a > 0$, any $(-a)^\frac{n}{n}$ is a meaningless piece of writing. The way that this expression violates the definition of a rational exponent depends on the exact definition in use.

  • $\begingroup$ I would say instead that the statement you quoted should be corrected by restricting it to cases in which $n$ is even and has no common factors with $m$. $\endgroup$
    – David
    Jan 27, 2016 at 8:26
  • $\begingroup$ @David: I see why you say that; I spent in inordinate amount of time trying to figure out whether that was implicit in the statement or not. But even if it was, then the expression $(-8)^\frac{2}{6}$ seems to violate the given definition in the first place with its common factors. $\endgroup$ Jan 27, 2016 at 16:20
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    $\begingroup$ @user21820: Thank you for drawing my attention to the incorrect syntax I used in Wolfram Alpha; I'm removing that reference from my answer. $\endgroup$ Jan 30, 2016 at 4:09
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    $\begingroup$ I'm downvoting this because I really don't want you to think that the problem is the second equality, which is an application of the substitution rule for $=$ applied to $1=\frac22$. If equality doesn't have the properties of equality, a lot of stuff goes wrong and you don't want to take that path. You had it right the first time with $((-1)^2)^{1/2}\ne(-1)^{2\cdot1/2}$. (Also I can confirm that in both WA and Mathematica, (-1)^(2/2) evaluates to -1.) $\endgroup$ Jan 31, 2016 at 0:05
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    $\begingroup$ There is no discussion that $2/2=1/1$ and nothing distinguishes these two rationals. If you want to distinguish between $a^{2/2}$ and $a^{1/1}$, then the notation is wrong. (F.i. use $a^{2,2}\not\equiv a^{1,1}$.) Actually the definition says nothing but $a^{m/n}:=\sqrt[m']{a^{n'}}=(\sqrt[n']a)^{m'}$ where $m'=m/gcd(m,n),n'=n/gcd(m,n)$. $\endgroup$
    – user65203
    Feb 1, 2016 at 9:46

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