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).

Daniel R. Collins
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    Rational exponents are not defined for negative real numbers. So, the paradoxa you describe do not exist. – Björn Friedrich Jan 27 '16 at 06:21
  • @BjörnFriedrich: No, the example that immediately follows Ratti's definition is: $(-8)^\frac{2}{6} = (-8)^\frac{1}{3} = \sqrt[3]{-8} = -2$. – Daniel R. Collins Jan 27 '16 at 06:26
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    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. – user21820 Jan 30 '16 at 03:13
  • @user21820: "he did not say what $a^\frac{m}{n} $ is when $gcd(m,n)≠1$". Indeed; making it an undefined expression; I disagree that your implication is necessarily inferred. – Daniel R. Collins Jan 30 '16 at 03:15
  • It is, because the standard interpretation of exponents is that it is an operation, and **always** you evaluate the operands **before** you evaluate the operation on them. With this in mind, he states something of the form $P \Rightarrow Q$, and I'm merely using it together with the standard interpretation of binary operations. – user21820 Jan 30 '16 at 03:22
  • I've deleted my answer because you do not appreciate correct mathematics. Inform me if you want to see it again. – user21820 Jan 30 '16 at 13:03
  • @DanielR.Collins The way I interpret it, he has actually defined $a^r$ for all real $a$ and rational $r$. Just because he has restricted to the case $a^{m/n}$ where $(m,n)=1$ doesn't mean that the other cases are undefined - they are covered by rules about canceling factors in a fraction. Any rational number can be written as $m/n$ with $(m,n)=1$, and your example $(-8)^{2/6}=(-8)^{1/3}=\sqrt[3]{-8}=2$ is perfectly valid with Ratti's definition, where the first equality is simply because $\frac26=\frac13$. For that not to be the case would fundamentally break substitutivity of $=$. – Mario Carneiro Jan 30 '16 at 23:52
  • @MarioCarneiro: Well, it can't literally be the case that we've "defined $a^r$ for all real $a$ and rational $r$" here. What about $(-1)^\frac{1}{2}$? Obviously *something* isn't defined or the restriction at the end of the properties would be empty ("provided that all of the expressions used are defined"), for example. – Daniel R. Collins Jan 31 '16 at 16:45
  • @DanielR.Collins My apologies, I meant to say that he has defined $a^r$ for all real $a$ and rational $r$ for which there is a real solution. The point is that the $(m,n)=1$ restriction does not actually cause holes in the domain of definition, although the additional "$\sqrt[n]a\in\Bbb R$" restriction does, and indeed $(-1)^{2/4}$ and $(-1)^{1/2}$ are not defined by this definition. – Mario Carneiro Jan 31 '16 at 20:16
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    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. – Mario Carneiro Jan 31 '16 at 20:32
  • @MarioCarneiro: I agree that's equivalent to the Ratti, et. al., definition, although I'm not 100% convinced of the inference afterward. And you probably meant $\mathbb{N}$, not $\mathbb{Z}$, yes? – Daniel R. Collins Feb 01 '16 at 00:10
  • While this is probably ill defined, I note that $y=-1$ is not in the primary branch of the square root function $y=\sqrt{x}$, so, you have to consider why it's not in the primary branch. In fact, it is in the second branch of the square root function. While $$(-1)^{\frac22}=\sqrt{(-1)^2}$$, we have to consider all branches in order to find the solution. So, in fact, $x^{\frac mn}=\sqrt[n]{x^m}$ has **one** branch, for all $m,n$, where the equality is true. All other branches, as you noticed, produce apparent paradoxes. – Simply Beautiful Art Feb 01 '16 at 02:36
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    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) – Simply Beautiful Art Feb 02 '16 at 21:57
  • Interesting. $(-1)^{2 \cdot \frac{1}{2}} = ((-1)^2)^\frac{1}{2}$ would not bother me, but $((-1)^2)^\frac{1}{2} = (1)^\frac{1}{2}$ definitely would. $(-1)^2 = 1$ jumps right at me as wrong. Without getting into precise definitions, it has been beaten in my head in school that with even positive exponents the result is always non-negative, and two opposite reals when squared produce the same positive result, so this clearly is not an equality. – Rusty Core Sep 14 '20 at 16:52

12 Answers12


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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  • That's pretty good, except that what you're saying seems to imply that it's the 2nd equality, $(-1)^1 = (-1)^\frac{2}{2}$, that is actually the initial error, in that the right hand side is an undefined expression, yes? – Daniel R. Collins Jan 27 '16 at 06:34
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    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. – David Jan 27 '16 at 06:38
  • I otherwise like your answer, but I'm coming around to closely reading the statement "When $n$ is even and $a < 0$, the symbol $a^\frac{m}{n}$ is not a real number", as saying that it's erroneous to even write the expression $(-1)^\frac{2}{2}$ in the first place. – Daniel R. Collins Jan 27 '16 at 06:42
  • My answer was focused primarily on the underlying mathematical issues, and less on interpreting the intent of that particular author. In my experience, books with titles like *Precalculus: A Right Triangle Approach* are seldom written with a high level of concern for mathematical accuracy. Since $2/6$ and $1/3$ are the same number, it is not meaningful or desirable to say that $(-1)^{1/3}$ is defined but $(-1)^{2/6}$ is not. It would be meaningful and correct, however, to say that when $a < 0$ and $m/n$ is *in lowest form* with $n$ even, the expression $a^{m/n}$ is undefined. – David Jan 27 '16 at 06:49
  • Could you suggest a book that does treat this particular topic with a proper "concern for mathematical accuracy"? I have two bookshelves full of math books here and I can't find one that properly clarifies this issue. – Daniel R. Collins Jan 27 '16 at 07:01
  • Most books I'm familiar with that have that level of precision choose the first convention. The second convention is encountered occasionally, but the reader is usually expected to sort out what is meant for himself and the conventions are rarely made explicit. Are you interested in books that follow the first convention? – David Jan 27 '16 at 07:03
  • Yes, I would... – Daniel R. Collins Jan 27 '16 at 07:10
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    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... – David Jan 27 '16 at 07:49
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    ...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... – David Jan 27 '16 at 07:56
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    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.) – David Jan 27 '16 at 07:58
  • Do you have a response to the observation that Wolfram Alpha states that the equality $(-1)^1 = (-1)^\frac{2}{2}$ is false? (Just added to my answer.) – Daniel R. Collins Jan 30 '16 at 03:54
  • I don't know much about how these systems work, but if a system follows the first convention, then it will allow $(-1)^n$ only when it knows $n$ is an integer. Often computer systems will have "types" for variables, and as soon as a number is the result of a calculation that could potentially result in a non-integer, it will make the number floating-point. For example, a system may not see $2/2$ as being of a different type than $2/2.000000001$. That being said, the expression $w^z$ can be defined as a multi-valued function of complex numbers, and it becomes single-valued if a determination... – David Jan 30 '16 at 03:59
  • ...of the complex logarithm is chosen. My point is that although in math it makes perfect sense to say that $1$ is an integer and 0.999999999999987 is not, the difference between these may be too small for a computer. Therefore a computer will only accept as an integer what it knows is an integer exactly. – David Jan 30 '16 at 04:07
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    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. – Daniel R. Collins Jan 30 '16 at 04:10
  • Okay, but my point is, math and computer math are not the same thing. A computer usually can't tell the difference between a rational number and an irrational one. For example, if, for complicated reasons, the result of some very difficult calculation is exactly $3.4$, a computer may not be able to tell the difference between that and another, possibly even irrational, number that is very close. $(-2)^{3.4}$ makes perfect sense according to the second convention, while $(-2)^{3.400000001}$ does not (since the exponent can't be written with an odd denominator). – David Jan 30 '16 at 04:13
  • Honestly, I don't think that's an issue in this case for a symbolic math engine of this type: e.g., if you ask it "Is sqrt(2) a rational number?" then it correctly answers that it is not, while if you ask "Is 1.41421356 a rational number?" it answers that it is. You have to explicitly ask for a decimal approximation for it to return an answer in that format. (Symbolic engines like this have been in use for at least three decades now.) Anyway, bad on my part to get confused by a syntax error, thanks for your observations. – Daniel R. Collins Jan 30 '16 at 05:02
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    You are probably right about Wolfram Alpha. – David Jan 30 '16 at 05:19
  • Just to clarify your conclusion at the end: "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" is a restriction that *should be added* to the Ratti/Sullivan/OpenStax class of texts, because it's clearly not included there as printed, yes? (Similar to @DanChristensen's answer that added restrictions should be placed on both factors in the exponent.) – Daniel R. Collins Jan 31 '16 at 18:50
  • When $a < 0$, the rule $(a^x)^y = a^{xy}$ must carry the restriction that $x$ and $y$ can be written with odd denominators. – David Jan 31 '16 at 19:56
  • Nice...........+1@David – Bhaskara-III Feb 04 '16 at 22:12
  • @David You discussed this in the final paragraph of your answer but I was wondering if you could elaborate. Let's say I want to circumvent these issues and I define $a^x$ for only for positive $a$, say using the $\log$ and $\exp$ functions. Will this definition ever be, well, too restrictive, or generally problematic? – MathematicsStudent1122 Feb 16 '16 at 07:24
  • @MathematicsStudent1122 No, it's not problematic. That's equivalent to the first convention I referred to (though arrived at through a method that simultaneously considers irrational exponents). Then for $x < 0$ and a specific value of $r$, you might need to write $\pm (-x)^r$ instead of $x^r$. – David Feb 16 '16 at 07:31

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

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.

Dan Christensen
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  • 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). – Daniel R. Collins Jan 30 '16 at 03:08
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    @DanielR.Collins That would be like throwing out the baby with the bathwater. We really need results like $(-8)^\frac{1}{3}=-2$. – Dan Christensen Jan 30 '16 at 15:36
  • @DanielR.Collins See my edited version above. – Dan Christensen Jan 30 '16 at 16:08
  • 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. – Daniel R. Collins Jan 30 '16 at 20:17
  • How do you go from "provided all expressions are defined in $(a^r)^s=a^{rs}$" to "$a^r\in\Bbb R$ and $a^s\in\Bbb R$"? The latter term is not a subterm of anything in the expression. Reading the definition at face value, it is perfectly possible to just have $a^r,(a^r)^s,a^{rs}\in\Bbb R$, so that $(a^r)^s=a^{rs}=a^{sr}$ but $(a^s)^r$ is not defined. – Mario Carneiro Jan 31 '16 at 00:09
  • @MarioCarneiro Again, since multiplication is commutative, you would have $(a^r)^s = (a^s)^r$. If $a^s \notin \mathbb{R}$, then the RHS would be undefined. – Dan Christensen Jan 31 '16 at 04:18
  • Commutativity of multiplication only gives $(a^r)^s=a^{rs}=a^{sr}$. To get the last step $a^{sr}=(a^s)^r$ would require a second application of the power rule, which is stated not to be valid in the case $a^s\notin\Bbb R$. Thus $(a^s)^r$ is undefined but $(a^r)^s=a^{rs}=a^{sr}$ is, and this does not contradict anything from the author. – Mario Carneiro Jan 31 '16 at 15:55
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    @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. – Dan Christensen Jan 31 '16 at 16:55
  • Re: "We really need results like $(−8)^\frac{1}{3} =−2$". Well, all of my real analysis books do fine restricting the base to nonnegatives only. Admittedly, it's a bit irritating to have a definition of principal root in reals that gives entirely contradictory results to the complex definition. – Daniel R. Collins Jan 31 '16 at 16:55
  • Re: Sloppiness of the definition in the last paragraph note: Apparently it's the case with the complex definition that $(z^m)^\frac{1}{n} = (z^\frac{1}{n})^m$ precisely when $m$ and $n$ have no common factors (Silverman, 1975)? Perhaps that gives some motivation for this real-valued definition. – Daniel R. Collins Feb 06 '16 at 15:20
  • Sorry for the major bump, but in final logical statement, if n is an element from the set of natural numbers, this by definition means it cannot be zero and it must be either odd or even, so why do you need to state this again later? – slew123 Oct 25 '20 at 17:55
  • @slew123 It is fairly common to include 0 in the natural numbers in more advanced courses after high school. – Dan Christensen Oct 26 '20 at 21:43
  • I think it would be better to use the set of non-negative integers in that case. – slew123 Feb 01 '21 at 22:07

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.

Derek Elkins left SE
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  • 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? – Daniel R. Collins Jan 27 '16 at 06:18
  • 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 $a$. – Derek Elkins left SE Jan 27 '16 at 06:29
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    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 $1 = \frac{n}{n}$. – Derek Elkins left SE Jan 27 '16 at 06:50
  • 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. –  Feb 01 '16 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.

Mike Earnest
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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.

Daniel R. Collins
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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.

Marc van Leeuwen
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  • 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. – Daniel R. Collins Feb 04 '16 at 16:16

$(a^r)^s=a^{rs}$ 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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    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. – Marc van Leeuwen Feb 04 '16 at 05:23
  • @MarcvanLeeuwen: you are right, I shouldn't have used "used". I have rephrased to avoid the misinterpretation. –  Feb 04 '16 at 09:10

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.

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

Michael Hardy
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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}

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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.

Gyro Gearloose
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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.

Daniel R. Collins
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  • 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 $m$. – David Jan 27 '16 at 08:26
  • @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. – Daniel R. Collins Jan 27 '16 at 16:20
  • Edited this answer to make it more concise and cover both oft-seen definitions. I would still argue here that Ratti should delete the example that shows $(-8)^\frac{2}{6} = (-8)^\frac{1}{3} = \sqrt[3]{-8} = -2$. Thanks to David and DerekElkins for helpful answers/comments. – Daniel R. Collins Jan 27 '16 at 16:47
  • The fact that you attempt to use Wolfram Alpha to justify your point makes you look even more foolish, not only because you do not know how WA works internally to know whether it is asserting what you think it does, but also because WA states some bald falsehoods, which I will not mention so that they will not fix them. Whoever relies on WA without checking its answers themselves is asking for trouble. – user21820 Jan 30 '16 at 03:52
  • @user28120: In this case, which do you think it is: My misunderstanding the assertion, or a bald falsehood by Wolfram Alpha? – Daniel R. Collins Jan 30 '16 at 03:58
  • In this case, it is the first reason, namely because you used WA wrongly. See `http://www.wolframalpha.com/input/?i=%28-1%29^{2%2F2}` and `http://www.wolframalpha.com/input/?i=%28-1%29^1+%3D+%28-1%29^%282%2F2%29` for proof. You can't expect $-1 = \{-1\}$... – user21820 Jan 30 '16 at 04:04
  • Anyway if you want me to further explain anything, I would like you to first retract your criticism of my answer until you understand the mathematics. Otherwise as I said I really feel like deleting my answer and going away. – user21820 Jan 30 '16 at 04:08
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    @user21820: Thank you for drawing my attention to the incorrect syntax I used in Wolfram Alpha; I'm removing that reference from my answer. – Daniel R. Collins Jan 30 '16 at 04:09
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    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`.) – Mario Carneiro Jan 31 '16 at 00:05
  • I wouldn't take Ratti's quirky definition of rational exponents on $\mathbb{R}$ as, well, definitive. It's just silly that $4^\frac{2}{4}$ would be undefined while $4^\frac{1}{2}=2$. There may be a tendency among writers not to devote a lot of thought to what they may see as "preliminaries" in a textbook. I would be very surprised if Ratti subsequently claimed that something like $4^\frac{2}{4}$ was undefined. – Dan Christensen Jan 31 '16 at 22:56
  • @DanChristensen: Well, I couldn't call this definition "quirky" because it seems to be shared by what looks to be a reasonably wide swath of college algebra textbooks by different authors (e.g., the Sullivan text is a top-10 seller at Amazon in the category). I agree that it seems suspect, but also pretty widespread. – Daniel R. Collins Jan 31 '16 at 23:52
  • I must admit I'm become more convinced that this answer is inadequate, because another question runs into the same paradox without ever writing $a^\frac{n}{n}$, namely: for all $x$: $\sqrt{x^2} = (x^2)^\frac{1}{2} = x^1 = x$. See: http://matheducators.stackexchange.com/questions/10488/student-converted-sqrtx2-and-ended-up-with-just-x-instead-of-x – Daniel R. Collins Feb 01 '16 at 03:30
  • @DanielR.Collins Maybe some off-the-shelf boiler-plate for the "boring" chapter 0 stuff that no-one has ever taken a close look at... until now? It must be tempting for writers. And there is no copyright on definitions and theories. – Dan Christensen Feb 01 '16 at 03:33
  • @DanChristensen: I don't know, that's not how I read it. I picked Ratti because that seems to have the most painstakingly careful development in this vein; takes up 10 pages in the book. Also, I just noticed that Wikipedia also uses the same definition ("Rational powers $u/v$, where $u/v$ is in lowest terms..."), and I assume that page has been beaten on hundreds or thousands of time to get to the current state. – Daniel R. Collins Feb 01 '16 at 03:42
  • @DanielR.Collins The other "paradox" can be resolved in the same way I proposed for yours. You should only be able to infer $(x^2)^{\frac{1}{2} = x^{2\times \frac{1}{2}}$ only if $x^2\in \mathbb{R}$ and $x^\frac{1}{2}\in \mathbb{R}$, that latter being true only if $x\geq 0$. – Dan Christensen Feb 01 '16 at 03:44
  • @DanChristensen: Right, I agree that bolsters the case for that being the best answer. Actually, I used that to provide the answer on the other question. :-) – Daniel R. Collins Feb 01 '16 at 03:45
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    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 $m'=m/gcd(m,n),n'=n/gcd(m,n)$. –  Feb 01 '16 at 09:46