What are the Laws of Rational Exponents? 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 &amp; McWaters, Precalculus: a right triangle approach, section P.6): 




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).
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 $a &gt; 0$ 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 $a &lt; 0$.

The rules for exponents break down if you start allowing $a &lt; 0$ 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 $a^2 &gt; 0$.

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 $a &gt; 0$. 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 $a &lt; 0$, following the second convention. The reason is that the formula $\frac{d}{dx}(x^r) = rx^{r-1}$ is perfectly valid for $x &lt; 0$ and $r$ with odd denominator.
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 $a &gt; 0$, 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.