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 & 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).

Solution 1:

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.

Solution 2:

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

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.

Solution 3:

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.