$n$-th root for a negative $n$

The $n$-th root of a number $x$, where $n$ is a positive integer, is a number $r$ which, when raised to the power $n$ yields $x$. [Wikipedia.org]

But I don't see any problem with the following example:

$$\sqrt[-1]{\frac{1}{2}} = 2$$

Why $n$ has to be a positive integer? Is there any good reason why we should restrict only to positive integers?

Some definitions around the web: Wikipedia, PlanetMath, ICoachMath, etc...

• Have you learned that $b^\frac 1 n = \sqrt[n]{b}$? The negative is dealt with in the usual fashion, as a reciprocal. – pjs36 Nov 30 '15 at 22:35
• I have learned that. In fact, I used that in the example above: $\sqrt[-1]{\frac{1}{2}} = \left(\frac{1}{2}\right)^{\left(\frac{1}{-1}\right)} = \left(\frac{1}{2}\right)^{-1} = 2$. But the definition of n-th root restricts the degree of the root to a positive integer for some reason. – Jeyekomon Nov 30 '15 at 22:52
• OK, so maybe we should interpret the definition as being silent on negative indices (not necessarily preventing them) and just telling us what to do with positive ones. – pjs36 Nov 30 '15 at 22:57
• The thing is that we had to solve this problem in our math class: $\sqrt[x]{0,25}=4$ and our teacher told us that it has no solution... – Jeyekomon Nov 30 '15 at 23:10
• The one reason I (I think) is because it is based on taking the roots of polynomials (or the inverse of polynomials). As we can see $\sqrt[-x]{x}$ does not have the inverse of a polynomial but a rational function. Then for fractional $n$ such as $\frac{1}{2}$ we are taking the inverse of a root function. – Arbuja Jan 11 '16 at 21:43

Ultimately this is a question of notation, and function domains. Let's start with the non-controversial territory:

• If $x \neq 0$ and $n$ is an integer, then $x^{-n} = 1/x^{n}$.

$x \geq 0$ is real, and $n$ is a positive integer:

• There exists a unique non-negative real number $y$ satisfying $y^{n} = x$, often written $y = \sqrt[n]{x}$. (Existence of $y$ is a non-trivial theorem of elementary real analysis.)

• The number $\sqrt[n]{x}$ is sometimes denoted $x^{1/n}$; this is often taken to be the definition of raising a non-negative real number to the power $1/n$.

• If $m$ is a non-negative integer, then $(x^{1/n})^{m} = (x^{m})^{1/n}$; the common value is usually denoted $x^{m/n}$. (I'm implicitly defining $x^{0} = 1$ for all $x \geq 0$, even if $x = 0$. If this rankles, feel free to assume $x > 0$ throughout this section.)

• If $m$ is negative, and if $x > 0$, then $(x^{1/n})^{m} = (x^{m})^{1/n}$, and again the common value is denoted $x^{m/n}$.

• Subject to the preceding assumptions, one has, for all rational numbers $r$ and $s$, $$x^{r + s} = x^{r} \cdot x^{s},\qquad (x^{r})^{s} = x^{rs} = (x^{s})^{r}.$$ Particularly, $x^{(km)/(kn)} = x^{m/n}$.

• If $r$ is real and $x > 0$, it's common to define $x^{r} = e^{r \log x}$. This definition agrees with the preceding algebraic definition for rational exponents, and is the only continuous "extension" of exponentiation to arbitrary real exponents.

$x < 0$ is real, and $n$ is a positive integer:

• If $n$ is odd, there exists a unique real $y$, necessarily negative, such that $y^{n} = x$. For every integer $m$, we have $x^{m/n} = (x^{1/n})^{m} = (x^{m})^{1/n}$

• If $n$ is even, there does not exist a real $y$ such that $y^{n} = x$.

Cautions:

• One often agrees that $(-1)^{2/6} = \bigl((-1)^{2}\bigr)^{1/6} = 1$, while $(-1)^{1/3} = -1$.

• More generally, if you want to take roots of negative numbers by allowing complex-valued roots, you have to be careful with branches of complex roots; equations such as $(x^{a})^{b} = x^{ab}$ are not true without qualification. (Related issues are discussed in great detail elsewhere at Math.SE.)

In my experience, the notation $\sqrt[n]{x}$ with $n$ not an integer greater than $1$ is at best rare, possibly non-existent. One can certainly define $\sqrt[r]{x} = x^{1/r}$ subject to the preceding conventions (e.g., for arbitrary non-zero real $r$ if $x > 0$), in which case $\sqrt[r]{0.25} = 4$ (i.e., $0.25^{1/r} = 4$ in conventional notation) does have a solution, $r = -1$. This seems like an unsatisfying way of proving one's teacher wrong, however.

• The reason I did not talk about the exponents with negative bases is because there is not much controversy with negative bases raised to a negative integer. Now exponents of a negative fraction can be an issue. – Arbuja Jan 11 '16 at 22:26
• Amazing answer (+1) maybe you could explain why fractional exponents in general (including positive fractions) are not included in the definition of the $n$th root. – Arbuja Jan 11 '16 at 23:21
• @Arbuja: Thanks. By "fractional exponents ... are not included", do you mean "Why isn't there a special radical notation for $x^{m/n}$?"; or "Why don't we write \sqrt[1/n]{x}$?"; or something else...? – Andrew D. Hwang Jan 12 '16 at 0:35 • I meant "you could explain why the definition of the$n$th root includes integers and not fractions." – Arbuja Jan 12 '16 at 1:31 • @AndrewD.Hwang Thank you! That's exactly what I hoped to see... It makes much more sense now. I also realized that the topic is actually much more complicated and controversial than I thought. BTW.: I didn't post the question to prove either my teacher or anyone else wrong, I'm just sometimes curious about the details that our teacher leaves unexplained. – Jeyekomon Jan 15 '16 at 16:17 One reason for not including negative integers is because we are solving an equation that is inverse of a polynomial. Specifically a monomial. When we are taking$y^{n}=x$, if$n$has to be positive integer, it must be because we are solving an equation that is the inverse of only a monomial function.In other words we are solving for the$n$th root by solving for the$y$of the inverse of$x^{n}=y$I believe the terminology of the roots of a function came from the definition of the$n$th root of monomials. The roots of the function or relation are the values where$f(x)=0$or$f(y)=0$. Suppose$x=a$, where$a$is real number, then the$n$th root means solving$y^{n}-a=0$. For more information look here. In the case of the negative integer$n$in$y^{n}=x$we are solving for the$y$of the inverse of a rational function (ex:$x=\frac{1}{y^2}$) but not the inverse of a monomial. Now if there is no connection of$n$th roots to solving the inverse of a monomial, we could have included fractional values of$n$(ex:$y^{1/2}=x$). However, then again, we are solving the inverse of a root function, not a monomial. Edit: If we take$y^{n}=x$and$n$is an unreduced negative fraction and$x$is negative number then there are certain problems with solving for$y$($y^{3}=-2$is not$y^{6/2}=-2$and$(-2)^{1/3}{\neq}(-2)^{2/6}$since$\sqrt[3]{-2}{\neq}\sqrt[6]{(-2)^{2}}$) . On the other hand if n is a reduced fraction then there are no issues. Thus if$n$is a reduced negative integer you can use the identity$x^{n}=\frac{1}{x^{-n}}\$.