For what $x$ does $\sqrt[x]{n}$ make sense?

Question

I am confusing myself, I think, with the use of the radical (nth root) symbol.

I'm clear on the meaning of $\sqrt{n}$ or ${\sqrt[3]{n}}$ or ${\sqrt[1000]{n}}$ etc.

But what about when we have a general $\sqrt[x]{n}$? For what values of x does this make sense?

According to the internet it's implied that x is a positive integer, and in a problem I'm working on I am finding myself writing:

"Choose $ \delta \le \min \{ \delta_1 , \sqrt[\alpha]{\epsilon\over K} \}.$ We know that $\sqrt[\alpha]{\epsilon\over K}$ exists since $\alpha > 0$"

Does that make sense when the only thing we know about alpha is that it is greater than 0 (it could be any positive real number, presumably)?

I'm aware that using fractional powers clear up ambiguity, but they also permit alpha to be negative, which doesn't work with my proof since I'm trying to show a choice of delta that depends on having a positive alpha.

Answer 1

This is actually a more subtle thing than many realize. First of all, assuming we begin with a nonnegative base $b\ge 0$, then exponentiation and taking $n$-th roots are really the same thing. That is,

$$b^n=b^{\frac{1}{1/n}}=\sqrt[1/n]b\quad\text{and}\quad \sqrt[n]b=b^{1/n}.$$

The question is how we define $b^n$ for an arbitrary real number $n$. First we convince ourselves of the basic properties / desired properties: for $b\ge 0$, we should have $$b^{x+y}=b^xb^y,\quad b^x=\dfrac{1}{b^{-x}},\quad (b^x)^y=b^{xy},\quad \text{and}\quad b^{1/x}=\sqrt[x]b.$$

Here is a summary of this process.

• Let $x\ge 0$. First we define $x^n$ for $n\in\{0,1,2,\ldots\}$ where $x^0=1$ and $x^n=\underbrace{x\cdot x\cdot\cdots \cdot x}_{\text{n factors}}$
• Next we define $x^{1/n}$ to be the real number $y$ such that $y^n=x$. Note that the completeness axiom implies that $y$ exists and is unique.
• From the properties of exponents, we have defined $b^q$ for any $q\in \Bbb Q$.
• Finally, we define $b^r$ for $x\in \Bbb R$ as follows: let $\{a_n\}$ be a sequence of rational numbers converging to $r$. Such a sequence exists, for example, but just taking longer and longer decimal approximations to $r$. Then we simply define $$b^r=\lim_{n\to \infty}b^{a_n}.$$

This is basically the entire story. There are some details to fill in (such as uniqueness of $b^r$ in the last step), but you get the point.

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Intuitively, real exponents are defined to be exactly what they must be in order to make $b^x$ a continuous function.

For example to obtain $2^\pi$, take the sequence $3,3.1,3.14, 3.141,\ldots$. We can successively approximate $2^\pi$ with the sequence

$$2^3,\ 2^{3.1},\ 2^{3.14},\ \ldots.$$

This gives approximately

$$8,\ 8.57,\ 8.81524,\ 8.82135$$

and in the end we have a limit

$$2^\pi=8.8249778270762876239\ldots.$$

You can't write this number much better than as "$2^\pi$," but that's ok because we know the number exists, and we can approximate the decimal expansion with arbitrary accuracy.

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Edit: I'm not sure if I interpreted your question correctly. Once you have exponentiation by any real number, expressions like $\sqrt[r]x$ are defined to mean $x^{1/r}$.

Answer 2

But what about when we have a general $\sqrt[x]{n}$? For what values of $x$ does this make sense?

• In real analysis, the commonly used notation $\sqrt[x]{y}$ is defined only for positive real number $y$ and positive integer $x$. To be precise, for given positive real number $a$, and positive integer $n$, $\sqrt[n]{a}$ is defined to be the unique positive real number $y$ such that $y^n=a$. (One also writes $\sqrt[n]{0}=0$ for any positive integer $n$.) The existence of such number is proved by the least upper bound property of real numbers. The number $\sqrt[n]{a}$ is called the $n$-th root of $a$.
• One uses the $n$-th root as one step in defining the exponential function $f(x)=a^x$ ($a>0$ and $a\neq 1$).
• For $a>0$, one could define $\sqrt[x]{a}$ for any $x\neq 0$; but in the case when $x$ is not a positive integer, one would use the notation $a^{\frac{1}{x}}$ instead. There is no point inventing a new notation like $\sqrt[\pi]{2}$ or $\sqrt[-3]{7}$ when what one really means are $2^{\frac{1}{\pi}}$ and $7^{\frac{1}{-3}}$.

and in a problem I'm working on ...

According to this comment of yours, you are seemingly asking an XY problem. What you want to show is the following:

Let $I$ be an interval on $\mathbb{R}$ and suppose $f:I\to\mathbb{R}$ is a function such that there exist $M,\alpha>0$, $|f(x)-f(y)|\leq M|x-y|^\alpha$ for all $x,y\in I$. Show that $f$ is continuous on $I$.

The estimate you want is for some $|x-y|<\delta$, $$ |x-y|^\alpha<\frac{\epsilon}{M}\tag{1} $$ When $|x-y|=0$, (1) is trivially true by convention that $0^\alpha=0$. When $0<|x-y|$, (1) is true if and only if $$ |x-y|<\left(\frac{\epsilon}{M}\right)^{1/\alpha}. $$ because $f(z)= z^{\alpha}=e^{\alpha\log z}$ is an increasing function on $(0,\infty)$.

Answer 3

For $\sqrt[x]c$, where $x$ and $c$ are complex numbers, $x$ makes sense for any nonzero complex number.

$\sqrt[0]c$ is undefined.