Can $x^{p/q}$ be rigorously proven/disproven to be extended to a larger subset of reals that includes negative real numbers?

Question

There have been numerous arguments that $x^{p/q}$, if $p/q$ is not an integer, should not be extended to a subset of reals that includes negative non-integers $x$-values. Many have concluded that this is context-based since definition of $x^{p/q}$ at $x<0$ is ambiguous.

Before I begin, I shall note some confusion regarding $x^{p/q}$ vs $\sqrt[q]{x^{p}}$. Mathematicians who use the computer algebra system see $x^{p/q}$ as a relation equivelant to $y^{q}=x^{p}$ whereas $\sqrt[q]{x^p}$ is the positive solution/function when solving for $y$ in the relation. So note that I, and other mathematicians, see $\sqrt[q]{x^p}$ and $x^{p/q}$ as equivalent functions.

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ISSUES WITH $x^{p/q}$ BEING DEFINED AT NEGATIVE VALUES WHERE $p/q$ IS NOT AN INTEGER.

The first issue is some mathematicians believe, since $x^{p/q}=e^{\ln{\left(x\right)}{(p/q)}}$ is true for only positive real numbers, that $x^{p/q}$ should be restricted to only positive real numbers. The problem is we must apply this rule to the function if $p/q$ is an integer.For example, we can't restrict $y=x$ to only positive integers.

The second issue is that $x^{p/q}$ is not continuous to complex numbers when $x<0$. For example $\lim_{\epsilon\to0}{\left(x+\epsilon{i}\right)}^{p/q}\neq{x^{p/q}}$ unless $p/q$ is an integer. This may be true, but it does not prove that $x^{p/q}$ can never be extended to negative real numbers rather to avoid it.

A similar issue is whether $\lim_{a{\to}{p/q}}x^{a}=x^{p/q}$. Suppose we choose a ${p_1}/{q_1}$ close to $p/q=2/3$. If we take ${p_1}/{q_1}={667}/{1000}$ the problem is since the denominator is even, the equation does not include negative numbers. Thus you have to choose any $p_1$. This can also be applied to $\lim_{a\to{y}}({-2)}^{a}$ shown Andre Nicolas answer in Find the domain of $x^{2/3}$

The last issue is that, $x^{p/q}{\neq}\sqrt[q]{x^{p}}{\neq}{\left(\sqrt[q]{{x}}\right)^{p}}$ if $p/q$ is not reduced. For example, if we set $x=-2$ with $p/q=-2/6$ we have $\sqrt[6]{{(-2)}^{2}}\neq{\left(\sqrt[6]{(-2)}\right)}^{2}$. Along with this if you take an reduced $p/q$ and an unreduced $p/q$ (${p_1}/{q_1}$) you can find that $\sqrt[q]{{x}^{p}}\neq\sqrt[q_1]{{x}^{p_1}}$. For example $\sqrt[3]{(-2)^{1}}\neq\sqrt[6]{(-2)^{2}}$. This brings me to a point whether ${p_1}/{q_1}$ must be simplifed before being evaluated? If this is significant, it can cause issues in our mathematical foundations.

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QUESTIONS

Can it be a rigorously proved that $x^{p/q}$ can be extended to negative real numbers if $p/q$ is a non-integer and $q$ is an odd integer?

If $x^{p/q}$ cannot be proved or disproved to be extended to larger subset of reals including negative numbers then in which branches of mathematics (topology, complex analysis, real analysis etc.) can such an extension exist?

For more understanding see the following posts.

$(-27)^{1/3}$ vs $\sqrt[3]{-27}$.

Find the domain of $x^{2/3}$

What is the Domain of $f(x)=x^{\frac{1}{x}}$

Answer 1

The domain of a function is part of the function definition. It makes no sense to ask, "must $f(x)$ have the negative numbers as part of its domain?" You look at the definition of $f(x)$, and see if the negative numbers are included or not, and that's the end of the story.

There is a tendency (particularly in lower-level math classes) to ask question of the form, "what is the domain of the function $f(x) = \sqrt{x}$" as if the domain of the function were somehow external to the function itself. This is a meaningless question and worse, sows confusion and misconceptions about the nature of functions. One might try to sharpen it by asking, "for what $x$ is the expression $f(x)$ well-defined," and at least this question is meaningful if all terms in the definition of $f$ are unambiguously defined. This is not the case for even simple mathematical notation like $f(x) = \sqrt{x}$: a case could be made for any of the following potentially correct answers:

• the square integers
• ratios of square integers
• positive real numbers
• all real numbers
• all complex numbers
• positive-definite matrices
• etc.

So now to your question: you can define your function to include the negative numbers, or to exclude them. There is certainly no requirement to include the negative numbers as part of the domain of your function. But you mention several properties that you want your function to have, which leads to what is perhaps your real question: you want a function $f(x)$ where

1. $f(x)$ is continuous;
2. the domain of $f$ is an open subset of the real numbers and a superset of the positive reals;
3. $f(x)^q = x^p$ when $x$ is a positive real number;
4. some other natural manipulation rules hold involving $f$ that you may not have spelled out above (it is up to you which you decide to include).

You can then ask: is it possible to extend the domain of $f$ to a larger subset of the reals that includes (some) negative real numbers? It is important to understand that the answer is context-dependent: if you change the rules above, you can change the answer. It is impossible to deduce any kind of canonical domain from the expression $x^{p/q}$ alone.

Answer 2

I do not know if this settles your doubts but, for me, the problem is not so much in the domain as it is in the range.

You could define $\sqrt[q]{x^p}$ for $x<0$ without any problem if p is even integer because $q$-th root of positive real number exists and is unique.

But in the case when $p$ is an odd integer then $\sqrt[q]{x^p}$ for $x<0$ is $q$-th root of a negative number and then we are starting to enter into the domain of complex numbers and then $\sqrt[q]{x^p}$ would have $q$ solutions.

So you would have a "function" that at some points take one value and at some other points more than one value, of course that you could still view it as some kind of function but it is really not particularly useful function of a real variable.