What's the domain of $x^{2/6}$? $\mathbb{R}$ or $[0, \infty)$?

Question

Is domain of $f(x) = x^{1/3}$ same as $f(x) = x^{2/6}$?

This is a exam problem in pre-calculus which asked us to write domain of $x^{2/6}$. I think the answer should be $[0, \infty)$ but the teacher said $\mathbb{R}$.

It's equivalent to how to define the domain of $x^{p/q}$ with $p, q \in \mathbb{N}$?

Method1(The teacher uses): We need to let $p, q$ coprime, then compute the domain. In this case, the domain should be same. That is $x \in \mathbb{R}$.

Method2(I use): $x^{p/q} \equiv \left(\sqrt[q]{x}\right)^p \equiv \sqrt[q]{x^p}$. Then domain of $f(x) = x^{1/3}$ is $\mathbb{R}$. For domain of $f(x) = x^{2/6}$, there is no universal definition for $x < 0$. $((-1)^2)^{1/6} = 1$ but $((-1)^{1/6})^2$ is not well-defined in $\mathbb{R}$ or multivalued in $\mathbb{C}$ $e^{i (\frac{1}{3}\pi + \frac{2n}{3}\pi)}$. Therefore, we need to exclude negative number, that is, domain is $[0, \infty)$. As shown in wiki and proofWiki, it doesn't require to make $p, q$ coprime first.

Which one is correct? That is what's the precedece of viewing the symbol $x^{2/6}$.

Answer 1

The issue here is two-fold:

1. Since $1/3 = 2/6$, the domains of $f(x) = x^{1/3}$ and $g(x) = x^{2/6}$ must agree, as they are the same function. To say otherwise is to say $1/3 \neq 2/6$ because they have different numerators and denominators, and then we're not actually talking about rational exponents anymore.
2. The "domain" we talk about in a class like pre-calculus is more appropriately called an "implied domain". Roughly speaking, an implied domain is the largest set of input values such that the function makes sense at those values.

Together, this means that $f(x) = x^{2/6}$ is identically $f(x) = x^{1/3}$, and we agree [hopefully] that $f(x) = x^{1/3}$ makes sense for any real number (in the sense of every real number has a unique real cube root), so the implied domain at least includes $\mathbb{R}$. As that is the largest domain we consider in this type of class, the implied domain of $f(x) = x^{2/6}$ is $\mathbb{R}$.

Answer 2

@BrianMoehring's post covers the correct answer and rationale why $x^{1/3} \equiv x^{2/6}$, so they have the same (implied) domain.

The following is more about the confusion that prompted the question in the first place. It is essentially the same confusion as in How does the exponent of a function effect the result? (whether $x^{2/2} = x$) or the ever recurring Why $\sqrt{-1 \cdot {-1}} \neq \sqrt{-1}^2$? and What are the Laws of Rational Exponents?.

Method2 (I use): $x^{p/q} \equiv \left(\sqrt[q]{x}\right)^p \equiv \sqrt[q]{x^p}$.
...
As shown in wiki and proofWiki, it doesn't require to make $p, q$ coprime first.

The first equalities both derive from the power rule for exponents $\,x^{ab}=\left(x^a\right)^b\,$. That rule is ingrained in math memory since early algebra, since it's always true for integer exponents. For rational (or real) exponents $a,b$, however, the identity is not necessarily true, or even defined, when $x \lt 0$. The only case where the identity is guaranteed to always work is when $x \gt 0$.

Both the wiki and proofWiki pages linked by the OP make the assumption of positive $x$ very clear.

Which one is correct? That is what's the precedece of viewing the symbol $x^{2/6}$.

The power notation is $\text{base}^\text{exponent}$. The expression $x^{p/q}$ always means the base $x$ raised to power $p/q$. That is the one and only possible interpretation, which is why extra parentheses like $(x)^{(p/q)}$ are redundant and never used. It is not even a matter of "precedence", it's just implicit in the notation that the base and exponent are evaluated separately before exponentiation.

The temptation to write that $x^{p/q}=\left(x^p\right)^{1/q}$ or $=\left(x^{1/q}\right)^{p}$ comes from memories of the power rule for exponents, and confusion about when it is legal and justified to use it.

Answer 3

I would suggest that you just have to ask your teacher for their definition of "$x^{2/6}$". It sounds like you were originally given a narrow definition for "$x^{p/q}$" where $p,q$ are relatively prime. It is then not clear if that fraction bar is really division, or if the "$p/q$" is just textual notation. And if it was just a suggestive notation, then the question becomes if "$x^{2/6}$" is simply undefined, or if you are expected to pick up the same definition that was given for "$x^{p/q}$" except leaving out the relatively prime requirement. These all lead to different answers, and without guidance from your instructor about what they meant, more than one answer is valid.