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What is the domain of $\sqrt[x]{a}$, and is $\sqrt[x]{a}=a^{1/x}$ always true??

I was told that the domain of $\sqrt[x]{a}$ is natural numbers and the domain of $a^{1/x}$ is real numbers, so they are not identical. Is it true??

Martin Sleziak
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Harshit Joshi
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  • Did you mean $\sqrt[x]{a}$ or $\sqrt[a]{x}$? – Mandelbrot Sep 01 '18 at 12:22
  • $\sqrt[x]{a} = a^{1/x}$. By making the substitution $y = \frac{1}{x}$, we can write this as $a^y$, which is the exponential function with base $a$. For $a > 0$, this is defined for $y \in \mathbb{R}$. Going back to $x$, we have $x = \frac{1}{y}$, which is defined for all nonzero $x$. If $a < 0$, things are more complicated. – Xander Henderson Sep 01 '18 at 12:22
  • @Bruce No I meant what I wrote. – Harshit Joshi Sep 01 '18 at 12:50
  • @Xander Henderson What i am asking is whether $\sqrt[x]{a} = a^{1/x}$ is always true or does the radical sign introduces new domain restrictions like $x$ only belonging to natural numbers?? – Harshit Joshi Sep 01 '18 at 12:54
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    @HarshitJoshi What does the notation $\sqrt[x]{a}$ mean? Without a definition, it is just notation. There are a number of ways of interpreting this notation, and the answer to your notation almost certainly depend upon which interpretation you take. – Xander Henderson Sep 01 '18 at 15:54
  • As Xander said, you will have to define your notation before using it. In my opinion, "$\sqrt[n]{x}$" should always be defined the same as "$x^{1/n}$. But there are 2 commonly used but incompatible ways to do so. See here for one way, which makes $(-1)^{1/3} = -1$. Another way is $x^y = \exp(y·\ln(x))$ with the principal branch of $\ln$ (continuously extended to the boundary except at $0$), which makes $(-1)^{1/3} = \exp(iπ/3)$. – user21820 Sep 01 '18 at 16:02

2 Answers2

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It is not common to use the notation $\sqrt[a]{b}$ for operands other than

  • $a$ is an integer $\ge 1$, and
  • $b$ is a non-negative real.

Then $\sqrt[a]{b}$ means the unique non-negative real $x$ that solves $x^a=b$.

Since $\sqrt[1]{b}$ and $\sqrt[2]{b}$ have simpler notations, you would rarely write them explicitly -- but if you have an $\sqrt[n]{\phantom{a}}$ with a variable $n$, it should be unproblematic to allow $n$ to be $1$ or $2$.

You can certainly decide to use the notation for a wider range of operands. For $a$ and $b$ where you have a definition of $b^{1/a}$ that makes sense, it is generally harmless to define $\sqrt[a]{b}$ to mean that. It would be a nice touch to warn your readers that you're using this extended meaning, though.

(If you have $a$ and $b$ where $b^{1/a}$ doesn't make sense to you, then you could still define some meaning for $\sqrt[a]{b}$ if you want to -- but then you definitely need to present this meaning explicitly. And if you define things such that $\sqrt[a]{b}$ and $b^{1/a}$ both exist but are not equal, then the resulting chaos and confusion will be on you).

hmakholm left over Monica
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  • This question arose because I was doing a question for finding number of $x$ when $log(4)+(1+\frac{1}{2x})log(3) = log(\sqrt[x]{3} + 27)$ and was getting t $x = 1/2$ and $x=1/4$ so the answer is 2. But the answer is mentioned to be 0 as "x can only be a natural number", so I was wondering whether it's just my book or is that the convention. – Harshit Joshi Sep 01 '18 at 16:31
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    @HarshitJoshi: It's just your book. It's definitely not a mathematical reason to say that an equation can only have such-and-such solutions just because the "$\sqrt[·]{·}$" notation only allows natural number root... In fact, I would rather say that a question is ill-defined if it asks you to solve for $x$ without specifying what kind of creature $x$ is. =) – user21820 Sep 01 '18 at 16:41
  • @user21820 Okay thanks for all the answers everyone. – Harshit Joshi Sep 01 '18 at 16:42
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Were you also told to not split infinitives and that sentences shouldn't end with prepositions like "by" or "with"?

By convention we agree that $\root x \of a$ refers to the principal $x$th root of $a$. If $x$ is a positive real integer, then $a$ has precisely $x$ roots, and each of the other roots can be obtained by multiplying the principal root by the appropriate $x$th root of 1.

For example, $\root 4 \of 4$ is the principal root of $x^4 - 4$, which of course can be simplified to $\sqrt 2$, roughly 1.414213562373. The other roots are $-\sqrt 2$, $i \sqrt 2$ and $-i \sqrt 2$, which consist of $\sqrt 2$ multiplied by the quartic roots of 1 in turn: $-1$, $i$, $-i$.

And just to make sure I get some flack for this answer, I'm going to say these four roots can just as validly be expressed as $\sqrt 2$, $-\sqrt 2$, $\sqrt{-2}$ and $-\sqrt{-2}$.

In the case of $\sqrt{-2}$, it has two roots, and we can probably come to the agreement that it represents the principal root of 2 (which is $\sqrt 2$) multiplied by $i$.

Some people will be pedantic and tell you that $\sqrt{-2}$ is undefined and you really should write $i \sqrt 2$. But if everyone understands that's what you mean, then what is the problem?

If we agree that the radical symbol stands for a principal root, we should also agree that $a^{\frac{1}{x}}$ also stands for a principal root.

Lastly, I'd like to mention that, depending on your TeX installation, you may or may not be able to use \root x \of a instead of \sqrt[x]{a}. I personally prefer the former to the latter.

Robert Soupe
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