Prove that for any positive whole integer n that is not equal to $1$, the $n$'th root of $n$ is always irrational.

Question

Disclaimer, I am currently in year 9 and may have missed some key fundamentals and have made a mistake in my notation or phrasing.

For almost 6 months I have been attempting to make a proof that the nth root of n is always irrational when n is not equal to 1.

$\forall n\in \mathbb{Z}, n\neq 1, \sqrt[n]{n}\notin \mathbb{Q}$

So far I have had almost no luck at my attempts of proving this and I am wondering if someone else could help prove this statement.

Do you know the proof for the specific case of $n=2$? If so, could you adapt it to $n=3$? That is not too hard. Then $n=4$ has a wrinkle you may be able to work through. After that perhaps you will have sone ideas for how to make a more general argument. — MJD, Jun 22 '25 at 04:05
Did you mean $\not\in\mathbb Q$ where you typed $\not\in\mathbb R$? — J. W. Tanner, Jun 22 '25 at 04:08
Your title is somewhat contradictory with your body. In the title it says any positive whole integer $n$ whereas you wrote $\exists n\in \mathbb{Z}$ in the body. — Dstarred, Jun 22 '25 at 04:11
Can you use the rational root theorem? Why did you tag [tag:solution-verification] if you couldn’t find a solution? — J. W. Tanner, Jun 22 '25 at 04:13
You also probably meant to type $\forall n \in \mathbb Z$ when you typed $\exists n \in \mathbb Z$. — Torsten Schoeneberg, Jun 22 '25 at 05:11
Duplicate of How to prove: if $a,b \in \mathbb N$, then $a^{1/b}$ is an integer or an irrational number? and many others. Please search for answers before asking questions. — Bill Dubuque, Jun 22 '25 at 16:37
For a solution-verification question to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be used as a proof checking machine. — Bill Dubuque, Jun 22 '25 at 16:39

score 5 · Accepted Answer · edited Jun 22 '25 at 08:03

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I found a proof by contradiction. Let $n \in \mathbb{Z},n>1$.

Assume $\sqrt[n]{n}$ is rational. $\sqrt[n]{n}=\frac{a}{b}\Rightarrow n=\frac{a^n}{b^n}$ where $\frac{a}{b}$ is irreducible, and $a,b \in \mathbb{N}$. Since $\frac{a}{b}$ is irreducible, $\frac{a^n}{b^n}$ is also irreducible. Since n is a integer, b must equal $1\Rightarrow$ $n=a^n$. This equation has no real solutions for $a\geq2$. Since a is a natural number less than 2, $a=1 \Rightarrow n=1$, which is a contradiction since $n>1$.

edited Jun 22 '25 at 08:03

J. W. Tanner

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answered Jun 22 '25 at 05:07

d ds

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To see why there's no solutions to $n=a^n$ for $a\geq 2$, you can show $n<2^n$ by induction, and then $2^n\leq a^n$, I mention it as it wasn't explained in the solution – Bruno Andrades Jun 22 '25 at 05:29
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Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. $\ \ $ – Bill Dubuque Jun 22 '25 at 16:45
Sorry, I didn't know this was a dupe. – d ds Jun 22 '25 at 17:50
The site is a decade and a half old, so most common exercises like this are duplicates many times over. – Bill Dubuque Jun 22 '25 at 18:37

Prove that for any positive whole integer n that is not equal to $1$, the $n$'th root of $n$ is always irrational.

1 Answers1