# Is $n^\frac{1}{n}$ ever rational?

Sorry if this is a duplicate, as usual I'm struggling with how to search for this.

I was wondering to myself how to prove that you can't get a square number that is twice another square number, I.e. $$m^2=2n^2$$ and I quickly came up with a neat proof using the fact: $$\frac{m}{n}=\sqrt{2}$$ The next obvious step is cubes that are thrice another cube, etc. etc. I then realised you can use this approach to prove that any power of p cannot be p times another power of p if $p^\frac{1}{p}$ is never rational. I suspect this true, but I need to go to sleep, so can somebody help me out with a proof?

• Sure, if $n = 1$. I'm guessing, you want $n>1$? – Ennar Nov 25 '17 at 13:44
• Well yes, I do indeed – Phill Nov 25 '17 at 13:45
• Show that $\sqrt[n]{m}$ is either an integer or irrational. That's similar to the proof of irrationality of $\sqrt{2}$ or $\sqrt{3}$. Then note that $2^n > n$. – Daniel Fischer Nov 25 '17 at 13:45
• @DanielFischer This allows the stronger statement : $m^{\frac{1}{n}}$ with positive integers $m,n$ is either an integer or irrational. – Peter Nov 25 '17 at 13:51

$n^{\frac{1}{n}}$ cannot be rational for any positive integer $n>1$ (No matter whether $n$ is prime or composite)

This is because the number $n^{\frac{1}{n}}$ is a root of the polynomial $x^n-n$.

The leading coefficient is $1$, hence any rational root woule be an integer. If we denote $m:=n^{\frac{1}{n}}$, we get $m^n=n$. $m$ is clearly positive, so it would have to be a posiive integer, if it were rational.

We would have $m\ne 1$, hence $m\ge 2$, but then $m^n\ge 2^n>n$ for $n>1$, hence we arrive at a contradiction.

• Wonderful. Thank you. – Phill Nov 25 '17 at 13:59

$$1 < n^{1/n} < 2 \quad \forall n >1 , n\in \Bbb N$$

Also (I think more hint is required as downvotes are too fast) note that $n^{1/m}$ can be rational iff it is an integer.

• What exactly does this prove? – user370967 Nov 25 '17 at 13:47
• Why so many downvotes? Peter used almost the same argument in his answer. – Ennar Nov 25 '17 at 13:49
• @Math_QED It was a hint. – Jaideep Khare Nov 25 '17 at 13:52
• It is not obvious that $n^{\frac{1}{n}}$ must be an integer or irrational. (To clarify : I did not downvote) – Peter Nov 25 '17 at 14:26

If $n=p$ is prime and $p^{1/p}=\frac{m}{l}$ was rational it follows that $l^p*p=m^p$. Now use the uniqueness of prime factorization:

Let the prime number $p$ occur on the left site $x$ times and $y$ times on the right site. Then $y$ is divisible by $p$ whereas $x$ isn't. Contradiction.

Consider the polynomial $p$, such that $p(x):= x^n -n$ assume $\sqrt[n]{n}$ as a root.

By the rational root theorem we know that if $p$ has a rational root, it will be the one of the dividers $d_1,\cdots,d_m$ of $n$ (because the coefficient of monomial $x^n$ is $1$). But NONE of then will be a root of $p$. Therefore, the real roots of $p$ are all irracional roots, including $\sqrt[n]n$.

• The rational root theorem can be used to show that certain number is irration, since this number is algebraic. There, you can creat a polinomial $p, p\in\mathbb{Z}[\,x\,]$. – Gustavo Mezzovilla Nov 25 '17 at 14:23

Using part of the answer Jaideep Khare alredy posted (to give a full solution)

Lemma 1

if $a \in \mathbb{Q} \setminus \mathbb{Z}$, then $a^n \in \mathbb{Q} \setminus \mathbb{Z}$ for all $n \in \mathbb{N} \setminus \{ 0 \}$

lemma 2

if $x^m - m =0$ has a rational solution, then it must be an integer one, for all $m \in \mathbb{N}$

lemma 3

$m^{\frac{1}{m}} \in ~ ]1,2[$

The first lemma is easy to prove, the second is a consequence of the first, and the third may be also shown with ease

Suppose $n=(a/b)^n$ with $n,a,b\in\mathbb{N}$. We may assume that $a$ and $b$ have no prime factors in common. Suppose $p\mid a$, where $p$ is prime. Then $p^n\mid a^n=nb^n$. Since $p\not\mid b$, we must have $p^n\mid n$. But $p^n\ge2^n\gt n$, which is a contradiction. Hence $a$ has no prime factors, i.e., $a=1$. But $nb^n=1^n=1$ implies $n=b=1$ as well. Thus the only $n\in\mathbb{N}$ for which $n^{1/n}$ is rational is $n=1$.

Remarks: The inequality $2^n\gt n$ requires its own proof by induction. The step in which $p^n\mid nb^n$ and $p\not\mid b$ imply $p^n\mid n$ also, technically speaking, requires a touch of induction, starting from Euclid's Lemma ($p\mid xy$ implies $p\mid x$ or $p\mid y$) as the base case.