TITLE: How to prove that nth root of n is irrational
QUESTION [5 upvotes]: Can someone please tell me if this is correct or not?
Assume $n^{1/n}$ is rational, i.e $n^{1/n} = a/b$; $a, b$ integers and $b > 1, \gcd(a,b)=1$, and $a^n = nb^n$.
If b exists and $b>1$, it must have at least one prime factor $p$ (Fundamental Theorem of Arithmetic).
It follows that $p|b$ and $p|a^n$.
But $p|a^n$ implies $p|a$ (by Theorem: if $p$ prime and $p|uv$, then $p|u$ or $p|v$. Let $v=u^{n-1} \Rightarrow p|u^n$ implies $p|u$).
Therefore $p$ is a common factor of $a, b$ and we have a contradiction.
PS: I am very new at proofs, so please be kind and only reply with elementary proofs/concepts.

REPLY [3 votes]: As pointed out, you must show that ${n^{\frac{1}{n}}}$ is not an integer, hence ${b>1}$ for this proof to be complete. @Integrand gave an awesome proof for this. Another way you could prove this is by using the Intermediate Value Theorem.
Firstly, if ${n^{\frac{1}{n}}=k}$ for some integer ${k}$, then ${n=k^n}$. Replace $k$ with a variable, ${x}$, and consider the function
$${f(x)=x^n-n}$$
Notice that ${f(x)=0}$ gives you solutions for ${n=k^n}$. We only care about ${x\geq 0}$ (since ${n > 0}$). For the next part of the proof, we assume ${n\geq 2}$.
Notice that ${f'(x)=nx^{n-1}>0\ \forall\ x > 0}$. So ${f}$ is increasing on ${(0,\infty)}$. Next, notice that
$${f(1)=1-n < 0\ \forall\ n \geq 2}$$
and
$${f(2)=2^{n}-n > 0\ \forall\ n \geq 2}$$
So by the Intermediate Value Theorem, ${\exists\ c \in (1,2)}$ such that ${f(c)=0 \Leftrightarrow c^n=n}$. But ${c}$ is not an integer since it's between ${1}$ and ${2}$. To finish the proof, we have to argue this is the only positive root. Well, since ${f(x)}$ was shown to be strictly increasing on the domain we care about, indeed, it has to be. And we are done. We have shown ${n^{\frac{1}{n}}}$ cannot be an integer for ${n\geq 2}$.