Is $n^{th}$ root of $2$ an irrational number? 
Possible Duplicate:
$a^{1/2}$ is either an integer or an irrational number. 

Will every $n^{th}$ root of $2$ be an irrational number? If yes, how can I prove that?
 A: If $\rm\ \sqrt[n] c = a/b,\ \ gcd(a,b) = 1\ $ then $\rm\ c\:\! b^n = a^n \Rightarrow \ b\:|\: a^n.\,$ $\rm\ gcd(a,b) = 1\Rightarrow gcd(a^n,b) = 1\ $ by here, so $\,\rm b = 1\ $ hence $\rm\:\ c \  =\ a^n.\, $ In particular $\rm\ c = 2 \Rightarrow\ n = 1.$
There are many possible variations on such irrationality proofs, e.g. using the Rational Root Test or directly using Unique Factorization of integers, or using the principality of denominator ideals. Perhaps the most elegant is to employ Dedekind's notion of the conductor  ideal - which yields a one-line proof that a  $\rm PID$ is integrally closed, i.e. satisfies the monic case of the rational root test.
A: Yes. In fact, for every integer $k$ and every $n\gt 1$, the $n$th root of $k$ is either an integer or irrational.
One way to prove it is to use exactly the same idea as for proving the square root of $2$ is irrational: suppose $\sqrt[n]{k} =\frac{p}{q}$, with $p$ and $q$ integers, relatively prime. Then $q^nk = p^n$. Now think about the prime factorizations: every prime that divides $q$ must divide $p$, but $p$ and $q$ are relatively prime, so $q=1$. That means that you must have $k=p^n$ with $p$ an integer. That is, the only way for the $n$th root of $k$ to be a rational is if $k$ is an $n$th power of an integer.
Or you can use the Rational Root Test: an $n$th root of $k$ is a root of the polynomial $x^n - k$. But a rational root of a polynomial with integer coefficients that is written in lowest term $\frac{p}{q}$ must have denominator $q$ that divides the leading coefficient and numerator $q$ that divides the constant coefficient. So any rational root of $x^n-k$ must be an integer.
Getting this back to your question, since $2$ is not an $n$th power of an integer for any $n\gt 1$, $\sqrt[n]{2}$ is not a rational for any integer $n\gt 1$. 
