It is well known that $\sqrt{2}$ is irrational, and by modifying the proof (replacing 'even' with 'divisible by $3$'), one can prove that $\sqrt{3}$ is irrational, as well.  On the other hand, clearly $\sqrt{n^2} = n$ for any positive integer $n$.  It seems that any positive integer has a square root that is either an integer or irrational number.


  
  How do we prove that if $a \in  \mathbb N$, then $\sqrt a$ is an integer or an irrational number?
  


I also notice that I can modify the proof that $\sqrt{2}$ is irrational to prove that $\sqrt[3]{2}, \sqrt[4]{2}, \cdots$ are all irrational.  This suggests we can extend the previous result to other radicals.


  
  Can we extend 1? That is, can we show that for any $a, b \in \mathbb{N}$, $a^{1/b}$ is either an integer or irrational?
  


Theorem: If $a$ and $b$ are positive integers, then $a^{1/b}$ is either irrational or an integer.
If $a^{1/b}=x/y$ where $y$ does not divide $x$, then $a=(a^{1/b})^b=x^b/y^b$ is not an integer (since $y^b$ does not divide $x^b$), giving a contradiction.
I subsequently found a variant of this proof on Wikipedia, under Proof by unique factorization.
The bracketed claim is proved below.
Lemma:  If $y$ does not divide $x$, then $y^b$ does not divide $x^b$.
Unique prime factorisation implies that there exists a prime $p$ and positive integer $t$ such that $p^t$ divides $y$ while $p^t$ does not divide $x$. Therefore $p^{bt}$ divides $y^b$ while $p^{bt}$ does not divide $x^b$ (since otherwise $p^t$ would divide $x$).  Hence $y^b$ does not divide $x^b$.
[OOC: This answer has been through several revisions (some of the comments below might not relate to this version)]
