There is a very simple proof by means of divisibility that $\sqrt 2$ is irrational. I have to prove that $\sqrt 3$ is irrational too, as a homework. I have done it as follows, ad absurdum:  Suppose $$\sqrt 3= \frac p q$$  with $p/q$ irreducible, then  $$\begin{align} &amp; 3q^2=p^2 \\ &amp; 2q^2=p^2-q^2 \\ &amp;2q^2=(p+q)(p-q) \\ \end{align}$$  Now I exploit the fact that $p$ and $q$ can't be both even, so it is the case that they are either both odd, or have different parity. Suppose then that $p=2n+1$ and $q=2m+1$  Then it is the case that  $$\begin{align} &amp;p-q=2(n-m) \\ &amp;p+q=2(m+n+1) \\ \end{align}$$  Which means that  $$\begin{align} &amp;2q^2=4(m-n)(m+n+1) \\ &amp;q^2=2(m-n)(m+n+1) \\ \end{align}$$  Then $q^2$ is even, and so is then $q$, which is absurd. Similarly, suppose $q=2n$ and $p=2m+1$.  Then $p+q=2(m+n)+1$ and $p-q=2(m-n)+1$. So it is the case that  $$\begin{align} &amp;2q^2=(2(m-n)+1)(2(m+n)+1)\\ &amp;2q^2=4(m^2+m-n^2)+1 \\ \end{align}$$  So $2q^2$ is odd, which is then absurd.    Is this valid?
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?
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