Irrational Numbers and their squares If $s$ is irrational is $s^2$ irrational? 
Looking at example (a) $s= \sqrt 2$  then $s^2= 2$, which is rational but looking at example (b) $s= 5^{1/3}$, then $s^2= 5^{2/3}$ which is irrational or $\pi^2$ is also irrational. So what's the answer? $s^2$ irrational or not? Or does it depend on the value of $s$? 
 A: In general if $a,b,c,d$ are positive integers with $a/b$ and $c/d$  in lowest terms then $(a/b)^{c/d}$ is rational if and only if both $a$ and $b$ are $d$th powers of integers.
Suppose $(a/b)^{c/d}=e/f$ where $e,f$ are positive integers with $e/f$  in lowest terms. Then $$a^c f^d=b^c e^d.$$  Suppose $p$ is any prime divisor of $a,$ and let $p^k$ be the  largest power of $p$ that divides $a.$ Then the largest power of $p$ that divides $a^c$ is $p^{k c}.$ 
Now suppose  $p^{k c+1}| a^c f^d$. Then $p|f.$ But since $a^c f^d=b^c e^d,$ we also have $p|a\implies p|b^c e^d\implies p|e$ (because $p$ does not divide $b$), but then we would have $p|e$ and $p|f,$ which is false. So $p^{k c+1}$ does not divide $a^c f^d$
Hence $p^{k c}$ is the largest power of $p$ to divide $a^c f^d=b^c e^d.$ And $p$ does not divide $b,$ so $p^{k c}$ is the largest power of $p$ to divide $e^d.$ But the largest power of $p$ to divide $e^d$ is $p^{l d}$ for some positive integer $l.$
Therefore $k c=l d.$
Finally since $c/d$ is in lowest terms, we have $k c=ld\implies d|k c\implies d|k.$ 
Therefore if $p$ is any prime divisor of $a,$ and $p^k$ is the largest power of $p$ to divide $a,$ then $k$ is a multiple of $d.$ This implies that $a$ is the $d$th power of a natural number.
To show that $b$ is also a $d$th power, let $a'=b, b'=a, e'=f, f'=e,\,$ and apply the above argument's effect on the equation $(a'/b')^{c/d}=e'/f'.$ 
A: Lets show that $s=\sqrt{\sqrt{2}}$ is irrational, which will give you a concrete example of $s,s^2$ both being irrational. 
Suppose $s=a/b$. Then $\sqrt{2}=a^2/b^2$. But this would mean $\sqrt{2}$ is rational, a contradiction.
In general, the square root of an irrational number must be irrational as well. This doesn't hold the other way around, with $\sqrt{2},2$ being an example.
