Square of an irrational number can be a rational number e.g. $\sqrt{2}$ is irrational but its square is 2 which is rational.
But is there a irrational number square root of which is a rational number?
Is it safe to assume, in general, that $n^{th}$-root of irrational will always give irrational numbers?
Best Answer
Obviously, if p is rational, then p2 must also be rational (trivial to prove).
$$ p \in \mathbb Q \Rightarrow p^2 \in \mathbb Q. $$
Take the contraposition, we see that if x is irrational, then √x must also be irrational.
$$ p^2 \notin \mathbb Q \Rightarrow p \notin \mathbb Q. $$
By negative power I assume you mean (1/n)-th power (it is obvious that $(\sqrt2)^{-2} = \frac12\in\mathbb Q$). It is true by the statement above — just replace 2 by n.