Prove that $(n+\sqrt n)^\frac{1}{2}$ is irrational for every $n\in\Bbb N.$
I guess the only thing I can do here is assume the opposite, i.e. $(n+\sqrt n)^\frac{1}{2}=\frac{p}{r}$ and somehow see the contradiction. If I square both sides I get $n+\sqrt n=\frac{p^{2}}{r^{2}}$ and now I don't know what to do next.
If $n$ is a perfect square ($n=k^2$), the expression reduces to $\sqrt{k^2+k}$ for $k\in\Bbb N$. However, $k^2<k^2+k<(k+1)^2$, so $k^2+k$ is not a perfect square and $\sqrt{n+\sqrt n}$ is irrational.
If $n$ is not a perfect square, assume $\sqrt{n+\sqrt n}$ to be rational. Then $n+\sqrt n$ and therefore $\sqrt n$ are rational, since rational numbers are closed under the four operations. But since $n$ is not a perfect square, $\sqrt n$ is irrational and a contradiction arises. Hence $\sqrt{n+\sqrt n}$ is irrational.