irrationality proof of $\sqrt{n}+\sqrt{n+1}$ for any $n>0$ irrationality proof of $\sqrt{n}+\sqrt{n+1}$ for any $n>0$
My attempt:
$\sqrt{n}+\sqrt{n+1}=\frac{p}{q}$
$2n+1+2\sqrt{n(n+1)}=\frac{p^2}{q^2}$
Now we have to show $2n+1+2\sqrt{n(n+1)}$ cannot be a perfect squere but I don't know how.

$2n + 1 + 2\sqrt{n(n+1)}$ is irrational if and only if $\sqrt{n(n+1)} = \sqrt{n^2 + n}$ is irrational. Then observe that
$$n^2 < n^2 + n < n^2 + 2n + 1 = (n+1)^2,$$
which implies that indeed $n(n+1)$ is not a perfect square. But the square roots of positive integers are either positive integers or irrational, by the rational root theorem or Gauss' lemma. So for $x=\sqrt{n} + \sqrt{n+1}$, we see $x^2$ is irrational, and hence $x$ is irrational.
