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.
 A: $2n + 1 + 2\sqrt{n(n+1)}$ 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.
A: If $\sqrt{n+1}+\sqrt{n}$ is rational, so is
$$
\frac{1}{\sqrt{n+1}+\sqrt{n}}=\sqrt{n+1}-\sqrt{n}
$$
By summing and subtracting, we get that both $\sqrt{n+1}$ and $\sqrt{n}$ are rational.
It's not difficult to show that $\sqrt{n}$ is rational if and only if $n$ is a perfect square, but the only consecutive squares in the natural numbers are $0$ and $1$.
A: Notice that $\sqrt{n}+\sqrt{n+1}$ is a root of $$x^4 -2 (2 n+1) x^2+1.$$ Now use the rational root theorem (assuming that $n$ is an integer) to find that the only possible rational roots are $\pm 1$.
A: Assuming that $\sqrt{n}+\sqrt{n+1}$ is a rational number, it follows by squaring that $\sqrt{n(n+1)}$ is a rational number. Given a positive integer $m$, $\sqrt{m}$ is a rational number only if $m$ is a square, but $n(n+1)$ cannot be a square: if we assume it is a square, it follows that $$4n(n+1) = (2n+1)^2-1$$
is also a square, but the only consecutive squares are $0$ and $1$.
