prove that there is no positive integer n for which root n - 1 + root n + 1 is rational.
Suppose $\displaystyle a=\sqrt{n-1}+\sqrt{n+1}$ is rational.
Then
$\displaystyle a^2=2n+2\sqrt{(n+1)(n-1)}$
$\displaystyle \frac{a^2-2n}2=\sqrt{(n+1)(n-1)}$
Hence $\displaystyle \sqrt{(n+1)(n-1)}$ is rational, which implies that $\displaystyle (n+1)(n-1)=n^2-1$ is a perfect square.
We get $\displaystyle n^2-1=m^2$ for some integer m.
But the only solutions in integere are $\displaystyle n=\pm1$ and $\displaystyle m=0$.
But for n=1 we have $\displaystyle a=\sqrt2$, which is irrational.
EDIT: Corrected typo n(n-1) to (n+1)(n-1) on one place.
$\displaystyle a=\sqrt{n-1}+\sqrt{n}+1$ is rational if and only $\displaystyle b=a-1=\sqrt{n-1}+\sqrt{n}$ is rational.
You get:
$\displaystyle b=\sqrt{n-1}+\sqrt{n}$
$\displaystyle b^2=2n-1+2\sqrt{(n-1)n}$
$\displaystyle \frac{b^2-2n+1}2=\sqrt{(n-1)n}$
meaning that $\displaystyle \sqrt{(n-1)n}$ is rational and therefore (n-1)n is a perfect square.
Now since n-1 and n are coprime, both n-1 and n are perfect squares. So again you have two squares which differ by 1 and you can continue int he same way as I did in my first post. You get that the only solution is n=1.
BTW do you know the result, that a square root of an integer is rational if and only if this integer is a perfect square? If you did not learn this, you were probably supposed to look for a completely different solution. (Perhaps there is a simpler one and I have overlooked it...)
This one is extremely easy. But this expression is rational for many integers. You have:
$\displaystyle a=2\sqrt{n}$
$\displaystyle \sqrt{n}=\frac a2$
Hence $\displaystyle \sqrt{n}$ is rational if and only if n is a perfect square.
You can try it yourself for $\displaystyle 1=1^2$, $\displaystyle 4=2^2$, $\displaystyle 9=3^2$, $\displaystyle 16=4^2$, etc.