Prove $\sqrt{n}+\sqrt{n+k}$ is irrational For what $k\in\mathbb N$, $\sqrt{n}+\sqrt{n+k}$ is irrational? ($\forall n\in\mathbb N$)
Well, a possible but perhaps in the long run not exhaustive method, since you must find for what $k \in \mathbb{N}$ this is true; consider
$$(\sqrt{n}+\sqrt{n+k})(-\sqrt{n}+\sqrt{n+k})=-n+(n+k)$$
Which of course gives $k$.

now, since $k$ is rational, it means that $\sqrt{n}$ and $\sqrt{n+k}$ is rational thus
\begin{align}
n&amp;=&amp;p^{2} \\
n+k&amp;=&amp; q^{2}
\end{align}
Perhaps you can see where I am going with this and finish it off?
Suppose that $$\sqrt n +\sqrt {n+k}=r, \quad r\in\mathbb{Q}$$

Now we have: $$\sqrt {n+k} =r-\sqrt n $$
$$n+k=r^2-2r\sqrt n +n$$ 
$$2r\sqrt n=r^2-k$$
$$\sqrt n=\frac{r^2-k}{2p}$$

This is a contradiction because the left is $\sqrt n\in\mathbb{I=\mathbb{Q^c}}$, and the right is $\frac{r^2-k}{2p}\in\mathbb{Q}$. The contradiction is due to a wrong assumption that $\sqrt n+\sqrt {n+k}$ was a rational number.