TITLE: Prove that $\sqrt{2}+\sqrt{3}+\sqrt{5}$ is irrational. Generalise this.
QUESTION [10 upvotes]: I'm reading R. Courant & H. Robbins' "What is Mathematics: An Elementary Approach to Ideas and Methods" for fun. I'm on page $60$ and $61$ of the second addition. There are three exercises on proving numbers irrational spanning these pages, the last is as follows.

Exercise $3$: Prove that $\phi=\sqrt{2}+\sqrt{3}+\sqrt{5}$ is irrational. Try to make up similar and more general examples.

My Attempt:

Lemma: The number $\sqrt{2}+\sqrt{3}$ is irrational. (This is part of Exercise 2.)

Proof: Suppose $\sqrt{2}+\sqrt{3}=r$ is rational. Then
$$\begin{align}
2&=(r-\sqrt{3})^2 \\
&=r^2-2r\sqrt{3}+3
\end{align}$$ is rational, so that
$$\sqrt{3}=\frac{r^2+1}{2r}$$ is rational, a contradiction. $\square$
Let $\psi=\sqrt{2}+\sqrt{3}$. Then, considering $\phi$,
$$\begin{align}
5&=(\phi-\psi)^2 \\
&=\phi^2-2\psi\phi+5+2\sqrt{6}.
\end{align}$$
I don't know what else to do from here.
My plan is/was to use the Lemma above as the focus for a contradiction, showing $\psi$ is rational somehow.
Please help :)
Thoughts:
The "try to make up similar and more general examples" bit is a little vague.

The question is not answered here as far as I can tell.

REPLY [3 votes]: An alternative solution. Assume that $\alpha=\sqrt{2}+\sqrt{3}+\sqrt{5}=\frac{a}{b}\in\mathbb{Q}$. By quadratic reciprocity, there is some prime $p>b$ such that $3$ and $5$ are quadratic residues $\!\!\pmod{p}$ while $2$ is not. That implies that $\alpha$ is an algebraic number over $\mathbb{K}=\mathbb{F}_p$ with degree $2$, since $\sqrt{2}$ does not belong to $\mathbb{K}$ but belongs to a quadratic extension of $\mathbb{K}$. On the other hand $b<p$ ensures that $b$ is an invertible element $\!\!\pmod{p}$ and $\alpha\in\mathbb{K}$, contradiction.
This proof has a straightforward generalization: the sum of the square roots of some prime numbers is never a rational number.