Why is $n_1 \sqrt{2} +n_2 \sqrt{3} + n_3 \sqrt{5} + n_4 \sqrt{7} $ never zero? Here $n_i$ are integral numbers, and not all of them are zero.
It is natural to conjecture that similar statement holds for even more prime numbers. Namely, 
$$ n_1 \sqrt{2} +n_2 \sqrt{3} + n_3 \sqrt{5} + n_4 \sqrt{7} + n_5 \sqrt{11} +n_6 \sqrt{13} $$ is never zero too.
I am asking because this is used in some numerical algorithm in physics

We will prove a more general statement:

If $a_i\in \mathbb{Z}-\{0\}$ and $b_i$ are distinct positive integers which are square-free then $a_1\sqrt{b_1}+\cdots+a_k\sqrt{b_k}$ is irrational.

Proof: We'll prove by induction on $n\ge1$ that $a_1\sqrt{b_1}+\cdots+a_k\sqrt{b_k}$ is not irrational of second degree i.e. it's not a root of a polynomial of second degree with integer coefficients.
For $n=2$ it can be easily checked.
Let it's true for $n\leqslant k-1$. Then $\alpha=a_1\sqrt{b_1}+\cdots+a_k\sqrt{b_k}$ cannot be integer.
Suppose that $\alpha$ is a root of $p_2(x)=x^2+bx+c$ with integer coefficients.
We'll consider $P(x)=\prod(x-\alpha^*)$ where product taken by all numbers of form $\alpha^*=\varepsilon_1a_1\sqrt{b_1}+\cdots+\varepsilon_ka_k\sqrt{b_k}$
 where $\varepsilon_i \in \{-1,+1\}$.
The next two statements can be proved easily:
1) $P(x)$ is polynomial degree $2^k$ with integer coefficients.
2) $P(x)$ divides into $p_2(x)$.
From these statements we conclude that some $\alpha^*$ is also root of $p_2(x)$. By Vieta's theorem $-b=\alpha+\alpha^*=2(a_{i_1}\sqrt{b_{i_1}}+\cdots+a_{i_m}\sqrt{b_{i_m}})$ where $m<k$ and we get contradiction.
