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
Assume that: $$ n_1\sqrt{2}+n_2\sqrt{3}+n_3\sqrt{5}+n_4\sqrt{7}=0.\tag{1} $$ That implies: $$ (2n_1^2+3n_2^2)+2n_1 n_2 \sqrt{6} = (5n_3^2+7n_4^2)+2n_3 n_4 \sqrt{35}\tag{2} $$ as well as (just keep rearranging and squaring): $$ (2n_1^2+3n_2^2-5n_3^2-7n_4^2)^2 = 24 n_1^2 n_2^2 + 140 n_3^2 n_4^2 - 8n_1 n_2 n_3 n_4 \sqrt{2\cdot 3\cdot 5\cdot 7} \tag{3}$$ but the last line implies $\sqrt{2\cdot 3\cdot 5\cdot 7}\in\mathbb{Q}$, which we know to be false.
The same approach works also in the second case.
If $p_1,\dots,p_n$ are prime numbers with $p_i\ne p_j$ for $i\ne j$, then the field extension $\mathbb Q\subset\mathbb Q(\sqrt{p_1},\dots,\sqrt{p_n})$ has degree $2^n$ and a basis is given by the set $$\{1,\sqrt{p_{i_1}\cdots p_{i_k}}:1\le i_1<\cdots<i_k\le n, 1\le k\le n\}.$$ See Proving that $\left(\mathbb Q[\sqrt p_1,\dots,\sqrt p_n]:\mathbb Q\right)=2^n$ for distinct primes $p_i$.
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.
