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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

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marked as duplicate by user26857, quid, egreg abstract-algebra Oct 25 '15 at 10:36

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ I am asking because this is used in some numerical algorithm in physics $\endgroup$ – kaiser Oct 25 '15 at 10:10
  • $\begingroup$ As far as I can tell, this is answered here as a quite general answer to a much simpler question. $\endgroup$ – Martin R Oct 25 '15 at 10:14
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    $\begingroup$ If $n_1\sqrt{2}+n_2\sqrt{3}=0$ then $\sqrt{\frac{3}{2}}=-\frac{n_1}{n_2}$ is rational, a contradiction. $\endgroup$ – Suhail Oct 25 '15 at 10:15
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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.

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  • $\begingroup$ "The same approach works also in the second case." Not so sure. Of course we can try it, but at some moment can get lost in calculations. $\endgroup$ – user26857 Oct 25 '15 at 11:12
  • $\begingroup$ @user26857: you do not need to really perform them, you just have to convince yourself that by rearranging and squaring we may obtain many integers and a single square root, the square root of $2\cdot 3\cdot 5\cdot\ldots$. $\endgroup$ – Jack D'Aurizio Oct 25 '15 at 11:17
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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 $p_i$.

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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.

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  • $\begingroup$ Could you prove two statements? It is not easy for me. $\endgroup$ – H.T.H Feb 17 '18 at 14:46

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