Is there a way to show the sum of any different square root of prime numbers is irrational? [duplicate]

Is there a way to show the sum of any different square root of prime numbers is irrational? For example, $$\sqrt2+\sqrt3+\sqrt5 +\sqrt7+\sqrt{11}+\sqrt{13}+\sqrt{17}+\sqrt{19}$$ should be a irrational number.

One approach I used is to let the sum be a solution of an even polynomial $f(x)$with integer coefficients and prove by induction that by adding another $\sqrt{p_{k+1}}$. The new polynomial can be written as $$f(x+\sqrt{p_{k+1}})f(x-\sqrt{p_{k+1}})$$

where $$f(x+-\sqrt{p_{k+1}})=P(x)+- Q(x)\sqrt{p_{k+1}},$$

where $P(x)$ is an even plynomial and $Q(x)$ is an odd polynomial.

The new polynomial can be written as $$P^{2}(x)- Q^{2}(x)p_{k+1}.$$

Assume it has a rational solution $a$, we must have$$P(a)=Q(a)=0.$$

My calculation stopped here since I can't find any contradiction result from this. Can anyone continue this proof, or has other better way to solve this? Thanks!

marked as duplicate by Ross Millikan, JMoravitz, dxiv, erfink, projectilemotionMay 28 '17 at 21:28

• – Tob Ernack May 28 '17 at 17:32
• "For example... should be a prime number." You mean "should be an irrational number?" – Thomas Andrews May 28 '17 at 17:42
• – Jack D'Aurizio May 28 '17 at 17:58
• Ok, then. Thanks! – Zhenyuan Lu May 28 '17 at 21:06

For $n\ge 0$ let $E_n=\Bbb Q(\sqrt {p_1},\ldots, \sqrt{p_n})$ be the smallest field extension of $\Bbb Q$ containing $\sqrt{p_k}$ for $1\le k\le n$.
Claim. For every $(\epsilon_1,\ldots, \epsilon_n)\in\{-1,1\}^n$, there is an automorphism $\phi$ of $E_n$ with $\phi(\sqrt{p_i})=\epsilon_i\sqrt{p_i}$, $1\le i\le n$.
Proof. [By induction]. The claim is vacuously true for $n=0$.
Let $n\ge 0$ and assume $\sqrt{p_{n+1}}\in E_{n}$. Clearly, $E_{n}$ is spanned as a $\Bbb Q$-vector space by all products of some of the $\sqrt{p_i}$ (including the empty product, $1$). Thus we can write $$\tag1\sqrt{p_{n+1}}=\sum_{S\subseteq\{1,\ldots,n\}}q_S\prod_{i\in S}\sqrt{p_i}$$ with $q_S\in \Bbb Q$. Among all such representations, pick one with the minimal number of non-zero coefficients. Assume there are at least two non-zero coefficients $q_A, q_B$. Pick $k\in A\mathop{\Delta}B$. By induction hypothesis, there is an automorphism $\phi$ of $E_n$ that maps $\sqrt{p_k}\mapsto-\sqrt{p_k}$ and for $i\ne k$ maps $\sqrt{p_i}\mapsto\sqrt{p_i}$. Also, $\phi(\sqrt{p_{n+1}})=\pm\sqrt{p_{n+1}}$ so that one of $\frac{\sqrt{p_{n+1}}\pm \phi(\sqrt{p_{n+1}})}2$ equals $\sqrt{p_{n+1}}$. This way, we obtain another representation of $\sqrt{p_{n+1}}$ of the form $(1)$, but with less non-zero coefficients because at least either $q_A$ or $q_B$ (depending on the "$\pm$") is replaced with a $0$. From this contradiction, we conclude that in a minimal representation, at most one $q_S$ in $(1)$ is non-zero. Thus $\sqrt{p_{n+1}}=q_S\prod_{i\in S}\sqrt{p_i}$, contradicting the irrationality of $\prod_{i\in S}\sqrt{p_i}\sqrt{p_{n+1}}$. We conclude that $\sqrt{p_{n+1}}\notin E_n$, hence $E_{n+1}$ is a quadratic extension of $E_n$. Also, $\sqrt{p_{n+1}}\mapsto -\sqrt{p_{n+1}}$ is an automorphism of $E_{n+1}$ over $E_n$ and the claim follows for $n+1$. $\square$