In my analysis class these questions have kept coming up over and over again about sums of radicals of primes being irrational. I wanted to just prove the general case so I never had to worry again. The argument should extend to the case where $p_i$ are allowed to be any non-squares in $\mathbb{N}$. Could anyone tell me if my argument holds? Sorry that it's a little long.
First we prove that $\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_{n}})$ is a degree $2^n$ extension of $\mathbb{Q}$, and $\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_{n}})$ is the splitting field for $\{x^2-p_1,...,x^2-p_{n}\}$
Proof:
Choose $p_1,p_2$ primes. Notice that if deg$(\sqrt{p_1},\mathbb{Q})=2$ for $x^2-p_1$ is irreducible by Eisenstein's Criterion, similarly deg$(\sqrt{p_2},\mathbb{Q})=2$. If $\sqrt{p_2}=q_0+q_1\sqrt{p_1}$, then since $\sqrt{p_2}\not\in\mathbb{Q}$ we find $q_1\not=0$. We compute $$p_1=q_0^2+2q_0q_1\sqrt{p_1}-p_1^2,$$ since $p_2\not=-p_1%2$ we must have $q_0\not=0$. By field operations this contradicts $\sqrt{p_1}\not\in\mathbb{Q}$, therefore $\sqrt{p_2}\not\in\mathbb{Q}(\sqrt{p_1})$. It follows that $\mathbb{Q}(\sqrt{p_1},\sqrt{p_2})=2^2$. Moreover, it is a splitting field for the $\{x^2-p_1,x^2-p_2\}$.
Suppose that $n\geq 1$, and for all $1\leq i\leq n$ it follows that $\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_i})=2^i$ for distinct primes $p_1,...,p_i$. Moreover, suppose $\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_i})=2^i$ is the spliting field for $\{x^2-p_1,...,x^2-p_i\}$. Suppose $p_1,...,p_{n+1}$ are distinct primes. We wish to show that $\sqrt{p_{n+1}}\not\in\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_n})$.
Suppose that $$\sqrt{p_{n+1}}=q_1p'+q_2p''+r$$ where $p',p''\in\{\sqrt{\prod_{i=1}^jp_{\sigma{i}}}|\sigma\in\text{S}_n,1\leq j\leq n\}$ are distinct, $r\in\text{Span}(\{\sqrt{\prod_{i=1}^jp_{\sigma{i}}}|\sigma\in\text{S}_n,0\leq j\leq n\}\backslash\{p',p''\})$, and $q_1,q_2\in\mathbb{Q}\backslash\{0\}$. By Eisenstein's $x^2-(p'')^2\in\mathbb{Q}[x]$ is irreducible, therefore $\mathbb{Q}(p'')$ is a degree 2 extension, and is the splitting field for $\{x^2-{p''}^2\}$ over $\mathbb{Q}$. By linear independence $p'\not\in\mathbb{Q}(p'')$, and by Eisenstein's $x^2-(p')^2\in\mathbb{Q}[x]$ is irreducible, therefore $\mathbb{Q}(p',p'')$ is a degree 4 extension of $\mathbb{Q}$, and is the splitting field for $\{x^2-{p''}^2,x^2-(p')^2\}$ over $\mathbb{Q}$. By the isomorphism extension and conjugate isomorphism theorems we can find $\sigma_{(-1)^n,(-1)^m}\in\text{Gal}(\mathbb{Q}(p',p'')/\mathbb{Q})$ such that $\sigma_{(-1)^n,(-1)^m}(p')=(-1)^np'$ and $\sigma_{(-1)^n,(-1)^m}(p'')=(-1)^mp''$ for all $n,m\in\{0,1\}$. By the isomorphism extension theorem these can be extended to isomorphism, $\tau_{(-1)^n,(-1)^m}$, of $\mathbb{Q}(p',p'',r)$ such that $\tau_{(-1)^n,(-1)^m}|_{\{p',p''\}}=\sigma{(-1)^n,(-1)^m}$ and $\tau_{(-1)^n,(-1)^m}(r)=r$. From the conjugate isomorphism theorems it follows that $(-1)^nq_1p'+(-1)^mq_2p''+r$ is a conjugate of $\sqrt{p_{n+1}}$, but $\sqrt{p_{n+1}}$ has two conjugates, a contradiction. Therefore $\sqrt{p_{n+1}}\not=q_1p'+q_2p''+r$.
It follows that if $\sqrt{p_{n+1}}\in\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_n})$, then $\sqrt{p_{n+1}}=q_0+q_1p'$ where $p'\{\sqrt{\prod_{i=1}^jp_{\sigma{i}}}|\sigma\in\text{S}_n,1\leq j\leq n\}$, and $q_0,q_1\in\mathbb{Q}$ with $q_1\not=0$. We compute $$\sqrt{p_{n+1}}=q_0+q_1p'$$ $$p_{n+1}=q_0^2+2q_0q_1p'+q_1^2(p')^2.$$ Since $p'$ is irrational we have $q_0=0$ and $p_{n+1}=(q_1p')^2$. Note, $(p')^2$ is a product of distinct primes, so since the denominator of $q_1^2$ is a perfect square, and $q_1^2(p')^2\in\mathbb{N}$ we have $q_1^2\in\mathbb{N}$. Then every prime dividing $(p')^2$ divides $p_{n+1}$, a contradiction, so $\sqrt{p_{n+1}}\not=q_0+q_1p'$, and $\sqrt{p_{n+1}}\not\in\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_n})$. Since $x^2-p_{n+1}$ is irreducible of degree 2 with roots $\pm\sqrt{p_{n+1}}$ it follows that $\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_{n+1}})$ is a degree $2^{n+1}$ extension of $\mathbb{Q}$. Moreover, $\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_{n+1}})$ is the splitting field for $\{x^2-p_1,...,x^2-p_{n+1}\}$. By P.M.I. this completes the proof. $\square$
Now we prove $\sum_{i=1}^nq_i\sqrt{p_i}+q_0$ is irrational for $q_0,...,q_n\in\mathbb{Q}$, $q_1,...,q_n\not=0$, and $p_i$ distinct primes.
Proof:
Suppose $r_1,...,r_n$ are distinct products of primes, and $q_1,...,q_n\in\mathbb{Q}$ are non-zero. Moreover, suppose no square divides any $r_i$. Assume $p_1,...,p_t$ are all distinct primes dividing $r_1\cdot...\cdot r_n$. It follows that $\sqrt{r_1},...,\sqrt{r_n}$ are distinct elements of $\{\sqrt{\prod_{i=1}^jp_{\sigma{i}}}|\sigma\in\text{S}_n,0\leq j\leq n\}$, which is a basis for $\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_n})$ over $\mathbb{Q}$, therefore by linear independence $\sum_{i=1}^nq_1\sqrt{r_i}\not=q$ for any $q\in\mathbb{Q}$. Our claim is a special case of what we've just proven, therefore this completes the proof.$\square$
