Prove $\sqrt{2}+\sqrt{3}+\sqrt{5}+...+\sqrt{p_{n}}$ is irrational, where $p_{n}$ is the nth prime.

Question

My motivation is making general proof , instead of trying to prove special cases.

To which branch of mathematics does my question belong?

I am highly interested in irrational numbers.

Is it good idea , proving that the number in my question has algebraic degree higher than one(and why is degree greater than one implies number is irrational)?

I searched and I didn't find good references to my wondering.

Answer 1

Call $p_1:=2$, $p_2:=3$ and so on, $p_n$ is the $n$th prime number. Consider the set

$$E_n:=\{\varepsilon_1 \sqrt{p_1}+\ldots+\varepsilon_n \sqrt{p_n}:\ \varepsilon_i∈\{-1,1\}\}$$

so $E_n$ contains all $2^n$ possible combinations of sums and differences of the square roots $\sqrt2$, $\sqrt3$, $\ldots$, $\sqrt{p_n}$.

Now, consider the polynomial $$p_n(x):=\prod_{\alpha\in E_n} (x-\alpha)$$

This is called the $n$th Swinnerton-Dyer polynomial, it is an irreducible polynomial with coefficient in $\mathbb Z$. So all its roots are irrational numbers with degree $2^n$