For example, can the square root of some prime and another square root of another prime be rational (just an example)?
Not defined by each other as in not pi over 4 plus one minus pi over 4.
The specific problem was to prove the square root of 3 plus the cube root of two to be irrational.
We are going to prove that its irrational by contradiction. Lets assume its rational.
Let $\displaystyle \sqrt{3+\sqrt[3]{2}} \ =\ \frac{p}{q} \ where\ gcd( p,q) =1$ So $\displaystyle 3+\sqrt[3]{2} \ =\ \frac{p^{2}}{q^{2}}$
So $\displaystyle \sqrt[3]{2} \ =\ \frac{p^{2}}{q^{2}} -3$
But in LHS we have a irrational term but in rhs we have a rational quantity.
So that means that $\displaystyle \frac{p^{2}}{q^{2}}$ cannot be rational which subsequently means $\displaystyle \frac{p}{q}$ cant be rational.
But this is a contradiction as we assumed it was rational. So $\displaystyle \sqrt{3+\sqrt[3]{2}} \ $ is irrational.
Regarding your question: No Sum of square roots of two "prime" numbers cannot be rational, as each of them itself are irrational. Answered Previously