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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.

Bill Dubuque
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רואי בורלא
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    What exactly do you mean by "not defined by eachother"? To answer this question in a sensible manner, we will need a rigorous definition of that phrase. – Arthur Nov 28 '22 at 10:38
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    If $\sqrt{a} + \sqrt{b}$ is rational, then so is $(\sqrt{a} + \sqrt{b})^2 = a + b + 2\sqrt{ab}$. So, in particular, $\sqrt{ab}$ must also be rational. – Mathmo123 Nov 28 '22 at 10:41
  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Nov 28 '22 at 10:46
  • This is too vague. I suggest focussing on the specific, given problem. – lulu Nov 28 '22 at 11:14
  • Consider the Equation $x^2+2x-1=0$ , then the two roots are not "Defined" by each other but by the given Equation. It will turn out that both the roots are irrational where their sum is rational ! – Prem Nov 28 '22 at 11:17
  • Also, regarding your specification of "not defined by each other": If we have $s + t = r$ with $s, t$ irrational and $r$ rational, then we always necessarily have $s = r-t$, which is exactly what you ay you don't want. We could describe $s$ and $t$ in two different ways so that this connection is disguised, but you still can't escape this fact. – Arthur Nov 28 '22 at 11:23
  • It's known that the square roots of the primes are linearly independent over the rationals. – Gerry Myerson Nov 28 '22 at 12:10

1 Answers1

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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

mrtechtroid
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