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Can there exist two (or more) different prime numbers, such that their roots added together equal a rational number?

This came up while trying to prove that two square roots, added together, can only equal an integer if perfect squares were used in the first place, but I realized I can't think of any instance where the sum of square roots would even be rational, unless those square roots were themselves integers. (Now, you don't know me, so you will have to take my word for it that me not being able to think of something does not, in fact, mean that it can not exist.)

Is there some kind of proof that two irrational numbers added together will never equal a rational number (other than cheating like $\pi + (1 - \pi)$)?

CWilson
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    As the proofs in the question I've linked show, not only can we not have primes such that $\sqrt{p_1}+\sqrt{p_2}\in\mathbb{Q}$, nor can we have $q_1\sqrt{p_1}+q_2\sqrt{p_2}+q_3\sqrt{p_3}+\ldots\in\mathbb{Q}$ (for nonzero $q_i\in\mathbb{Q}$) either. Regarding your rational sums of irrationals question, I think you'll have to be more specific with what's 'cheating'. – Jam Nov 16 '19 at 17:55
  • Questions like these are studied in Field Theory and the answer to "is the sum of irrationals irrational too?" is non-trivial and all you can really say is "sometimes" since it depends on the constraints you put of the numbers. Indeed, if $a,b\notin\mathbb{Q}$ but $a+b\in\mathbb{Q}$ then by definition, there are integers, $m,n$ such that $b=\frac{m}{n}-a$ but I don't think that's saying much and wouldn't call it cheating. That part of your question is discussed here. – Jam Nov 16 '19 at 18:19
  • @Jam Yep, well done. Those two links together solved my problem. Sorry my Google skills were sub par. That said, the dupe target might or might not be the same question, but I am not smart enough to understand how its answer applies at all. The second linked question is a different question, but the highest voted answer does address my question fully and completely. So, thank you... and should I close this with the current dupe, or let you change the dupe target first? – CWilson Nov 16 '19 at 18:46
  • If you could just leave it up, that'd be great. It helps link the site together. The first question I linked asks whether there you can have a "linear combination" of roots of prime numbers $\sqrt{2},\sqrt{3},\ldots$ equal $0$. i.e. can we get $0$ by adding together the roots, by using the right rational numbers as coefficients (e.g. something like $\frac{5}{7}\sqrt{2}-\frac{2}{3}\sqrt{3}+\ldots=0$). The proofs (which are rather advanced) show that this is impossible. – Jam Nov 16 '19 at 19:10
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    That property is called "linear independence". Your question was the special case when you're only choosing $2$ primes but it must be impossible too since it falls under the same umbrella. – Jam Nov 16 '19 at 19:11

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