Can square roots of imperfect square such as $ \sqrt{2}$ , $ \sqrt{3}$.....$ \sqrt{n}$ be written as sum of other real numbers or other imperfect square roots which are not linear combinations with multiple of $ \sqrt{n}$ as one of its term, where n is the imperfect square whose root need to be represented. I believe it can't be, but is there any theorem which states that ? To put it even simply. Does there exist $ a,b \in R $ such that

$$ a+b= \sqrt{n}$$ where $ n$ is an imperfect square and a,b are not linear combinations using multiples of $\sqrt{n}$ one of their terms.

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    $\begingroup$ What cannot happen, if you like such results : if $n_1,n_2,...,n_k$ are distinct integers such that none of $\frac{n_i}{n_j}$ is a perfect square of a rational number, then no rational linear combination of $\sqrt {n_i}$ can be equal to an integer. Of course, if you are fixing $n$, then $\sqrt n$ can't be equal to a rational linear combination of $\sqrt{n_i}$ where each of $\frac{n}{n_i}$ is not the square of a rational number. See here : qchu.wordpress.com/2009/07/02/… $\endgroup$ – Teresa Lisbon Aug 26 at 15:13
  • $\begingroup$ See this answer for some general results about linear independence of square-roots. $\endgroup$ – Bill Dubuque Aug 26 at 21:38

If "real numbers" is your only restriction, then this is simple (and almost feels like cheating, if that makes any sense). Let $a = \pi$ and $b = \sqrt n - \pi$. Then $$ a + b = \pi + \sqrt n - \pi = \sqrt n $$

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  • $\begingroup$ I just changed the constraints $\endgroup$ – Siddharth Prakash Aug 26 at 15:19
  • $\begingroup$ @SiddharthPrakash Well, in some sense, my solution is the only solution. Pick $a$ to be whatever real number you want. $b$ will necessarily be $\sqrt n - a$. You can hide the $\sqrt n$ away so that it's difficult to see, but it will be there, no matter what you try. $\endgroup$ – Arthur Aug 26 at 15:23
  • $\begingroup$ @SiddharthPrakash And the fact that you aren't satisfied with my answer is exactly what I was refering to when I said it "almost feels like cheating". Your first instinct was to change the question to make sure that such an uninteresting solution was disallowed. But that's misguided. There is no other kind of solution. $\endgroup$ – Arthur Aug 26 at 15:34
  • $\begingroup$ Yeah I also feel like it there is not. But I wanted to know if there is any theorem and a proof which could concretely explain it $\endgroup$ – Siddharth Prakash Aug 26 at 15:44

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