# Roots of imperfect square as sum of other real numbers

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.

• 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/… – Teresa Lisbon Aug 26 at 15:13
• See this answer for some general results about linear independence of square-roots. – 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 $$a + b = \pi + \sqrt n - \pi = \sqrt n$$
• @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. – Arthur Aug 26 at 15:23