I use $\mathbb{N}^0$ to denote the non-negative integers. Let $r$ be a real number. Clearly, for some $r$, the equation $$r=\sqrt n + \frac1r$$ does not have a solution $n \in \mathbb{N^0}$ (more specifically, $r-\frac1r$ must be a square root of some non-negative integer for the above to be true. In the case of $r=0$, we exclude $\frac1r$.) I am wondering the following:
Does every $r \in \mathbb{R^+}$ satisfy: $$r=u_0 + \frac{1}{u_1+\frac{1}{u_2+\frac{1}{\ddots}}}$$ For not necessarily distinct integer square roots $u_0,u_1, u_2 \dots$, with $u_0 \geq0$ and $u_1,u_2\dots >0$? The above fraction is infinite.
To me, this problem smells like Khinchin or Ramanujan, although I cannot find anything addressing this question specifically, and I certainly cannot approach it like my first example. I would like either a reference that discusses this question or help in answering it.
I say the answer to my question is 'no,' but my only justification is my intuition saying that the bound between $\mathbb{R^+}$ and $\mathbb{N^0}$ should take more than some square roots and fractions to leap. But if the answer is no, then what conditions must $r$ satisfy for it to be true?
My attempt: if $r$ is to satisfy my equation, then (I think?) it must follow that: $$r(u_1u_2u_3\dots) = u_0u_1u_2u_3\dots+1 \implies r=u_0+\frac{1}{u_1u_2\dots}$$ So, in order for $r$ to have such a representation, the RHS must be true. In other words, if $r$ is to have such a representation, it must be expressible as the sum of a non-negative integer square root and the reciprocal of the square root of some positive integer. I believe this encompasses all integers? Actually yes, I am almost certain this is the case. Leaving this here and going to work on it more in the meantime.
All positive integers are integer square roots. And every positive real number has a continued fraction expression over the positive integers. So the answer to your question is yes.