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I am trying to prove the last statement used in Adam's answer for this question (the last sentence).

Namely that, given any $x, y > 0$ with $y > x$, and that there is an $n \in \mathbb{N}$ with $\psi = \frac{1}{\sqrt{n+1}+\sqrt{n}} < y - x$, we can find a $k \in \mathbb{Z}$ such that $x < k\psi < y$.

This statement is obvious, of course, because $\psi$ can be made arbitrarily small and $k$ arbitrarily large. For example, I think that $k = \lceil \frac{x}{\psi} \rceil$ could be one such $k$ (providing the proper $n$ is used). Therefore, my concern is mainly proving the existence of this $k$.

Can someone kindly give me a hint on how to prove such a statement?

Thanks in advance.

1 Answers1

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Choose $n$ such that $\psi < (y-x)/2$. (This can always be done by choosing $n$ large enough, as you say.) Then $x-\psi < \lfloor \frac{x}{\psi} \rfloor \cdot \psi \leq x$ and so $x < (\lfloor \frac{x}{\psi} \rfloor + 1) \cdot \psi \leq x+\psi < y$.