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Trying to prove or disprove this (pretty sure it's correct): Let $\mathcal{F}=\{f\mid f:\mathbb{N}\to\mathbb{R}^+\}$

$$\forall f\in\mathcal{F}: \left\lfloor \sqrt{\lfloor f(n)\rfloor }\right\rfloor \in O\left(\sqrt{f(n)}\right)\;.$$

How can I get rid of both of the floor functions? Because I think that'll make it simpler

Progress: Now I am trying to prove $\lfloor \sqrt{\lfloor x)\rfloor }\rfloor = \lfloor \sqrt{x} \rfloor$ which will then make the rest of proof pretty easy.

Right now I have: $$Let \ k = \lfloor \sqrt{\lfloor x\rfloor }\rfloor$$

$$Then \ k \le \sqrt{\lfloor x\rfloor } < k +1$$ $$Then \ k^2 \le \lfloor x\rfloor < (k+1)^2$$ But I am not too sure how to reach a point where it says $$k \le \sqrt[]{x} < k+1$$ Which is the definition of floor function and conclude they are equal

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In general, for $x\geq 0$, $\lfloor\sqrt{\lfloor x\rfloor}\rfloor \leq \sqrt{x}$.

This follows because $\lfloor z\rfloor \leq z$ and $\sqrt{z}$ is an increasing function.

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  • $\begingroup$ Well we can say that ⌊f(n)⌋ <= f(n), but how can we introduce the square root after this and finally the floor of the last one? Thanks $\endgroup$ – Amir Mousavi Mar 25 '16 at 21:44

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