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Can you make all natural number from 3 with only four function that

$$x!,\; \sqrt{x},\;\lceil x\rceil,\;\lfloor x \rfloor $$

?

ex) $1=\lfloor \sqrt3 \rfloor$

$\;\;\;\;\;2= \lceil \sqrt3 \rceil$

$\;\;\;\;\;3=3$

$\;\;\;\;\;4=\lceil\left(\sqrt{3!}\right)!\rceil$

$\;\;\;\;\;5=\lfloor \sqrt{\sqrt{(3!)!}} \rfloor$

I know that all natural number s.t. $n \geq 3$ can make 3.

Sintist
  • 190
  • My knowledge here is extremely limited, but could there be a proof by induction? – Ola May 19 '18 at 08:07
  • Apparently, we are allowed to use $\Gamma(x+1)$ instead of $x!$ as the example for $4$ shows. – Peter May 19 '18 at 08:28
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    Just note that you can often hit two consecutive numbers by changing floor/ceiling functions at the end, for example $(8,9) = (\lfloor\sqrt{\sqrt{7!}}\rfloor, \lceil\sqrt{\sqrt{7!}}\rceil)$, $(10,11)=(\lfloor\sqrt{5!}\rfloor, \lceil\sqrt{5!}\rceil)$, $(14,15)=(\lfloor\sqrt{\sqrt{8!}}\rfloor, \lceil\sqrt{\sqrt{8!}}\rceil)$. – Sil May 19 '18 at 09:09

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