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I came over a recent post that asked if you can make all natural numbers from the number 3 using factorials and square roots. This is a question in the same vein that seems more approachable, and which may give clues as how to solve the related problem.

All varibles used in the post are natural numbers.

Using these fuctions, is it possible to make all natural numbers from the number 3?

$$x^2,\; \frac{x}{2},\;\lceil x\rceil,\;\lfloor x \rfloor$$ Examples:

$1=\lfloor \frac{3}{2} \rfloor$

$2=\lceil \frac{3}{2} \rceil$

$3=3$

$4=\lfloor \frac{3^2}{2} \rfloor$

$5=\lceil \frac{3^2}{2} \rceil$

Notes:

  • The answer is yes if all natural numbers can be expressed as $\lfloor \frac{3^{2^a}}{2^b} \rfloor$ or $\lceil \frac{3^{2^a}}{2^b} \rceil$.
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  • $\begingroup$ how will you create 10 $\endgroup$ – Abhishek May 20 '18 at 11:19
  • $\begingroup$ @Abhishek $10=\lfloor (\frac{3^2}{2})^2 \rfloor$ $\endgroup$ – Ola May 20 '18 at 11:24
  • $\begingroup$ its not 10 it gives 12 $\endgroup$ – Abhishek May 20 '18 at 11:30
  • $\begingroup$ @Abhishek It does give 10. See wolframalpha.com/input/?i=floor((((3%5E2)%2F2)%5E2)%2F2) $\endgroup$ – Ola May 20 '18 at 11:33
  • $\begingroup$ you forgot to write $\frac{1}{2}$ inside floor function $\endgroup$ – Abhishek May 20 '18 at 11:39

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