Is the following true

$ \left \lfloor{\frac{\left \lfloor{\frac{n}{2}}\right \rfloor}{2}}\right \rfloor= \left \lfloor{\frac{n}{2^2}}\right \rfloor$

such that, $n \in \mathbb{I}$

  • $\begingroup$ Is n an integer? $\endgroup$ – samjoe May 31 '18 at 7:02
  • $\begingroup$ @samjoe yes it is $\endgroup$ – glockm15 May 31 '18 at 7:03
  • $\begingroup$ Then say so in the original question! $\endgroup$ – samjoe May 31 '18 at 7:33


Test with $n=4a,4a+1,4a+2,4a+3$ where $a$ is an integer

For all the case

$$\left\lfloor\dfrac n4\right\rfloor=a$$

$$\left \lfloor{\frac{\left \lfloor{\frac{n}{2}}\right \rfloor}{2}}\right \rfloor=?$$

  • $\begingroup$ Alright, I guess it is not true for any integer, then how does $ \left \lfloor{\frac{\left \lfloor{\frac{n}{2}}\right \rfloor}{2}}\right \rfloor $ simplify if at all $\endgroup$ – glockm15 May 31 '18 at 7:08
  • $\begingroup$ @glockm15, Why not any? If $n=4a+3$ $$\left\lfloor\dfrac n2\right\rfloor=2a+1$$ and $$\left\lfloor\dfrac n4\right\rfloor=a$$ $\endgroup$ – lab bhattacharjee May 31 '18 at 7:13
  • $\begingroup$ Ops, my bad, I used $a$ for the second floor function instead of $n$. Got it thanks, so the equation is true $\endgroup$ – glockm15 May 31 '18 at 7:16

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