# Simplifying floor > $\left \lfloor{\frac{\left \lfloor{\frac{n}{2}}\right \rfloor}{2}}\right \rfloor= \left \lfloor{\frac{n}{2^2}}\right \rfloor$

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, $n \in \mathbb{I}$

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

Hint:

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

For all the case

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

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

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