Quotients in Ceilings and Floors How would I simplify the expression
$\lceil\frac{2x + 1}{2}\rceil - \lceil\frac{2x + 1}{4}\rceil + \lfloor\frac{2x + 1}{4}\rfloor$
I've tried writing the expression without floors or ceilings, but with no success. I also tried some casework on the parity of x. 
 A: We would need to first look at the last two terms. The floor and ceil of a number are equal if and only if it is an integer. So if $\frac{2x + 1}{4} \in \mathbb{Z}$, i.e. if $x = \frac{4n - 1}{2}$ where $n$ is an integer, then the last two terms cancel out. Otherwise, the floor of the number is $1$ more than the ceil of the number. 
Now, consider the first term. If $x = \frac{4n - 1}{2}$ where $n\in\mathbb{Z}$, satisfying the cancellation of the last two terms, then the first term would equal $2n$ because $2n$ is an integer. Thus, when the last two terms cancel, i.e. $x = \frac{4n - 1}{2}$ for some $n\in \mathbb{Z}$ then the expression simplifies to $2n$.
If $x \neq \frac{4n - 1}{2}$ for any $n\in\mathbb{Z}$, then the last two terms amount to $-1$. For the first term, if $x = \frac{2k - 1}{2}$ for some $k\in \mathbb{Z}$, then the first term would equal $k$. Otherwise, it would be equal to $k+1$. Hence, in this case, the whole expression would be equal to $k-1$ if the first condition is satisfied or equal to $k$ if the second condition satisfied.
A: If $2x+1=4t+u,0<u<4$  where $t$ is any integer
$\lceil\frac{2x + 1}2\rceil - \lceil\frac{2x + 1}4\rceil + \lfloor\frac{2x + 1}4\rfloor=\lceil\dfrac u2\rceil-\lceil\dfrac u4\rceil-\lfloor\dfrac u4\rfloor$
Now $0<u<4\iff0<\dfrac u2<2$
For $0<\dfrac u2\le1,\lceil\dfrac u2\rceil=1$ and for 
$1<\dfrac u2<2,\lceil\dfrac u2\rceil=2$
$0<u<4\iff0<\dfrac u4<1\implies\lceil\dfrac u4\rceil=1$ and $\lfloor\dfrac u4\rfloor=0$
A: Let me try this way - 
Take $y=\frac{2x+1}4$
If {$y$} $<0.5$, It would reduce to $$(2[y]+1) - ([y]+1) + ([y])   = 0$$
If {$y$} $>0.5$, It would be $$(2[y] +2) - ([y]+1)+([y]) = 1$$
So, now, {$y$}$= 2${$x$}$+0.25$ if {$x$} $<0.375$ 
{$y$}$= 2${$x$}$-0.75$ if $0.875>${$x$} $>0.375$ 
{$y$}$= 2${$x$}$-1.75$ if {$x$} $>0.875$
You will now get the regions of $x$ where this hold good
For $0<${$x$}$<0.125$ The answer is $0$
For $0.125<${$x$}$<0.375$ The answer is $1$
For $0.375<${$x$}$<0.625$ The answer is $0$
For $0.625<${$x$}$<0.875$ The answer is $1$
For $0.875<${$x$}$<1$ The answer is $0$
