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. 

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.
