The question I am trying to solve is
$\lfloor x - \lceil \frac{x}{2}\rceil \rfloor =3.$
If anyone could explain/give me advice on how to solve this question, that would be great.
Edited: I initially equated ceiling(x/2) to n/2, n/2<x≤n/2+1, but I do not understand how this could be solved with the presence of another x. Another thing that confuses me is how there is a ceiling function inside of the floor function.
Thank you in advance.
Don't overcomplicated it. A floor and ceil are just things that modify their arguments.
$\lfloor y\rfloor = k$ is just a all the reals $k \leq y < k+1$. This is by definition!
It's just a short hand notation.
So use your expression for y and k:
$3 \leq x - \lceil \frac{x}{2} \rceil < 4$
So you have to find an x value that when you subtract around half from it gives you something between 3 and 4.
e.g., what if $x = 6$? Then
$3 \leq 6 - \lceil \frac{6}{2}\rceil = 6 - 3 = 3 < 4$
So $x = 6$ works. What about x = 8?
$3 \leq 8 - \lceil \frac{8}{2}\rceil = 8 - 4 = 4 < 4$
and $x = 8$ almost works.
$3 \leq x - \left(\frac{x}{2} - \epsilon\right) < 4$
We can replace the floor or ceil with a compensation factor because, say $\lceil 4.3 \rceil = 5 = 5.3 - .7$. Here $\epsilon = 0.7$.
$3 \leq \frac{x}{2} + \epsilon < 4$
$6 - 2\epsilon \leq x < 8 - 2\epsilon$
Now $0 < \epsilon < 1$
$6 - 2\epsilon \leq x < 8 - 2\epsilon$