About the ceiling function Let $⌈x⌉$ be the ceiling function of $x$. Given that $a<b$ are fixed real numbers find all $x$ satisfying the following conditions:
1) $a<⌈x⌉<b$ 
2) $a≤⌈x⌉<b$ 
3) $a<⌈x⌉≤b$ 
4) $a≤⌈x⌉≤b$ 
I have no idea to start.
 A: If $\alpha$ and $\beta$ are integers then $\alpha < \lceil x \rceil \le \beta \implies \alpha < x \le \beta$. 
In order to find the solution of your problem, we try to transform the given conditions into $\alpha < \lceil x \rceil \le \beta$, where $\alpha$ and $\beta$ are integers.
To this end, we note that 
$$a < \lceil x \rceil \iff \lfloor a \rfloor<\lceil x \rceil,$$
$$a \le \lceil x \rceil \iff \lceil a \rceil-1<\lceil x \rceil,$$
$$\lceil x \rceil < b\iff \lceil x \rceil \le \lceil b \rceil-1,$$
and
$$\lceil x \rceil \le b\iff \lceil x \rceil \le \lfloor b \rfloor.$$
With the above-mentioned relationships, we arrive at


*

*$a < \lceil x \rceil < b \implies \lfloor a \rfloor < x \le \lceil b \rceil-1,$

*$a \le \lceil x \rceil < b \implies \lceil a \rceil-1 < x \le \lceil b \rceil-1,$

*$a < \lceil x \rceil \le b \implies \lfloor a \rfloor < x \le \lfloor b \rfloor,$

*$a \le \lceil x \rceil \le b \implies \lceil a \rceil-1 < x \le \lfloor b \rfloor.$

A: Let $a=\alpha_0.\alpha_1\alpha_2\ldots$ in its decimal form, where $\alpha_0\in\mathbb{N}$ and $\alpha_1,\alpha_2,\ldots$ are single digits. Then to satisfy the inequality $a<\lceil x\rceil$, the ceiling of $x$ has to be at least $\alpha_0+1$, so $x>\alpha_0=\lfloor a\rfloor$.
Satisfying $\lceil x\rceil<b$ is alittle bit more tricky. If we similarly write $b=\beta_0.\beta_1\beta_2\ldots$, then we have two cases.


*

*Case 1: $b$ is not integer. Then the integer part of $x$ can be as high as $\beta_0-1=\lfloor b\rfloor-1$ (or anything less, of course), because then $\lceil x\rceil\le\beta_0<b$. And $x=\beta_0$ also works, because $\lceil x\rceil=\lceil\beta_0\rceil=\beta_0<b$. But any number of the form $y=\beta_0.\beta_1'\beta_2'\ldots$ with nonzero fractional part wouldn't work, because its ceiling is $\lceil y\rceil=\beta_0+1>b$. In summary, $x\le\lfloor b\rfloor=\lceil b\rceil-1$ in this case.

*Case 2: $b$ is an integer, i.e. $b=\beta_0$. Then the maximal possible value of $x$ is $\beta_0-1$, so that $\lceil x\rceil\le\beta_0-1<b$. So in this case we again find that $x\le\lceil b\rceil-1$.
The bottom line is that $a<\lceil x\rceil<b$ is satisfied for all $x$ in $\lfloor a\rfloor<x\le\lceil b\rceil-1$ (where we interpret it as the empty set when the left boundary is higher than the right boundary).
I'm sure this solution can be shortened to avoid considering different cases, but this is the first thing that came to my mind. And you can handle the other inequalities similarly.
Note: this solution has been edited since I posted it originally, because the original solution was wrong. Sorry…
A: All $x$ that map to the same $\lceil x\rceil=n$ are in the range
$$(n-1,n],$$ i.e.
$$n<x+1,x\le n.$$
Then
$$a<\lceil x\rceil<b$$ is achieved by
$$a<n<x+1,x\le n<b$$
or
$$a-1<x<b.$$
Equality on the left is never achieved, equality on the right is preserved.
