Can all natural numbers be expressed as $\lceil \frac{3^a}{2^b} \rceil$ or $\lfloor \frac{3^a}{2^b} \rfloor$? Can any natural number $n$ be expressed as
$n=\lceil \frac{3^a}{2^b} \rceil$ or $n=\lfloor \frac{3^a}{2^b} \rfloor$ where $ a,b \in \mathbb{N}$ ?
I have found no approach to this problem. All suggestions are appeciated.
Examples:
$0=\lfloor \frac{3^0}{2^1} \rfloor$
$1=\lceil \frac{3^0}{2^1} \rceil$
$2=\lceil \frac{3^1}{2^1} \rceil$
$3=\lceil \frac{3^1}{2^0} \rceil$
$4=\lfloor \frac{3^2}{2^1} \rfloor$
$5=\lceil \frac{3^2}{2^1} \rceil$
 A: Yes, all natural numbers $n$ can be represented as $n=\left\lceil \frac{3^a}{2^b} \right\rceil$ or as $n=\left\lfloor \frac{3^a}{2^b} \right\rfloor$ (whichever one you want). Let's do floors, for concreteness.
For $n=\left\lfloor \frac{3^a}{2^b} \right\rfloor$ you want positive integers $a$ and $b$ such that $n \le \frac{3^a}{2^b} < n + 1$, or equivalently $\log n \le a \log 3 - b \log 2 < \log(n + 1)$, or in other words 
$$\frac{\log n}{\log 2} \le \left(a \frac{\log 3}{\log 2} - b\right) < \frac{\log(n+1)}{\log 2}.$$
That is, if you call the ratio $\frac{\log 3}{\log 2}$ as $\theta$, then you want to find an integer $a$ such that the fractional part of $a\theta$ lies between the fractional part of $\frac{\log n}{\log 2}$ and that of $\frac{\log(n+1)}{\log 2}$.
This problem is known as “inhomogeneous diophantine approximation”. As $\theta$ is irrational, its fractional parts are dense and equidistributed in $[0, 1)$ (this is a theorem), so you can make the fractional part of $a\theta$ arbitrary close to any number you choose. There are many questions on this site about it; two I've answered are at Fractional part of $b \log a$ and at Prefix of Fibonacci number but you can also click on the “Linked” and “Related” on those questions for more.
