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:

$0=\lfloor \frac{3^0}{2^1} \rfloor$

$1=\lceil \frac{3^0}{2^1} \rceil$

$2=\lceil \frac{3^1}{2^1} \rceil$

$3=\lceil \frac{3^1}{2^0} \rceil$

$4=\lfloor \frac{3^2}{2^1} \rfloor$

$5=\lceil \frac{3^2}{2^1} \rceil$

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.

• (Well, to be pedantic, $0$ can't be represented as the ceiling of such a ratio… but otherwise it holds for all other nonnegative integers.) – ShreevatsaR May 21 '18 at 8:43
• To be even more pedantic, not everyone considers $0$ a natural number. There exists a very popular distinction that only positive integers are natural, and nonnegative integers are whole. – gen-z ready to perish May 21 '18 at 8:51
• @ChaseRyanTaylor Yes I know, but the question gave $0$ as an example — so considering that the OP may care about that case too, I chose to ignore the ambiguous terminology of “natural number” (which I repeated in the answer) and instead use “nonnegative integer” in the “pedantic” comment. – ShreevatsaR May 21 '18 at 8:53
• Touché ${}{}{}{}{}{}{}{}$ – gen-z ready to perish May 21 '18 at 9:06
• @ChaseRyanTaylor: Arguably, "whole" numbers could equally well be $\Bbb Z$, $\Bbb Z^+$, or $\Bbb N$. I prefer "counting numbers" for $\Bbb Z^+$, and think that the natural numbers $\Bbb N$ include $0$. – John Bentin Dec 19 '18 at 13:38