Given any integer $n$ sufficiently large, I want to prove (or disprove) that there exists another integer $m\ge n$ with the form $m=2^a3^b$ ($a,b$ are no negative integers) such that $mn=o(n)$, i.e., $n$ can be approximated by $m$.

2$\begingroup$ I added a vote to close as the question is not intrinsically a research level question, and no additional context is given that might link it to one. $\endgroup$– user9072Mar 16, 2016 at 15:06

2$\begingroup$ The growth rate of the semigroup <2,3> was explicitly estimated in the paper by BourgainLindenstraussMichelVenkatesh regarding effective versions of Furstenberg's theorem. The estimation uses the BakerWustholz theorem. $\endgroup$– AsafMar 16, 2016 at 16:01
2 Answers
Fix $\varepsilon>0$. We have to prove (for large enough $n$) that there exist nonnegative integers $a,b$ such that $\log_2 n\leqslant a+b\log_2 3<\log_2 n+\varepsilon$. This follows from the fact that fractional parts of $b\alpha$, $\alpha:=\log_2 3$, are dense in $(0,1)$. Indeed, choose nonnegative integers $b_1,\dots,b_k$ so that fractional parts of $b_i\alpha$ form an $\varepsilon/2$net of $[0,1]$. Then we may choose $b\in\{b_1,\dots,b_k\}$ and appropriate integer $a$. This $a$ is positive provided that $n$ is large enough.

4$\begingroup$ If you want an effective result about the size of the error, you can obtain it from a paper of George Rhin, who bounded the irrationality exponent of $\log_2 3$. $\endgroup$ Mar 16, 2016 at 13:38

$\begingroup$ Is there explicit bound? Just to check the quality of the abc triples. $\endgroup$– joroMar 16, 2016 at 13:49
I believe replacing $o(n)$ by $O(n^a)$ for $0 < a < 1$ contradicts the $abc$ conjecture. Fix small natural $b$ coprime to $6$ (say $5$) and set $n=b^k$ for natural $k$.
In the abc triple $(mn,n,m)$ the radical is $O(n^a)$ while $c$ is $b^k$, giving fixed quality of $\frac1a > 1$ infinitely often as $k$ varies (2,3 and $b$ dont contribute essentially anything to the radical).
This doesn't rule bound of the form something like $n/\log{n}$, but I suspect the answer to the question is negative.

7$\begingroup$ This contradicts not only abc, but common sense: there are not enough many numbers of the form $2^a3^b$ for such strong approximations. $\endgroup$ Mar 16, 2016 at 13:07