prove that there exist $a,b > N, y - x < a\log 3 -b\log 2 < y-x+\epsilon$ 
Prove or disprove that for any $\epsilon > 0, N>0,x\leq y \in \mathbb{R},$ there exist integers $a,b > N$ so that $ y - x < a\log 3 -b\log 2 < y-x+\epsilon$.


Prove or disprove that for any $\epsilon > 0, N>0,x\leq y \in \mathbb{R},$ there exist integers $a,b > N$ so that $ y - x > a\log 2 -b\log 3 > y-x-\epsilon$.

I know that the set $\{a \log 3 - b\log 2:a,b\in\mathbb{Z}\}$ forms an additive group that is dense in $\mathbb{R}$ because it is nontrivial and not cyclic. However, I'm not sure how to prove the given claim. Also, the second claim seems very similar to the first one; it seems all that's necessary is that $\log 2/\log 3$ isn't rational. So there's likely a generalization. I think one might be able to consider the smallest positive number that can be written in the form $a\log 3 - b\log 2$ for nonnegative integers $a$ and $b$. Note that if no such positive number exists, then for any $t > 0,$ there must exist infinitely many terms $a\log 3 - b\log 2 \in (0,t)$. Then we can choose $t = \epsilon$ and two terms $t_1 < t_2 \in (0,t)$. But the issue is that it might not be possible to write $t_2 - t_1$ as $a\log 3 - b\log 2$ for nonnegative integers $a,b$. Since there are infinitely many such terms in the interval, we may exclude all such terms with $a,b \leq N$.
 A: We only need to prove that for any $\epsilon > 0, N > 0$, there exist integers $a', b' > N$ so that $0 < a'\log 3-b'\log 2 < \epsilon$.
Thus, if $k$ is the largest integer such that $k(a'\log 3-b'\log 2) < y-x$, then $y-x < (k+1)(a'\log 3-b'\log 2) < y-x+\epsilon$. Taking $a=(k+1)a', b=(k+1)b'$ will prove the first question.
If $0 < a_n\log 3-b_n\log 2 = m < \log 3$, we can prove there exist $a_{n+1} \geq a_n, b_{n+1} \geq b_n$ such that $0 < a_{n+1}\log 3-b_{n+1}\log 2 < \frac{m}{2}$.
Let $s$ be the largest integer such that $s(a_n\log 3-b_n\log 2) < \log 3$. If $(s+1)(a_n\log 3-b_n\log 2) - \log 3 < \frac{m}{2}$, then choose $a_{n+1} = (s+1)a_n-1, b_{n+1} = (s+1)b_n$. Otherwise we have $\log 3-s(a_n\log 3-b_n\log 2) < \frac{m}{2}$, let $t$ be the largest integer such that $t(\log 3-s(a_n\log 3-b_n\log 2)) < \log 3$, we can choose $a_{n+1} = 1+t(sa_n-1), b_{n+1} = stb_n$.
Select $a_1, b_1 > N$ satisfy $0 < a_1\log 3-b_1\log 2 < \log 3$, there exist integers $a'=a_n, b'=b_n$ so that $0 < a'\log 3-b'\log 2 < \epsilon$.
The second question can be proved in a similar way.
