Is the difference of the natural logarithms of two integers always irrational or 0?
If $\log(a)-\log(b)$ is rational, then $\log(a)-\log(b)=p/q$ for some integers $p$ and $q$, hence $\mathrm e^p=r$ where $r=(a/b)^q$ is rational. If $p\ne0$, then $\mathrm e=r^{1/p}$ is algebraic since $\mathrm e$ solves $x^p-r=0$. This is absurd hence $p=0$, and $a=b$.