Wsrtjtyk

Can the ratio of $q$-smooth numbers ever approach one? Asked more rigorously, can we ever have
$$1in overline{{a/b; :; a,b ;text{ are } qtext{-smooth}}}=overline{{prod_{p_ile q}p_i^{alpha_i}; :; (a_1, dots, a_{pi(q)})inmathbb{Z}^{pi(q)}}};?$$
If $q=2$, the answer is obviously no because
$$1notin overline{{2^{alpha}; :; alphainmathbb{Z}}}={0}cup {2^{alpha}; :; alphainmathbb{Z}}$$
however beyond this the problem is not easy. The reason I would like to know this is because this would imply that
$$G(q)=inf {a/b; :; a>b,; a,b ; qtext{-smooth}}>1.$$
I suspect that we may be able to say even further that
$$overline{{prod_{p_ile q}p_i^{alpha_i}; :; (a_1, dots, a_{pi(q)})inmathbb{Z}^{pi(q)}}}={0}cup {prod_{p_ile q}p_i^{alpha_i}; :; (a_1, dots, a_{pi(q)})inmathbb{Z}^{pi(q)}}$$
which would clearly imply the result.

Yes, if $qge 3$.

This is because $log_2(3)$ is irrational, so the fractional parts of $log_2(3^n)$ lie densely in $[0,1)$. In particular they can become arbitrarily small.

Yes. We can just consider fractions of the form $frac {3^a}{2^b}$. The base $2$ log of this is $alog_2(3)-b$. Because $log_2(3)$ is irrational, we can choose $a,b$ to make this difference arbitrarily close to $0$, which means the ratio can be arbitrarily close to $1$.