Can ratio of smooth numbers approach 1?

Question

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

Answer 1

Yes, if $q\ge 3$.

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.

Answer 2

Yes. We can just consider fractions of the form $\frac {3^a}{2^b}$. The base $2$ log of this is $a\log_2(3)-b$. 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 $1$.