Does the set $\{2^q : q\in \mathbb{Q}, 0\le q < 1\}$ form a linearly independent set over the field $\mathbb{Q}$

Question

My intuition says yes, but I'm having a lot of difficulty proving it.

Answer 1

Suppose $$\tag1 \sum_{i=1}^n c_i2^{a_i/b_i}=0$$ with $c_i\ne 0$. Then let $\alpha=2^{1/b}$ where $b=\operatorname{lcm}(b_1,\ldots, b_n)$. We find that $(1)$ turns into a polynomial equation for $\alpha$: $$\sum c_i\alpha^{m_i}=0 $$ with $m_i=a_i\frac b{b_i}\in\Bbb N_0$. This polynomial must be a multiple of the minimal (because irreducible) polynomial $X^b-2$, but $m_i<b$ for all $i$, contradiction.