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

extremeaxe5
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  • Note, though, that this has nothing to do with Hamel bases. Even the existence of an uncountable set of reals linearly independent over $\mathbb{Q}$ - which is provable without the axiom of choice, and in fact not hard to construct - has nothing to do with Hamel bases. – Noah Schweber Sep 29 '18 at 15:58

1 Answers1

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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.

Hagen von Eitzen
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