# Idea for a proof that a Hamel basis is infinite without using a countability argument

I am trying to prove that a Hamel basis is infinite without using the countability argument. My idea goes like this:

Assume the basis is finite with irrational elements $a_1,a_2,a_3, \dots, a_n$. I want to show that the element $(a_1+a_2+a_3+ \dots a_n)^{\frac{1}{2}}$ is not present in the span of the set (if it is so that is). I can then keep adding basis elements to the given basis using this method, proving that it is infinite.

Note: I am aware that the basis is uncountable.

-

Unfortunately this promising-looking approach doesn't work. If $n=2$ and $a_1=\pi$, $a_2=\pi^2-\pi$, then $\sqrt{\pi+(\pi^2-\pi)} = \pi$ is definitely in the span of $\{a_1,a_2\}$.
It's not clear to me whether you are given a Hamel basis and want to prove that it's infinite, or whether you want to construct an infinite Hamel basis, or whether you want to prove that any Hamel basis must be infinite. (You might or might not have at your disposal the fact that all bases have the same cardinality.) The following simplification of your idea might be helpful for some of these options: let $a_1$ be transcendental, and let $a_{n+1} = \sqrt{a_n}$ for $n\ge1$. (Or just let $a_n = a_1^n$ for $n\ge1$.)