Hamel Basis. Consequences?

Question

I am not sure to have the concept clear. Where is my reasoning flaw?

According to the proven existence of a Hamel Basis of $\mathbb{R}$ as a vector space over $\mathbb{Q}$, then it is possible to write any $x \in \mathbb{R}$ as finite sum of elements of $\mathbb{Q}$. So, there exist actual $x_1, x_2, ..., x_n \in \mathbb{Q}$ that sum up to $\pi$? How could it possibly be ever right?

I know that this would rely on using the axiom of choice. But anyway...

Answer 1

The elements of the Hamel Basis $H$ are elements of the vector space, $\mathbb{R},$ in your case.

What is means that $H$ is a basis is that for every $x\in\mathbb{R}$, there are finitely many $h_1,...,h_n\in H$ and rationals $q_1,...,q_n\in\mathbb{Q}$, such that $x=q_1h_1+...+q_nh_n$.

For the particular case of $x=\pi$, you wouldn't get $\pi$ equal to a finite sum of rationals, but a finite sum of elements of $H$, each multiplied by a rational.

Naturally, all or all but one of the elements of $H$ are not rational themselves. This is a consequence of them being linearly independent. If $h_1,h_2\in H\cap \mathbb{Q}$, then $0=h_2h_1-h_1h_2$ would be a non-trivial $\mathbb{Q}$-linear combination of $h_1,h_2$ that is equal to zero.

Answer 2

The sum is $x_1r_1+\dots+x_nr_n$ where the $r_i$ are the real basis elements who are mostly irrational.