Does the group of rationals under addition have a basis?

Question

I'm trying to determine whether $\mathbb Q$ under addition has a basis. My naive idea, based on what I have covered so far, would be to try and show that if we assume $\mathbb Q$ has a basis then it would be finitely generated which I believe would then lead to a contradiction. But I don't think this is a valid approach and I'm not sure what other approaches I could take.

I'd appreciate some ideas/approaches about determining whether $\mathbb Q$ has a basis or not :)

Answer 1

You haven't defined what a "basis" is.

Here's, however, a statement showing that $\mathbf{Q}$ is far to having a basis as abelian group for any reasonable notion of basis as group:

For every generating subset $S$ of $\mathbf{Q}$, for every finite subset $F$ of $S$, the subset $S-F$ generates $\mathbf{Q}$.

(I assume that a basis would at least have the property that it's a generating subset in which no element is redundant; the above shows it can't exist.)

The claim is true: indeed, the quotient $H=\langle S\rangle/\langle S-F\rangle$ is a finitely generated quotient of $\mathbf{Q}$. Hence, if nonzero, it admits for some prime $p$ a cyclic group $C_p$ of order $p$ as quotient. But since $p\mathbf{Q}=\mathbf{Q}$, it cannot admit $C_p$ as quotient, so $H=0$, that is, $\langle S-F\rangle=\langle S\rangle$.