Prove that $\mathbb R$ seen as a $\mathbb Q$-vector space cannot have a countable basis.
I think that it is obvious that the main idea to use is that $\mathbb Q$ is countable but $\mathbb R$ is not. But I don't know how to write it a rigorous way.
Existance of a countable base for $\mathbb R$
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$\begingroup$
linear-algebra
first-countable
1 Answers
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Your notion is right, so I'd just give you a hint:
Suppose $\mathbb{R}$ has a countable $\mathbb{Q}$ basis $B = (b_1, b_2 ,\dots)$ Then every element $x$ in $\mathbb{R}$ is a linear combination of finitely many basis vectors, i.e. $$x = \sum_{i=1}^{n} a_i b_{k_i}$$ for some $n \in \mathbb N$, $a_i \in \mathbb Q, b_{k_i}$ in $B$.
So every real number can be seen as an element of $\mathbb Q^n \times B^n$ for some natural $n$.
Now, what can you say about the cardinality of $\mathbb Q^n \times B^n$? And what does this imply for the cardinality of $\mathbb R$?