Take the 2-minute tour ×  It would seem that one way of proving
this would be to show the existence of non-algebraic numbers. Is
there a simpler way to show this?  share|improve this question A
finite dimensional vector space over $\mathbb{Q}$ is countable.
–  user641 Oct 7 '10 at 2:19 @Steve: Please add that as an
answer so people can upvote. –  Aryabhata Oct 7 '10 at 2:50 Does
that mean that a vector space over $\mathbb{Q}$ is finite-
dimensional iff the set of the vector space is countable? If so,
please prove it. –  Isaac Solomon Oct 7 '10 at 2:55 @Isaac Your
question doesn't require the 'only if' anyway. Steve's
observation answers your original question. –  yasmar Oct 7 '10
at 3:04 @Isaac: how is the fact that $\mathbb{R}$ is finite
dimensional over itself relevant? –  Arturo Magidin Oct 7 '10 at
3:31 show 2 more comments  4 Answers  up vote 52 down vote
accepted  As Steve D. noted, a finite dimensional vector space
over a countable field is necessarily countable: if
$v_1,\ldots,v_n$ is a basis, then eve