Let \(V\) be the set of real numbers. Regard \(V\) as a vector space over the field of rational numbers, with the usual operations. Prove that this vector space is not finite-dimensional.
Solution
Suppose for contradiction that \(\RR\) were finite-dimensional over \(\QQ\), say with basis \(v_1,\ldots,v_n\). Then every real number would have the form \[ q_1v_1+\cdots+q_nv_n, \qquad q_i\in\QQ. \] Thus the map \[ \QQ^n\longrightarrow\RR, \qquad (q_1,\ldots,q_n)\longmapsto \sum_{i=1}^n q_iv_i \] would be surjective. But \(\QQ\) is countable, so the finite Cartesian product \(\QQ^n\) is countable; hence its image would be countable. This would make \(\RR\) countable, contradicting the uncountability of the real numbers. Therefore \(\RR\) is not finite-dimensional as a vector space over \(\QQ\).