Dimension of a vector space below

I have to prove that the dimension of the vector space of real numbers over Q (rational numbers) is infinity. How can I prove? I have no idea.

• Do you know that the cardinal of $\;\Bbb R\;$ is really bigger than that of $\;\Bbb Q\;$ ? In fact, we have that $$|\Bbb Q|=\aleph_0\;,\;\;|\Bbb R|=2^{\aleph_0}=\mathcal c$$ If you know the above, the reult follows at once from cardinality considerations. – Timbuc Dec 12 '14 at 15:28
• Is there a finite linear combination with coefficients in $\mathbb Q$ that gives $\pi$? I guess if you have to prove the answer though it might take you some time... – GPerez Dec 12 '14 at 16:16

Hint: Show that all $\log(p)$ for $p$ prime are linearly independent over $\mathbb{Q}$. Apply the exponential function to the equation $\lambda_1\log(p_1)+\cdots +\lambda_n\log(p_n)=0$, and conclude that all $\lambda_i$ are zero. It follows that $\dim_{\mathbb{Q}}\mathbb{R}\ge n$ for all $n\ge 1$.
In your case, the "vectors" are real numbers and the "scalars" are rational numbers. So the dimension being finite means there is a finite set $r_1,\ldots,r_n$ of real numbers such that every real number $r$ can be written in the form $r=q_1r_1+q_2r_2+\cdots q_nr_n,$ for some rational numbers $q_1,\ldots,q_n$. Can you see the contradiction here by thinking about countability vs uncountability?