Show that real field $\mathbb R$ is a vector space of infinite dimensions over the field of rational numbers $\mathbb Q$. I am doing it in the following way. Is it correct?
Set $S = \{1,\pi,\pi^2,\pi^3,...,\pi^n\}$ is LI over $\mathbb Q$
Suppose 
$a_0\times 1 + a_1\times\pi + ... a_n\times \pi^n = 0$ where all the a_i's are not $0$.
Then $\pi$ is a root of $a_0 + a_1x + ... + a_nx^n = 0$
which is imposible since $\pi$ is a transcendental number.
Therefore, S is LI. Hence $\mathbb R$ is of infinite dimension over $\mathbb Q$.
 A: A much simpler argument would be to use the fact that real numbers form an uncountably infinite set, whereas rationals form a countable set. 
Check that a vector space of countable dimension over the rationals would still be a countable set. Hence  the set of real numbers as a vector space over the rationals is of uncountable dimension.
A: It seems correct to me, but you have to correct some few things:


*

*You said "Set $S=\{1,\pi,\dots,\pi^n\}$ is LI over $\mathbb Q$" and then you proved that it is LI and that's a bad way to write things. Instead you can say: "Set $S=\{1,\pi,\dots,\pi^n\}$. Let's prove that $S$ is LI over $\mathbb Q$".

*To prove that $\mathbb R$ is infinite dimensional over $\mathbb Q$, you used the property that:

A vector space $V$ over a field $\mathbb F$ is infinite dimensional if and only if there is a sequence $v_1,v_2,\dots$ of vectors in $V$ such that $\{v_1,\dots ,v_n\}$ is linearly independent for every positive integer $n$.

but  you didn't emphasise on such a sequence. You could then choose the notation $S_n$ for $S$ and after your proof involving polynomials you say "Therefore $S_n$ is a sequence of vectors in $\mathbb R$ which is LI for all $n\in\mathbb N$. Thus $\mathbb R$ is infinite dimensional over $\mathbb Q$".
