Which of these statements about the field extension $\mathbb{R}/\mathbb{Q}$ are true? 
We know that $\mathbb R$ is an extension of $\mathbb Q$. Justify the following (true /false):

*

*$[\mathbb R:\mathbb Q]<\infty$


*$[\mathbb R:\mathbb Q]=$ countably infinite / uncountably infinite


*$[\mathbb R:\mathbb Q]$ is algebraic

My try:
If $[\mathbb R:\mathbb Q]<\infty$ then let $\{e_1,e_2,...,e_n\}$ be a basis of $[\mathbb R:\mathbb Q]$
then any $r\in \mathbb R$ can be expressed as $r=c_1e_1+...+c_nr_n$.
Since $\mathbb Q$ is countable so $\mathbb R$ is countable a contradiction.
But how to show the second case.
For 3, if $[\mathbb R:\mathbb Q]$ is algebraic then for any $r\in \mathbb R$ we have a polynomial $f(x)=a_0+\cdots+a_nx^n$ of which $r$ is  a root. Since there are countably such polynomials and each have finite number of roots so $\mathbb R$ is countable a contradiction.
Please say whether above is correct and how to do 2?
 A: To do 2, you can use an argument similar to your first one. Do you know how big $\bigoplus_{n \in \Bbb N} \mathbb{Q}$ is? If $[\Bbb R : \Bbb Q]$ were countable, this is what $R$ would look like as a vector space over $\Bbb Q$.
Edit: your answers to 1 and 3 look good to me.
A: Both (1) and (3) look sound. As for (2), consider that if $E = \{ e_{k} : k \in \mathbb{N} \}$ is a basis for $\mathbb{R}$, then we can write every real number $\mathbb{R}$ as a finite sum $\sum_{i = 1}^{N} e_{k_{i}} q_{1}$, where $q \in \mathbb{Q}$. Let $S_{N} = \{ \sum_{i = 1}^{N} e_{k_{i}} q_{i} : q_{i} \in \mathbb{Q} \}$. Then if we consider the map $f_{N} : (E \times \mathbb{Q})^{N} \to \mathbb{R}$ given by $f(<e_{k_{1}}, q_{1}>, \ldots, <e_{k_{N}}, q_{N}>) = \sum_{i = 1}^{N} e_{k_{i}}q_{i}$, then we have that $S_{N} = f_{N} ((E \times \mathbb{Q})^{N})$, so $S_{N}$ is the image of a countable set, and is thus countable. By assumption, $\mathbb{R} = \bigcup_{N \in \mathbb{N}} S_{N}$, and so $\mathbb{R}$ is a countable union of countable sets, and thus countable, a contradiction.
