Dimension of $\mathbb{R}$ over $\mathbb{Q}$ Show that the set $\mathbb{R}$ is a vector space on the set $\mathbb{Q}$. 
What is dimension of of $\mathbb{R}$ over $\mathbb{Q}$? 
First part is easy. Both $\mathbb{R}$ and $\mathbb{Q}$ are fields and $\mathbb{R}$ contain $\mathbb{Q}$, hence proved.
Assume the dimension is $n$ and the vector space has the ordered basis $B=(x_1,x_2,\dots,x_n)$ where $x_i\in \mathbb{R}$. Then for every $x\in \mathbb{R}$ we have exactly one ordered set of rational number $t_i$'s such that $x=x_1t_1+\cdots+x_nt_n$.
Now we define the map
$$f:\mathbb{R}\rightarrow \mathbb{Q^n}$$ 
$$f(x)=(t_1,\dots,t_n)\text{ where }x=x_1t_1+\cdots+x_nt_n\,.$$  
I think the function $f$ is well defined injective function and by:

Theorem: Let $f:X\rightarrow Y$ be an injection and $Y$ is countable, then $X$ is coutable as well.

We reach a contradiction because $\mathbb{R}$ is uncountable.
I'm not sure about the proof. It seems alright but this question was to be supposed a tough one. 
Please provide your thoughts about it. If it is wrong please point it out. 
 A: Your proof only shows that $\Bbb R$ is not of finite dimension over $\Bbb Q$, and by stretching it a bit more, you could even show that it implies that the dimension is not countable.
But being uncountable just means not countable. If you were asked how many elements are in $\{0,1,2,3,4\}$, then "more than one" is not the answer you're expected to give.
To solve the mystery, ask yourself, if $V$ is an infinite dimensional vector space over $\Bbb Q$, and $B$ is a basis for $V$. How would the cardinality of $B$ relate to the cardinality of $V$?
A: Assume that $\mathbb{R}$ has a countable basis. Let $\mathcal{B}=\{r_{1},r_{2},...\}$ be your basis for $\mathbb{R}$ over $\mathbb{Q}$.
Then Let $\mathcal{B_{n}}=\{r_{1},r_{2},...r_{n}\}$.
Now we can say that $$\mathbb{R}=\text{span}(\mathcal{B})=\bigcup_{n=1}^{\infty}\text{span}(\mathcal{B_{n}})$$.
But let $a\in\text{span}(\mathcal{B_{n})}$ be a real number.
Then $a=\sum_{i=1}^{n}q_{i}r_{i}$.
Then $f: \text{span}(\mathcal{B_{n})}\to \mathbb{Q}^{n}$ given by
$f(a)=(q_{1},q_{2},...,q_{n})^{T}$ is a one-one map.
Hence $\text{span}(\mathcal{B_{n}})$ is countable as $\mathbb{Q}^{n}$ is countable.
And hence $$\mathbb{R}=\bigcup_{n=1}^{\infty}\text{span}(\mathcal{B_{n}})$$ is countable as countable union of countable sets is countable.
This contradicts uncountability of $\mathbb{R}$.
A: Since all the real numbers are not rational, i.e. we have irrationals like $ e,\sqrt{2},...$ which can be expressed as a series of rational numbers. Precisely here in the expression $$e=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...$$ there are infinitely many rational numbers express a single real number. So the dimension of the vector space $\mathbb{R}$ over $\mathbb{Q}$ is infinite.
