# 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.

• As a related exercise, note that a $\mathbb{Q}$ vector space of dimension $\kappa$ has cardinality $\kappa$ when $\kappa$ is infinite. More generally, when both the field and dimension are infinite, the cardinality of the vector space is the max of those two cardinalities. – Mike Haskel Mar 25 '17 at 6:53
• you can use \text{ where } when being in math statement. – zwim Mar 25 '17 at 7:11
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 $V$?