Is this proof about why the vector space of $\mathbb{R}$ over $\mathbb{Q}$ is infinite-dimensional correct?

Question

The problem says that

Let $V$ be the set of real numbers. Regard $V$ as a vector space over the field of rational numbers, with the usual operations. Prove that this vector space is not finite-dimensional.

In my proof I use this Corollary:

"Let $V$ be a finite-dimensional vector space and let $n=\dim( V)$. Then any subset of $V$ which contains more than $n$ vectors is linearly dependent."

So, my proof goes like this:

"Suppose that $V$ is a finite-dimensional vector space and let $n=\dim(V)$. Notice that the set $\left \{ \sqrt[m]{2}:m\geq n,m\in \mathbb{N} \right \}$ have more than $n$ vectors and is an independent set which is a contradiction by the lemma above, thus $V$ is not a finite-dimensional vector space. "

I was looking for another proofs on google but I could not find something like this, so is this proof correct? Any help will be appreciated

Answer 1

Here is another proof. Suppose $\mathbb R$ is a finite-dimensional $\mathbb Q$ vector space, with basis $\{x_1,\dots,x_n\}$. But $span\{x_1,\dots,x_n\}\cong\mathbb Q^n$ has cardinality $\aleph_0$ while $\mathbb R$ has cardinality $2^{\aleph_0}$, which is absurd.

Answer 2

Your proof is sufficient for the question that was asked once you show that those vectors are independent. It could well be that you have already shown that, perhaps as a lead-in for this question. As Kenta S has shown, you can prove more, that the dimension is $\mathfrak c$, but you were not asked for that.

Answer 3

You can use the linear algebra lemma in the following way. Suppose that this vector space is of dimension $n$. Take any real number $\alpha$. Then the powers $1, \alpha,...,\alpha^n$ are linearly dependent (as in the lemma). Then for some $k\le n$ we have that $\alpha^k$ is a linear combination of $1,...,\alpha^{k-1}$ with rational coefficients. Hence $\alpha^k=a_0\cdot 1+...+a_{k-1}\alpha^{k-1}$, so $\alpha$ is a root of the polynomial $x^k-a_{k-1}x^{k-1}-...-a_0$, hence $\alpha$ is algebraic. But $\Bbb R$ contains transcendental numbers (say, $\pi$), a contradiction.