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

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

• I'm not convinced. How do you know this set is independent over $\mathbb{Q}$? Jan 10, 2022 at 5:03
• You really couldn't find another proof of this on the internet? Jan 10, 2022 at 5:06
• @William in this math.stackexchange.com/q/6517 they proved some similar Jan 10, 2022 at 5:09
• @DionelJaime No, my bad. I was trying to say that I could not find a proof using this corollary. Jan 10, 2022 at 5:10
• In that thread they show that the set of all $1/(2^n)$th roots of $2$ are independent. That's not what you're claiming. Hint: can you show that $2^{1/6}$ is in $\mathbb{Q}[2^{1/3},2^{1/2}]$, and thereby conclude your claim is hopeless? Jan 10, 2022 at 5:17

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.
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.
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.