# Proving that the vector space of $\mathbb R$ over field $\mathbb Q$ is not finite dimensional.

Suppose that $$\mathbb R$$ denotes the set of all real numbers, $$\mathbb Q$$; the set of all rational numbers.

Let $$V=\mathbb R(\mathbb Q)$$ denote the vector space of real numbers over field $$\mathbb Q$$.

Suppose on the contrary, that $$V$$ is finite dimensional. It follows that $$V$$ has a finite basis $$\mathbb B=\{b_i\in \mathbb R:1\le i\le n\},$$ where $$n$$ is a fixed natural number.

Every $$r\in \mathbb R$$ can be written uniquely as linear combination of vectors in $$\mathbb B$$.

That is, for any $$r\in \mathbb R$$, there exist unique $$q_{ir}\in \mathbb Q, 1\le i\le n$$ such that

$$r=q_{1r}b_1+q_{2r}b_2+\cdots+q_{nr}b_n$$

Let $$f: \mathbb R\to \mathbb Q^n$$ be a function defined as

$$f(r)=(q_{1r},q_{2r},\cdots,q_{nr})$$

$$f$$ is one-one and therefore $$\mathbb R$$ has cardinality not exceeding cardinality of $$\mathbb Q^n$$, i.e. aleph-null. This is a contradiction as $$\mathbb R$$ is uncountable.

Hence by contradiction, $$V$$ must be infinite dimensional vector space.

Is my proof correct? Thanks.

• Looks good to me. Sep 16, 2021 at 5:03
• @JairTaylor: Thanks a lot for reviewing the proof. :)
– Koro
Sep 16, 2021 at 5:07
• It's an alright proof, but there's no need to mention cardinality: all you had to use is the fact that $\mathbb{R}$ is not countable while $\mathbb{Q}^n$ is (of course, I recognize I'm saying the exact same thing you did, but in a simpler way). Sep 16, 2021 at 5:09
• @MatheusAndrade: Thanks for the comment. Yeah, I thought about that but that somehow involved cardinal arithmetic ($r=\sum q_i b_i$ so $|\mathbb R|\le |\mathbb Q|^n$ as $q_i$ can be chosen in $|\mathbb Q|$ ways) so I wanted to avoid that.
– Koro
Sep 16, 2021 at 5:13
• @Koro Sorry for the mistake, but $\mathbb{R}$ is absolutely not countable, I mistyped but corrected the mistake in a later edit (it's correct now). If there were an injection from $\mathbb{R}$ to a countable space then $\mathbb{R}$ would be countable, which is not the case. Sep 16, 2021 at 5:14