# Prove that this vector space is not finite dimensional. [duplicate]

Let $V$ be the set of real numbers. Regard V as a vector space over the field of rational numbers $F$ with the usual operations. Prove that this vector space is not finite dimensional. My attempt: Let $\beta$ be the basis of $V$ such that $\beta =\{\alpha_1, \alpha_2, ..., \alpha_n\}$. $\quad\therefore\;\forall\;\alpha\in span(\beta),\;\alpha = c_1\alpha_1 + c_2\alpha_2 + ..... c_n\alpha_n$ where $c_1, c_2, ...., c_n \in F$. Now I have to show there exists $\alpha \in V$ such that $\alpha \notin span(\beta)$ but I cant figure out which $\alpha$ will not belong in $span(\beta)$.

## marked as duplicate by sdcvvc, Community♦Apr 11 '15 at 8:42

• Do you know some basic set theory (cardinalities, countable, uncountable and stuff)? – Timbuc Apr 10 '15 at 18:11
• Yes, sets of integers and rational numbers are countable while the set of real numbers is not – In78 Apr 10 '15 at 18:14
• @ln Fine, then read below. – Timbuc Apr 10 '15 at 18:15

Consider $\{\log 2, \log 3, \log 5, \dots\}$, the logarithms of the primes. This infinite set is linearly independent over the rationals, by the Fundamental Theorem of Arithmetic (uniqueness of prime factorizations).

Another solution, not involving cardinalities, is to find an infinite linearly independent set.

Let $\alpha$ be any transcendental number ($\pi$ works). Consider the set $\{\alpha^{n}|n \in \mathbb{N}\}$. Then this set is linearly independent.

Suppose $\sum\limits_{i=1}^{n}a_i\alpha^i = 0$ for $a_i \in \mathbb{Q}$ then $a_i = 0$ for all $i$. Otherwise $\alpha$ is a root of the polynomial $p(x) = \sum\limits_{i=1}^{n}a_ix^i$, but this cannot be with $\alpha$ transcendental.

• It might take a little bit of finessing to prove, without appealing to cardinality, that transcendental numbers exist. – Milo Brandt Apr 10 '15 at 22:59

Remember that $\;|\Bbb Q|=\aleph_0<2^{\aleph_0}=|\Bbb R|\;$ , so for any countable set $\;S:=\{r_1,\ldots,r_n,\ldots\}\subset\Bbb R\;$ , we get that

$$\left|\text{Span}_{\Bbb Q}\{S\}\right|\le \aleph_0\cdot\aleph_0=\aleph_0<|\Bbb R|$$

So not only $\;\dim_{\Bbb Q}\Bbb R\;$ is not finite: it even is uncountable infinite

• I can't understand why $|Span\{S\}|$ is less than or equal to $\aleph_0.\aleph_0$ – In78 Apr 10 '15 at 18:45
• The cardinality of both $\;\Bbb Q\,,\,\,S\;$ is $\;\aleph_0\;$ , so you have at most $\;\aleph_0\cdot\aleph_0\;$ possible finite linear rational combinations of elements in $\;S\;$ . – Timbuc Apr 10 '15 at 18:49