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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 $span(\beta)$.

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

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

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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.

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

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  • $\begingroup$ I can't understand why $|Span\{S\}|$ is less than or equal to $\aleph_0.\aleph_0$ $\endgroup$ – In78 Apr 10 '15 at 18:45
  • $\begingroup$ 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 $\;S\;$ . $\endgroup$ – Timbuc Apr 10 '15 at 18:49

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