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It would seem that one way of proving this would be to show the existence of non-algebraic numbers. Is there a simpler way to show this?

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    $\begingroup$ A finite dimensional vector space over $\mathbb{Q}$ is countable. $\endgroup$ – user641 Oct 7 '10 at 2:19
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    $\begingroup$ @Steve: Please add that as an answer so people can upvote. $\endgroup$ – Aryabhata Oct 7 '10 at 2:50
  • $\begingroup$ Does that mean that a vector space over $\mathbb{Q}$ is finite-dimensional iff the set of the vector space is countable? If so, please prove it. $\endgroup$ – Elchanan Solomon Oct 7 '10 at 2:55
  • $\begingroup$ @Isaac Your question doesn't require the 'only if' anyway. Steve's observation answers your original question. $\endgroup$ – yasmar Oct 7 '10 at 3:04
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    $\begingroup$ @Isaac: how is the fact that $\mathbb{R}$ is finite dimensional over itself relevant? $\endgroup$ – Arturo Magidin Oct 7 '10 at 3:31
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As Steve D. noted, a finite dimensional vector space over a countable field is necessarily countable: if $v_1,\ldots,v_n$ is a basis, then every vector in $V$ can be written uniquely as $\alpha_1 v_1+\cdots+\alpha_n v_n$ for some scalars $\alpha_1,\ldots,\alpha_n\in F$, so the cardinality of the set of all vectors is exactly $|F|^n$. If $F$ is countable, then this is countable. Since $\mathbb{R}$ is uncountable and $\mathbb{Q}$ is countable, $\mathbb{R}$ cannot be finite dimensional over $\mathbb{Q}$. (Whether it has a basis or not depends on your set theory).

Your further question in the comments, whether a vector space over $\mathbb{Q}$ is finite dimensional if and only if the set of vectors is countable, has a negative answer. If the vector space is finite dimensional, then it is a countable set; but there are infinite-dimensional vector spaces over $\mathbb{Q}$ that are countable as sets. The simplest example is $\mathbb{Q}[x]$, the vector space of all polynomials with coefficients in $\mathbb{Q}$, which is a countable set, and has dimension $\aleph_0$, with basis $\{1,x,x^2,\ldots,x^n,\ldots\}$.

Added: Of course, if $V$ is a vector space over $\mathbb{Q}$, then it has countable dimension (finite or denumerable infinite) if and only if $V$ is countable as a set. So the counting argument in fact shows that not only is $\mathbb{R}$ infinite dimensional over $\mathbb{Q}$, but that (if you are working in an appropriate set theory) it is uncountably-dimensional over $\mathbb{Q}$.

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    $\begingroup$ Related to the note about uncountable dimension, there are explicit examples of continuum-sized linearly independent sets, as seen in this MathOverflow answer by François G. Dorais: mathoverflow.net/questions/23202/… $\endgroup$ – Jonas Meyer Oct 7 '10 at 4:18
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    $\begingroup$ Yes: but can one show that there is a basis for $\mathbb{R}$ over $\mathbb{Q}$ without some form of the Axiom of Choice? (There is a difference between exhibiting a large linearly independent subset and exhibiting a basis). $\endgroup$ – Arturo Magidin Oct 7 '10 at 14:51
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    $\begingroup$ No, one cannot, but without the Axiom of Choice the notion of dimension breaks down (except for finite vs. infinite). Assuming AC (as I virtually always do), the size of a linearly independent set gives a lower bound on the dimension of the vector space, and I think it is wonderful that in this case such "explicit" proof exists that the real numbers have continuum dimension, as opposed to the nice qualitative proof one could give by extending your argument to larger cardinals. $\endgroup$ – Jonas Meyer Oct 7 '10 at 16:02
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    $\begingroup$ @Jonas: No argument there (with any of your points). $\endgroup$ – Arturo Magidin Oct 7 '10 at 17:23
  • $\begingroup$ good link @Jonas, thanks $\endgroup$ – Leon Sep 23 '11 at 0:23
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The cardinality argument mentioned by Arturo is probably the simplest. Here is an alternative: an explicit example of an infinite $\, \mathbb Q$-independent set of reals. Consider the set consisting of the logs of all primes $\, p_i.\,$ If $ \, c_1 \log p_1 +\,\cdots\, + c_n\log p_n =\, 0,\ c_i\in\mathbb Q,\,$ multiplying by a common denominator we can assume that all $\ c_i \in \mathbb Z\,$ so, exponentiating, we obtain $\, p_1^{\large c_1}\cdots p_n^{\large c_n}\! = 1\,\Rightarrow\ c_i = 0\,$ for all $\,i,\,$ by the uniqueness of prime factorizations.

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    $\begingroup$ @Bill +1 What a nice example. $\endgroup$ – Adrián Barquero Oct 7 '10 at 4:50
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    $\begingroup$ @Bill. Wow! :-) $\endgroup$ – Agustí Roig Oct 7 '10 at 6:08
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    $\begingroup$ Is this proof unique to Q or does it generalize to provide explicit examples of reals linearly independent over, e.g., Q(sqrt(2))? Above, Q appears to be "hard-wired" into the proof, as the group of exponents in the prime factorization. $\endgroup$ – T.. Oct 12 '10 at 7:14
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    $\begingroup$ Can you extend the idea of this proof to get the right dimension? Currently, you only have a countably infinite independent set, but the dimension is size continuum. $\endgroup$ – JDH Jun 23 '11 at 2:22
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    $\begingroup$ @student this can not happen: the primes are distinct. Put the primes with negative exponent on the other side (exponents become positive). You would find two distinct prime factorizations of the same number. $\endgroup$ – Student Feb 6 at 19:44
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No transcendental numbers are needed for this question. Any set of algebraic numbers of unbounded degree spans a vector space of infinite dimension. Explicit examples of linearly independent sets of algebraic numbers are also relatively easy to write down.

The set $\sqrt{2}, \sqrt{\sqrt{2}}, \dots, = \bigcup_{n>0} 2^{2^{-n}} $ is linearly independent over $\mathbb Q$. (Proof: Any expression of the $n$th iterated square root $a_n$ as a linear combination of earlier terms $a_i, i < n$ of the sequence could also be read as a rational polynomial of degree dividing $2^{n-1}$ with $a_n$ as a root and this contradicts the irreducibility of $X^m - 2$, here with $m=2^n$).

The square roots of the prime numbers are linearly independent over $\mathbb Q$. (Proof: this is immediate given the ability to extend the function "number of powers of $p$ dividing $x$" from the rational numbers to algebraic numbers. $\sqrt{p}$ is "divisible by $p^{1/2}$" while any finite linear combination of square roots of other primes is divisible by an integer power of $p$, i.e., is contained in an extension of $\mathbb Q$ unramified at $p$).

Generally any infinite set of algebraic numbers that you can easily write down and is not dependent for trivial reasons usually is independent. This because the only algebraic numbers for which we have a simple notation are fractional powers, and valuation (order of divisibility) arguments work well in this case. Any set of algebraic numbers where, of the ones ramified at any prime $p$, the amount of ramification is different for different elements of the set, will be linearly independent. (Proof: take the most ramified element in a given linear combination, express it in terms of the others, and compare valuations.)

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  • $\begingroup$ I realize you answered this question a loooong time ago, but if you're still around, how do you know that you can write the $n-$th root of $2$ as a linear combination of the $1st$ through $n-1$st roots of $2$? $\endgroup$ – ALannister Aug 17 '17 at 22:41
  • $\begingroup$ @ALannister I realize you questioned quite a long time ago, but s/he is assuming that there is such $n$ and showing that such assumption leads to contradiction. So in fact, s/he is proving exactly that $n$-th root of $2$ cannot be expressed by a linear combination of previous roots $\endgroup$ – user160738 May 9 '18 at 17:24
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For the sake of completeness, I'm adding a worked-out solution due to F.G. Dorais from his post.

We'll need two propositions from Grillet's Abstract Algebra, page 335 and 640:

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Proposition: $[\mathbb{R}:\mathbb{Q}]=\mathrm{dim}_\mathbb{Q}{}\mathbb{R}=|\mathbb{R}|$

Proof: Let $(q_n)_{n\in\mathbb{N_0}}$ be an enumeration of $\mathbb{Q}$. For $r\in\mathbb{R}$, take $$A_r:=\sum_{q_n<r}\frac{1}{n!}\;\;\;\;\text{ and }\;\;\;\;A:=\{A_r;\,r\in\mathbb{R}\};$$ the series is convergent because $\sum_{q_n<r}\frac{1}{n!}\leq\sum_{n=0}^\infty\frac{1}{n!}=\exp(1)<\infty$ (recall that $\exp(x)=\sum_{n=0}^\infty\frac{x^n}{n!}$ for any $x\in\mathbb{R}$).

To prove $|A|=|\mathbb{R}|$, assume $A_r=A_{s}$ and $r\neq s$. Without loss of generality $r<s$, hence $A_s=\sum_{q_n<s}\frac{1}{n!}=\sum_{q_n<r}\frac{1}{n!}+\sum_{r\leq q_n<s}\frac{1}{n!}=A_r+\sum_{r\leq q_n<s}\frac{1}{n!}$, so $\sum_{r\leq q_n<s}\frac{1}{n!}=0$, which is a contradiction, because each interval $(r,s)$ contains a rational number.

To prove $A$ is $\mathbb{Q}$-independent, assume $\alpha_1A_{r_1}+\cdots+\alpha_kA_{r_k}=0\;(1)$ with $\alpha_i\in\mathbb{Q}$. We can assume $r_1>\cdots>r_k$ (otherwise rearrange the summands) and $\alpha_i\in\mathbb{Z}$ (otherwise multiply by the common denominator). Choose $n$ large enough that $r_1>q_n>r_2\;(2)$; we'll increase $n$ two more times. The equality $n!\cdot(1)$ reads $n!(\alpha_1\sum_{q_m<r_1}\frac{1}{m!}+\cdots+\alpha_k\sum_{q_m<r_k}\frac{1}{m!})=0$. Rearranged (via $(2)$ when $m=n$), it reads

$$-\alpha_1\sum_{\substack{m<n\\q_m<r_1}}\frac{n!}{m!}-\cdots-\alpha_k\sum_{\substack{m<n\\q_m<r_k}}\frac{n!}{m!}-\alpha_1 =\alpha_1\sum_{\substack{m>n\\q_m<r_1}}\frac{n!}{m!}+\cdots+\alpha_k\sum_{\substack{m>n\\q_m<r_k}}\frac{n!}{m!}. \tag*{(3)}$$

The left hand side (LHS) of $(3)$ is an integer for any $n$. If $n$ is large enough that $(|\alpha_1|+\cdots+|\alpha_k|)\sum_{m=n+1}^\infty\frac{n!}{m!}<1$ holds (such $n$ can be found since $\sum_{m=n+1}^\infty\frac{n!}{m!}=\frac{1}{n+1}\sum_{m=n+1}^\infty\frac{1}{(n+2)\cdot\ldots\cdot m}\leq\frac{1}{n+1}\sum_{m=n+1}^\infty\frac{1}{(m-n-1)!}\leq\frac{1}{n+1}\exp(1)\rightarrow 0$ when $n\rightarrow\infty$), then the absolute value of RHS of $(3)$ is $<1$, and yet an integer, hence $\text{RHS}(3)=0$. Thus $(3)$ reads $\alpha_1=-\sum_{i=1}^{k}\sum_{m<n,q_m<r_i}\alpha_i\frac{n!}{m!}=0\;(\mathrm{mod}\,n)$. If moreover $n>|\alpha_1|$, this means that $\alpha_1=0$. Repeat this argument to conclude that also $\alpha_2=\cdots=\alpha_k=0$.

Since $A$ is a $\mathbb{Q}$-independent subset, by proposition 5.3 there exists a basis $B$ of $\mathbb{R}$ that contains $A$. Then $A\subseteq B\subseteq\mathbb{R}$ and $|A|=|\mathbb{R}|$ and Cantor-Bernstein theorem imply $|B|=|\mathbb{R}|$, therefore $[\mathbb{R}:\mathbb{Q}]=\mathrm{dim}_\mathbb{Q}{}\mathbb{R}=|\mathbb{R}|$. $\quad\blacksquare$

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  • $\begingroup$ Is this uncountable linearly independent set basis ? $\endgroup$ – user195218 Jun 5 '15 at 2:53
  • $\begingroup$ @user195218 It's not a basis, as it doesn't span the reals, but it is uncountable and linearly independent. $\endgroup$ – Akiva Weinberger Jul 25 '17 at 19:08
  • $\begingroup$ The only proper answer to the question. Such a shame that it is not the most voted answer. $\endgroup$ – Akerbeltz May 1 at 21:42
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Here is a simple proof that a basis $B$ of $[ \mathbb{R}:\mathbb{Q}]$ has cardinality $|\mathbb{R}|$.

Clearly since $B$ is contained in $R$, $|B| \le| \mathbb{R}|$.

But also $\mathbb{R}=span(B)$ and thus $|\mathbb{R}| =|span(B)| \le |\mathbb{Q}^B|=|B|$. The last equality follows because $B$ is not finite (if it was, then $|\mathbb{R}|=|span(B)| \le |\mathbb{Q}^B|=|\mathbb{N}|$, a contradiction).

Hence $ |\mathbb{R}| \le |B| \le| \mathbb{R}|$, so $|\mathbb{R}|=|B|$

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  • $\begingroup$ There is an error in what you wrote: $|\mathbb{Q}^B|\geq |2|^{|B|} > |B|$. $\endgroup$ – Nex Aug 18 '17 at 8:19
  • $\begingroup$ @Next: hmm so is it salvageable what I wrote? I guess I wasn't paying too much attention. $\endgroup$ – Joshua Benabou Aug 18 '17 at 19:08
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Another simple proof:

Take $P=X^n-p$ for a prime $p$.

By Eisenstein's criterion, it is $\mathbb Q[X]$-irreductible. Therefore, the set of algebraic numbers is of infinite dimension over $\mathbb Q$.

Since $\mathbb R$ is bigger, it works for $\mathbb R$ too.

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protected by user26857 Nov 25 '15 at 9:30

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