Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
80  
A finite dimensional vector space over $\mathbb{Q}$ is countable. –  user641 Oct 7 '10 at 2:19
3  
@Steve: Please add that as an answer so people can upvote. –  Aryabhata Oct 7 '10 at 2:50
    
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. –  Isaac Solomon Oct 7 '10 at 2:55
    
@Isaac Your question doesn't require the 'only if' anyway. Steve's observation answers your original question. –  yasmar Oct 7 '10 at 3:04
12  
@Isaac: how is the fact that $\mathbb{R}$ is finite dimensional over itself relevant? –  Arturo Magidin Oct 7 '10 at 3:31

4 Answers 4

up vote 59 down vote accepted

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 $\mathbb{Q}$.

share|improve this answer
2  
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/… –  Jonas Meyer Oct 7 '10 at 4:18
2  
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). –  Arturo Magidin Oct 7 '10 at 14:51
1  
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. –  Jonas Meyer Oct 7 '10 at 16:02
1  
@Jonas: No argument there (with any of your points). –  Arturo Magidin Oct 7 '10 at 17:23
    
good link @Jonas, thanks –  Leon Lampret Sep 23 '11 at 0:23

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: enter image description here enter image description here

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 $\alpha_1A_{r_1}\!+\ldots+\alpha_kA_{r_k}\!\!=\!0\;(1)$ with $\alpha_i\!\in\!\mathbb{Q}$. We can assume $r_1\!>\!\ldots\!>\!r_k$ (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 $n!(\alpha_1\sum_{q_m<r_1}\frac{1}{m!}\!+\ldots+\alpha_k\sum_{q_m<r_k}\frac{1}{m!}\!)\!=\!0$. Rearranged (via $(2)$ when $m=n$), it reads

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

The left hand side (LHS) of $(3)$ is an integer for any $n$. If $n$ is large enough that $(|\alpha_1|+\ldots+|\alpha_k|)\sum_{m=n+1}^\infty\frac{n!}{m!}<1$ 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 $\alpha_2\!=\!\ldots\!=\!\alpha_k\!=\!0$.

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

share|improve this answer

The cardinality argument mentioned by Arturo is probably the simplest. Here is an alternative: an explicit example of an infinite $\rm\: \mathbb Q$-independent set of reals. Consider the set consisting of the logs of all primes $\rm\: p_i\:$. Then if $\rm\ \: 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 $\rm\ c_i \in \mathbb Z\:,\:$ and, exponentiating, we obtain $\rm\: p_1^{\:c_1}\cdots p_n^{\:c_n} = 1\ \Rightarrow\ c_i = 0,\:$ for all $\rm\:i\:$.

share|improve this answer
21  
@Bill +1 What a nice example. –  Adrián Barquero Oct 7 '10 at 4:50
7  
@Bill. Wow! :-) –  a.r. Oct 7 '10 at 6:08
2  
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. –  T.. Oct 12 '10 at 7:14
2  
@George S. "Then, any number field has a finite extension whose class number is 1." This would imply in particular that there are infinitely many number fields of class number one, which I am pretty sure is an open problem. I wonder what you are thinking here? –  Pete L. Clark Jan 22 '11 at 10:20
10  
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. –  JDH Jun 23 '11 at 2:22

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, = \cup_{n>0} 2^{2^{-n}} $ is linearly independent over $Q$. (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 $Q$. (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 $Q$ 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.

share|improve this answer
1  
@T..: Yes, of course. But, in my experience, students find the proof using logs of primes to be much simpler than any of the algebraic versions. That's why I presented it that way. –  Bill Dubuque Oct 11 '10 at 14:18
1  
@Bill: the question asked about proofs that do not "show the existence of non-algebraic numbers", and so it is of interest to give proofs that stay entirely within the algebraic (real) numbers. The set of sqrt(p) has minimal "depth" and the sqrt(sqrt...(2)) tower has minimal "width" among algebraic examples, making them somewhat canonical. The log(p) argument can be thought of as using an Archimedean valuation, complementing the approach with p-adic (non-Archimedean) valuations, so there seems to be a common mechanism in the proofs posted so far. –  T.. Oct 11 '10 at 15:26
6  
@T..: But the proof using logs of primes is certainly not the same as a proof "by showing the existence of a non-algebraic number". Indeed, the log-prime proof makes no (explicit) mention nor use of the notion of the notion of algebraic or transcendental numbers. It can be understood by a reader who has no knowledge of those concepts. So it certainly provides a fine answer to the question as posed. Moreover, it is much more elementary that any alternative algebraic approach. –  Bill Dubuque Oct 11 '10 at 15:35
2  
@T..: Sure, but the "stricter standard" of using only real algebraic numbers is your spin on the problem. No such constraint was imposed by the OP. Instead, he explicitly asked asked for a quick proof simpler than the obvious one - viz. that the existence of real number transcendental over Q implies dim R/Q infinite. I can't think of any other proof besides the log(p) proof that meets both criteria. That said, of course, I agree that it is interesting to look at the problem from many angles. –  Bill Dubuque Oct 11 '10 at 17:16
1  
@T..: Above, where you said "the log(p) proof does not meet the stricter standard of not using transcendental numbers in any way". That standard is of your own creation. It is not in the original problem. In any case the point of my initial remark was merely to convey to you my experience presenting such proofs - not to begin an extended discussion comparing the merits of various proofs. It was meant merely as friendly advice on the the pedagogical difficulty of the algebraic approach. It's far too easy to overlook such pedagogical matters if one hasn't actually taught such topics. –  Bill Dubuque Oct 11 '10 at 18:27

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.