# Is there a quick proof as to why the vector space of $\mathbb{R}$ over $\mathbb{Q}$ is infinite-dimensional?

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?

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

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.

• @Bill +1 What a nice example. Oct 7 '10 at 4:50
• @Bill. Wow! :-) Oct 7 '10 at 6:08
• 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
• 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
• @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. Feb 6 '19 at 19:44

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 $\mathbb{Q}$.

• 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/… Oct 7 '10 at 4:18
• 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). Oct 7 '10 at 14:51
• 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. Oct 7 '10 at 16:02
• @Jonas: No argument there (with any of your points). Oct 7 '10 at 17:23
– Leo
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:  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 $[\mathbb{R}:\mathbb{Q}]=\mathrm{dim}_\mathbb{Q}{}\mathbb{R}=|\mathbb{R}|$. $\quad\blacksquare$

• Is this uncountable linearly independent set basis ?
– user195218
Jun 5 '15 at 2:53
• @user195218 It's not a basis, as it doesn't span the reals, but it is uncountable and linearly independent. Jul 25 '17 at 19:08
• The only proper answer to the question. Such a shame that it is not the most voted answer. May 1 '19 at 21:42
• Way to nail it down, nice work. Dec 28 '20 at 2:05

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.)

• 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 $2$?
– user100463
Aug 17 '17 at 22:41
• @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 May 9 '18 at 17:24

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.

As $$\pi$$ is trascendent over $$\mathbb{Q}$$. Then the set $$\{1, \pi, \pi^{2},\cdots\}$$ is linearly independent.

Let $$n$$ be a positive integer.

Let $$(\alpha_i)_{1 \le i \le n}$$ be a family of $$n$$ nonzero real numbers.

Proposition: The set $$V$$ containing all sums of the form

$$\tag 1 \displaystyle{\sum_{i=1}^n q_i \alpha_i} \quad \text{where } q_i \in \Bbb Q$$

is properly contained in $$\Bbb R$$.
Proof
Construct a family of finite subsets of $$\Bbb Q$$, $$(F_k)_{\,k \in \Bbb N}$$, such that

$$\tag 2 \displaystyle{\bigcup_{\,k \in \Bbb N} F_k = \Bbb Q} \; \text{ and }\; \{-1,+1\} \subset F_0 \; \text{ and } \; F_k \subset F_{k+1}$$

Define the set $$V_k$$ to be all sums of the form

$$\tag 3 \displaystyle{\sum_{i=1}^n q_i \alpha_i} \quad \text{where } q_i \in F_k$$

so that $$V$$ is the union of the $$V_k$$ family of sets with $$V_k \subset V_{k+1}$$.

We will construct a family of nested/shrinking closed intervals $$I_m = [a_m, b_m]$$ satisfying

$$\quad \cap\, I_m = \{\beta\} \text{ where } \beta \notin V$$

The nested interval theorem guarantees that the intersection of the closed intervals is a singleton $$\beta$$ while the algorithm constructing the $$[a_m, b_m]$$ must also take steps to exclude any element in $$V$$ from being in that intersection.

The algorithm (defined using recursion):

Set the initial closed interval to

$$\quad I_0 := [a_0,b_0] := [-\alpha_1, +\alpha_1]$$

Suppose $$I_m := [a_m, b_m]$$ has been set. There is a smallest $$k$$ such that

$$\quad \frac{a_m+b_m}{2} \in V_k$$

With that $$k$$,

if $$m+1$$ is odd set

$$\quad a_{m+1} = \text{max}\big(\{v \in V_k \mid v \lt b_m\}\big)$$
$$\quad b_{m+1} = b_m$$

if $$m+1$$ is even set

$$\quad b_{m+1} = \text{min}\big(\{v \in V_k \mid v \gt a_m\}\big)$$
$$\quad a_{m+1} = a_m$$

and define

$$\quad I_{m+1} := [a_{m+1}, b_{m+1}]$$

By the nested interval theorem the intersection of these intervals is a singleton set; call the element in that set $$\beta$$. Since every finite set $$V_k$$ gets 'consumed' by the algorithm,

$$\quad$$ for every $$k$$ we must have $$\beta \notin V_k$$.

and so $$\beta \notin V$$.

$$\blacksquare$$

• Here is a quick (off-beat) way of showing their are irrational numbers: 1) Construct $\Bbb R$ using, say, Dedekind cuts. $\quad$ 2) Prove the nested interval theorem. $\quad$ 3) Regard $\Bbb R$ as a vector space over $\Bbb Q$ and set $V$ to the vector subspace generated by $1$, and notice that $V = \Bbb Q$. $\quad$ 4) Invoke the above proposition. Jan 27 '20 at 21:57

Let $$(p_{n} )$$ be the sequence of all $$\textbf {prime numbers}$$. Then $$\forall a \neq 1$$ positive number, show the sequence $$( \log_a p_{n} )$$ in the vector space $$\mathbb{R}$$ over $$\mathbb{Q}$$ is linearly independent. So the vector space $$\mathbb{R}$$ over $$\mathbb{Q}$$ is infinite-dimensional.

• This example was already posted here $10$ years ago in my answer. Please don't post duplicate answers. Nov 12 at 9:03
• I didn't check the other answers, this was a solution in my teacher's book that I wrote. Nov 13 at 10:25
• You are supposed to check other answers before posting (esp. when answering a question 10 years old). The correct thing to do in cases like this is to delete this duplicate answer. Nov 13 at 16:47