I am looking for the easiest (elementary) proof that $\mathbb R$ is infinite dimensional as a $\mathbb Q$-vector space, without using cardinality. It should be understandable at highschool level.

So I guess the question could be reformulated as: what is the easiest infinite family of reals that can be showed to be independent ? So far square roots of primes seems a good candidate, but the proof is still a little intricate, is it the easiest possible ?

The goal of this is to show students that we can prove the result by different ways, and see that understanding cardinals is useful. But I still want the students to be able to understand the other proof.


marked as duplicate by Asaf Karagila, Rick Decker, hardmath, user940, Davide Giraudo Jul 16 '14 at 14:19

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


Consider the set:

$$\{\log p\}_p$$

of logs of prime numbers. They are linearly independent over $\Bbb Q$ by the Fundamental Theorem of Arithmetic--i.e. unique factorization of integers greater than $1$ into primes. As there are infinitely many primes, the set $\Bbb R$ contains an infinite dimensional subspace, and is so itself infinite dimensional. (This is the standard proof number theorists love).

  • $\begingroup$ Awesome, exactly what I was looking for, thanks ! It is almost better than the cardinality proof... $\endgroup$ – Denis Jul 16 '14 at 11:40
  • $\begingroup$ Glad to help! As a number theorist I love this kind of thing. :-) $\endgroup$ – Adam Hughes Jul 16 '14 at 11:41
  • 3
    $\begingroup$ in a strict sense you need FTA, because some of the $a_p$ can be negative, and then you have to say that the decomposition is unique. $\endgroup$ – Denis Jul 16 '14 at 11:44
  • 1
    $\begingroup$ I like recycling this fact a lot in algebra it takes the form of presenting $\Bbb Q$ as a $\Bbb Z$-module of the form: $$\Bbb Q\cong \bigoplus_p \Bbb Z$$ where the index represents the exponent of the prime. $\endgroup$ – Adam Hughes Jul 16 '14 at 11:47

Try $\{1, \pi, \pi^2, \pi^3,\ldots\}$ (or use any transcendental number instead of $\pi$). Since if $$\sum_{i=0}^na_i\pi^i=0, \text{ }a_i\in\mathbb Q$$then $\pi$ must be a root of $$a_0 + a_1x + \ldots + a_nx^n$$which would contradict the transcendence of $\pi$.

  • 1
    $\begingroup$ Thanks it is nice, but transcendance of $\pi$ is a hard result which must be admitted at this level... It is harder than the proof with square root of primes. $\endgroup$ – Denis Jul 16 '14 at 11:34
  • $\begingroup$ It would work with any transcendental number. Are there any that you could prove a result for? $\endgroup$ – Mathmo123 Jul 16 '14 at 11:35
  • 2
    $\begingroup$ The only argument I know for existence of transcendental numbers is cardinality ;) $\endgroup$ – Denis Jul 16 '14 at 11:35
  • $\begingroup$ I see the issue... there is Liouville's theorem (en.wikipedia.org/wiki/…) but that is hardly high school level. I'll have a think $\endgroup$ – Mathmo123 Jul 16 '14 at 11:37
  • $\begingroup$ Transcendence of $e$ is more easily proven than for $\pi$. $\endgroup$ – hardmath Jul 16 '14 at 14:12

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