# Dimension of R over Q without cardinality argument. [duplicate]

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.

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## marked as duplicate by Asaf Karagila, Rick Decker, hardmath, Byron Schmuland, Davide GiraudoJul 16 at 14:19

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).

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Awesome, exactly what I was looking for, thanks ! It is almost better than the cardinality proof... –  Denis Jul 16 at 11:40
Glad to help! As a number theorist I love this kind of thing. :-) –  Adam Hughes Jul 16 at 11:41
Very nice! Do you even need FTA? If $\sum_{p<N}a_p\log p = 0$, then $\log(\prod_{p<N}p^{a_p})=0$ so $\prod_{p<N}p^{a_p} = 1$ - which is only possible if $a_p = 0$ for all $p$ without FTA –  Mathmo123 Jul 16 at 11:42
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. –  Denis Jul 16 at 11:44
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. –  Adam Hughes Jul 16 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 $\pi$.

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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. –  Denis Jul 16 at 11:34
It would work with any transcendental number. Are there any that you could prove a result for? –  Mathmo123 Jul 16 at 11:35
The only argument I know for existence of transcendental numbers is cardinality ;) –  Denis Jul 16 at 11:35
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 –  Mathmo123 Jul 16 at 11:37
Transcendence of $e$ is more easily proven than for $\pi$. –  hardmath Jul 16 at 14:12