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How would one go about proving this? I'd like someone to just point in the right direction.

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Just check the axioms for a vector space. –  Brian M. Scott Dec 12 '12 at 23:00
    
What is the most difficult part to you? –  Sigur Dec 12 '12 at 23:00
    
I'm almost certain this is a duplicate of a fairly recent post, but I can't find it. –  Cameron Buie Dec 12 '12 at 23:42

3 Answers 3

To see that $\mathbb R$ is a vector space over the rational numbers one has to verify that:

  1. The addition of $\mathbb R$ is commutative, and there is a neutral element.
  2. There is a well-defined operation of multiplying a real number by a rational scalar.
  3. The scalar multiplication and the vector addition behave as they should. (That is, $(\alpha\beta)v=\alpha(\beta v)$, and both distributive laws.)

But most of those are quite trivial to verify because $\mathbb R$ is a field extending the rational numbers.

Do note that while it is tempting to try and prove by finding a basis for $\mathbb R$ over $\mathbb Q$ (i.e. a collection of real numbers which spans the entire field, and is linearly independent over $\mathbb Q$), such set is impossible to write "by hand". There is no formula expressing it directly, and we can only prove its existence using non-constructive axioms of mathematics.

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Good "do note"! Methodologically, or, when giving advice on methodology, I think it is as important as anything else to recognize the impulses one would reasonably have at whatever developmental stage, and provide explanation as to why the correspondingly-seemingly reasonable approaches may create needless trouble. I note that a stylized "axiomatic" approach declares such issues nonexistent or "personal", without addressing them. –  paul garrett Dec 12 '12 at 23:22
    
Thank you very much Paul. This has probably been one of the nicer things I've read in [any] comments recently! –  Asaf Karagila Dec 12 '12 at 23:24
    
You're welcome, @AsafKaragila... :) –  paul garrett Dec 13 '12 at 0:32
    
You don't have to ping me when it's my answer. Also I don't have to ping you because you're the only person (except of me) that commented here. The comments work in mysterious ways, but when you know how they truly work... you know it's just a bunch of voodoo... :-) –  Asaf Karagila Dec 13 '12 at 0:38
    
Aha!... Thx for info... This is arguably reasonable, but also not-so-arguably-canonical, so I will probably not remember... but maybe I will? :) –  paul garrett Dec 13 '12 at 0:46

Whenever $E\subseteq F$ is an inclusion of fields, $F$ is naturally a vector space over $E$. The axioms are easily checked.

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since every field is a vector space over its any subfield Hence F is a field over E which is subfield of F

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This was already pointed out in Andrea Mori's answer months ago. –  Henry T. Horton May 4 '13 at 4:36

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