How would one go about proving this? I'd like someone to just point in the right direction.
The real numbers form a vector space over the rationals (i.e. with Q as the scalar)
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0I'm almost certain this is a duplicate of a fairly recent post, but I can't find it. – 2012-12-12
3 Answers
To see that $\mathbb R$ is a vector space over the rational numbers one has to verify that:
- The addition of $\mathbb R$ is commutative, and there is a neutral element.
- There is a well-defined operation of multiplying a real number by a rational scalar.
- 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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0See also: http://drexel28.wordpress.com/2010/10/22/the-dimension-of-r-over-q/ – 2014-01-14
Whenever $E\subseteq F$ is an inclusion of fields, $F$ is naturally a vector space over $E$. The axioms are easily checked.
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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5This was already pointed out in Andrea Mori's answer months ago. – 2013-05-04