Q:

any field that is a superset of the real numbers is also a vector space over the real numbers?

This seems intuitively true since any such field would have to satisfy being closed over addition/multiplication, etc, which would translate directly to fulfilling vector space axioms. I'm a little confused as to the field only being over the real numbers though. It seems to me that such a field containing the real numbers would either have to be the field of real numbers or the field of complex numbers in order to be closed under addition & multiplication, but I'm not sure how to go about showing that.

A:

Okay, I got to the wikipedia article on vector spaces and consider the definition.
Okay so we need a set $V$ so that $F \supset \mathbb R$ is such that $F=\{av|a\in \mathbb R; v\in V\}$.
Well... let $V = F\setminus \mathbb R+\{0, 1\}$.
Then 
1-3) $V$ is a associative and commutative under addition has an additive identity.  Check, it inherits all those from the field $F$.
4) an additive inverse is in $V$.  Well, an additive inverse is in $F$.  if $v\in \mathbb R$ then $-v \in \mathbb R$ and $-(-v) = v$.  So $v \in \mathbb R \iff -v\in \mathbb R$.  So $v\in V; v\not \in \mathbb R \implies -v\not \in \mathbb R$ so $-v \in V$.  Check.
5-8) scalar (real) multiplication compatibility, multiplicative identity in $\mathbb R$, distributive over scalars (real numbers) over addition of elements of $V$ , distribution of elements of $V$ or addition of reals.  All check.  $F$ is a field.
That's it.
....
Actually, I'm not sure I see any problem with letting $V = F$.  I don't see anything in the definition that requires non-redunancy that some vectors can not also be scalars.

