Question : Is the set Q of rational numbers a real vector space (i.e. scalars are the real
numbers)? Explain your answer.
Yes, since all rational numbers are real numbers?
, can you multiply by, e.g.
o so no, since I get irrational numbers after certain axiom operations.
What axiom operations? You mean that there is no closure under the field action.
That I mean I engage in all those associativity, commutativity.. properties.
I'm a little confused, why are you thinking about associativity and commutativity? Those are fine,
is a field, so those will all work fine, it's closure that's the issue.