Say I have the set $V$ of rational numbers as vectors and the field $F$ of reals as scalars. Does $V$ form a vector space over $F$? I ask this because $V$ isn't closed under scalar multiplication.
On Wikipedia it doesn't state that being closed under scalar multiplication is an axiom, but on WolframAlpha it says "A vector space $V$ is a set that is closed under finite vector addition and scalar multiplication." If WolframAlpha were correct, then surely the rationals couldn't form a vector space over the reals?
Thanks for any replies.