rational number plane vector space or not?

Question

Two questions:
1. Is $\mathbb{Q}^2$ a vector space over the field $\mathbb{Q}$?
2. Is $\mathbb{Q}^2$ a vector space over the field $\mathbb{R}$?

My answer to the first question is yes. Because the multiplication of rational number is rational number. So it is closed under the multiplication.

However, the second one is not. Since it is possible that the multiplication of real number and a rational number is not a rational number, so it will not closed in the multiplication. Remember the vectoe space is $\mathbb{Q}^2$.

Is this proof right? or any strict proof?

Thanks

Answer 1

Another proof for point $2$: $\mathbb{Q}^2$ is countable, $\mathbb{R}$ is not, so $\mathbb{Q}^2$ can't be a vector space over $\mathbb{R}$.

Answer 2

Here is a definition that might help you to write out $(1)$ in a formal way.

Definition: Let $F$ be a field, a vector space over $F$ is a set $V$ together with

1. a binary operation $+$ on $V$ under which $V$ is an abelian group,

2. an action of $F$ on $V$ (that is, a map $F\times V \rightarrow V$) denoted by $rv$, for all $r\in F$ and for all $v\in V$ which satisfies $$(r+s)v = rv+sv,\quad \forall r,s\in F,\forall v\in V$$ $$(rs)v = r(sv),\quad \forall r,s\in F,\forall v\in V$$ $$r(v+u) = rv+ru,\quad \forall r\in F,\forall v, u\in V$$ $$1v = v \quad\forall v \in V.$$