# Prove $\mathbb R$ vector space over $\mathbb Q$

I am proving that $\mathbb R$ is a vector space over $\mathbb Q$. So far, I have stated that vector addition and scalar multiplication trivially hold in $\mathbb R$. I then showed that $(\mathbb R, +)$ is an abelian group.

Now I am trying to show distributivity of scalar multiplication, but I feel it's trivial to the point where I can't really prove it. This is what I have:

Let $\alpha, \beta \in \mathbb Q$ and $x,y \in \mathbb R$.

Then, $(\alpha+\beta)\cdot x = \alpha \cdot x+ \beta \cdot x$ and $\alpha \cdot(x+y)=\alpha \cdot x+ \alpha \cdot y$ by the usual operations in $\mathbb R$.

Is this correct? I feel it's inadequate, but I don't know how to break it down further.

Any feedback is appreciated.

• "vector addition and scalar multiplication trivially hold in $\mathbb R$"; I don't know what that means. What does it mean to say that vector addition (or scalar multiplication) holds in $\mathbb R$? Do you mean that you have defined your operations in terms of usual real number addition and multiplication? – Jonas Meyer Sep 19 '13 at 3:54
• @JonasMeyer I meant that vector addition is just regular addition in $\mathbb R$ and scalar multiplication is just multiplication in $\mathbb R$ but restricted to $\mathbb Q$ X $\mathbb R$ – Alti Sep 19 '13 at 3:58
• @JonasMeyer Yes, I assumed usual addition and multiplication. Is that an incorrect assumption? – Alti Sep 19 '13 at 3:59
• No, that is a good assumption. But you are then just saying what the operations are. Usually showing that something "holds" means that there is some property that you are proving is true. E.g., the associative property of addition "holds". – Jonas Meyer Sep 19 '13 at 4:07
• @JonasMeyer I'll make sure to keep note of that, thanks! – Alti Sep 19 '13 at 4:13

You're right that it's pretty trivial. You know that $\mathbb R$ under normal addition is an abelian group, and the rest of the axioms follow from the fact that $\mathbb R$ under normal addition and multiplication is a ring, and $x \in \mathbb Q$ implies $x \in \mathbb R$. So your proof of distributivity is correct.