Proof that $a(x − y) = ax − ay$ for x and y in vector spaces. Hi i'm confused about this homework question:
"Let $V$ be a vector space over the field $F$ and let $a ∈ F$ and $x, y ∈ V$ .
Show that $a(x − y) = ax − ay$ in $V$."
What I did is below but i'm confused because I feel as though it was too simple.
So from the vector space axioms there is a distributivity property that states that $a(\alpha+\beta)=a\alpha+a\beta$ where $a ∈ F$ and $\alpha, \beta$ $ ∈ V$ so I thought if I set $ \alpha=x$ and $\beta=-y$ then the equality above just becomes: $a(x+(-y))=ax+a(-y))$$=ax+(-a)y)$$=ax-ay$.
Is that ok what I've done?
Many thanks
 A: Assuming you've already proven that $-y = (-1)y$ (if not, prove it first), I'd recommend adding just a couple of extra steps (in red) so that you can justify each step by a single axiom/ previous lemma:
$$\begin{align}a(x − y) &= a(x+(-y)) & \text{Def of vector subtraction} \\ &= ax + a(-y) &\text{Distributivity over vector addition} \\ &\color{red}{= ax + a[(-1)y]} &\text{Previous lemma: $-y = (-1)y$} \\ &= ax + (a\cdot -1)y &\text{Compatibility of field and scalar multiplication$^\dagger$} \\ &\color{red}{= ax + (-1\cdot a)y} &\text{Commutativity of field multiplication} \\ &\color{red}{= ax + (-1)(ay)} &\text{Compatibility of field and scalar multiplication$^\dagger$} \\ &= ax - ay &\text{Def of vector subtraction}\end{align}$$
$\dagger:$ Meaning this axiom $a(by) = (ab)y$.
A: (Responding to the suggestions in the comments.)
Recall that anything multiplied by $0$ is $0$.
If unfamiliar, you may observe that $0x + 0x = (0 + 0)x = 0x$, and use a right cancellation law on the equation $0x + 0x = 0x$ to conclude $0x = 0$.
Next, observe that $(-1)y + y = (-1)y + 1y = (-1 + 1)y = 0y = 0$.
In particular, $(-1)y + y = 0$, so adding $-y$ on the right for both expressions, and again using a right cancellation law, we have $(-1)y = -y$ as desired.
