linear algebra - Vector space basics: scalar times a nonzero vector = zero implies scalar = zero?

I'm working through a linear algebra text, starting right from the axioms. So far I've understood and proved for myself that, for some vector space $V$ over a field $\mathbb{F}$:

• the additive identity (zero vector) $\mathbf{0}$ in a vector space is unique


• the additive inverses in a vector space are unique


• scalar zero times any vector is the zero vector: $\forall \mathbf{v} \in V: 0 \mathbf{v} = \mathbf{0}$


• any scalar times the zero vector is the zero vector: $\forall a \in \mathbb{F}: a \mathbf{0} = \mathbf{0}$

However, I'm stuck on the converse of the third statement: suppose $a \mathbf{v} = \mathbf{0}$ and $\mathbf{v} \neq \mathbf{0}$. Show that $a$ must be equal to $0$. In other words, show that the scalar zero is the only element of $\mathbb{F}$ that allows for rule of inference number 3 in the above list.

It seems like such a simple thing but I'm not used to proving super basic statements like this axiomatically. My attempt so far is something like:

$$\begin{align}a\mathbf{v} &= \mathbf{0} \quad &\text{(1)} \\
a \mathbf{v} + \mathbf{0} &= \mathbf{0} \quad &\text{(vector additive identity)} \\
a \mathbf{v} + a \mathbf{v} &= \mathbf{0} \quad &\text{(substitute from 1)} \\
(a + a) \mathbf{v} &= \mathbf{0} \quad &\text{(distributive property)} \\
(2a)\mathbf{v} &= \mathbf{0} \\
(2a)\mathbf{v} &= a \mathbf{v} \quad &\text{(substitute from 1)} \\
\mathrm{``therefore"} \quad 2a &= a \\
a &= 0
\end{align}
$$

But I'm not sure I'm "allowed" to do that second-to-last step yet, given the things proved so far. I think it might just be a circular argument. Is it?

EDIT:

I figured it out with some prodding; turned out I had all the pieces in front of me already but didn't realize it. Here it is for completeness:

Suppose $a \in \mathbb{F}$, $\mathbf{v} \in V$, and $a \mathbf{v} = \mathbf{0}$. Either $a = 0$ or $a \neq 0$. In the case where $a \neq 0$:

$$\begin{align}

a \mathbf{v} &= \mathbf{0} \\
\frac{1}{a} ( a \mathbf{v} ) &= \frac{1}{a} \mathbf{0} \\
(\frac{1}{a} a) \mathbf{v} & = \mathbf{0} \\
1\mathbf{v} &= \mathbf{0} \\
\mathbf{v} &= \mathbf{0}
\end{align}$$

But then suppose further that $\mathbf{v} \neq \mathbf{0}$. Then $a = 0$ by the contrapositive.

Answer

Hint: There was that axiom that said that $1v=v$, was there not?