I am reading Lang's Linear Algebra and trying to prove that $c0=0$ with the axioms given in $EV1-EV8$. I initially thought the proof was:
$$ c0\stackrel{?}{=}0 \\ c(A-A)\stackrel{?}{=}0\\ cA-cA\stackrel{?}{=}0$$
And from here I guess one could see that the relationship of inverseness would be preserved (by subtracting the inverse of $-cA$ from both sides?). Later, I thought about the following:
$$ c0\stackrel{?}{=}0 \\ \cfrac{1}{c}\cdot c0\stackrel{?}{=}\cfrac{1}{c} 0 \\ 1\cdot0\stackrel{?}{=}\cfrac{1}{c} 0 $$
And using that $1\cdot u=u$:
$$0\stackrel{?}{=}\cfrac{1}{c} 0 $$
But I guess I would be unable to cancel $1/c$ on the other side. I'm a bit confused, none of these proofs seems finished to me. The first one doesn't seems to come to a conclusion and the second one doesn't seems to use the axioms for vector spaces, they seem more to use the axioms for fields. I tried to do something which almost worked:
$$c0\stackrel{?}{=}0\\ (1+1+1+\dots + 1)0\stackrel{?}{=}0 \\1\cdot 0+1\cdot 0 + \dots + 1\cdot 0\stackrel{?}{=}0\\0+0+0+\dots + 0\stackrel{!}{=}0$$
But I noticed that if $c>\lfloor c \rfloor$ then I guess it wouldn't prove it totaly, because:
$$ 1\cdot 0+1\cdot 0 + \dots + 1\cdot 0+ (c-\lfloor c\rfloor)0\stackrel{?}{=}0\\0+0+0+\dots + (c-\lfloor c\rfloor)0\stackrel{?}{=}0 \\ (c-\lfloor c\rfloor)0\stackrel{?}{=}0$$