Problem
Show that in the definition of a vector space (1.19), the additive inverse condition can be replaced with the condition that $$0v = 0 \textrm{ for all } v \in V$$ Here the $0$ on the left side is the number $0$, and the $0$ on the right side is the additive identity of $V$
Solution
Let $v,w \in V$, such then: $$0 = 0v = (1 + (-1))v \stackrel{D1.19}{=} 1v + (-1)v$$ Now define $-v$ to equal $(-1)v$ and we see that any $v \in V$ has an additive inverse.