Prove that for any vector ${\bf v}$ its additive inverse $−{\bf v}$ is given by $(−1){\bf v}$.

Solution: We have

\begin{align*}

(-1){\bf v}=&\ (-1){\bf v}+{\bf 0}\\ =&\ (-1){\bf v}+({\bf v}+(-{\bf v}))\\

=&\ ((-1){\bf v}+{\bf v})+(-{\bf v})\\ =&\ ((-1){\bf v}+1{\bf v})+(-{\bf v})\\

=&\ (-1+1){\bf v}+(-{\bf v})\\ =&\ 0{\bf v}+(-{\bf v})\\ =&\ {\bf 0}+(-{\bf v})=-{\bf v}.

\end{align*}Here we used Exercise 1.1.7, namely $0{\bf v}={\bf 0}$.