[proofplan] Both parts follow directly from the vector space axioms. For (i), we exploit $0_{\mathbb{F}} = 0_{\mathbb{F}} + 0_{\mathbb{F}}$, apply distributivity, and cancel. For (ii), we show $(-1)v + v = \mathbf{0}$ using distributivity and part (i), then invoke uniqueness of additive inverses. [/proofplan] [step:Prove $0_{\mathbb{F}}v = \mathbf{0}$ using distributivity and cancellation] In the field $\mathbb{F}$, the additive identity satisfies $0_{\mathbb{F}} = 0_{\mathbb{F}} + 0_{\mathbb{F}}$. Applying the distributivity axiom (scalar addition distributes over scalar multiplication): \begin{align*} 0_{\mathbb{F}}v = (0_{\mathbb{F}} + 0_{\mathbb{F}})v = 0_{\mathbb{F}}v + 0_{\mathbb{F}}v. \end{align*} Adding $-(0_{\mathbb{F}}v)$, the additive inverse of $0_{\mathbb{F}}v$ in the abelian [group](/page/Group) $(V, +)$, to both sides: \begin{align*} \mathbf{0} = 0_{\mathbb{F}}v. \end{align*} [/step] [step:Prove $(-1)v = -v$ using distributivity and uniqueness of inverses] Using the distributivity axiom and the identity $1_{\mathbb{F}}v = v$ (the scalar multiplication identity axiom): \begin{align*} (-1)v + v = (-1)v + 1_{\mathbb{F}}v = (-1 + 1)v = 0_{\mathbb{F}}v = \mathbf{0}, \end{align*} where the last equality uses part (i). Since $(-1)v + v = \mathbf{0}$, the element $(-1)v$ is an additive inverse of $v$. The additive inverse in an abelian group is unique (if $w + v = \mathbf{0}$, then $w = w + \mathbf{0} = w + (v + (-v)) = (w + v) + (-v) = -v$). Hence $(-1)v = -v$. [/step]