# 英国代考|MATH0005 Algebra 1代数 1

Proof
(a) Suppose that $\mathbf{o}$ is also a zero element for $V$. Then $0+\mathbf{a}=\mathbf{a}$ for all $\mathbf{a} \in V$ by $V 2$, and $\mathbf{b}+\mathbf{0}=\mathbf{b}$ for all $\mathbf{b} \in V$, also by $V 2$. In particular, putting $\mathbf{a}=\mathbf{0}$ and $\mathbf{b}=\mathbf{0}$ we see that $\mathbf{0}+\mathbf{0}=\mathbf{0}$ and that $\mathbf{0}+\mathbf{0}=\mathbf{0}$. Hence $\mathbf{o}=\mathbf{0}$.
(c) Since $\alpha \boldsymbol{0} \in V$, ther is a vector $-(\alpha \mathbf{0}) \in V$, by V3. Thus
$\mathbf{0}=\alpha \mathbf{0}+-(\alpha \mathbf{0}) \quad$ by V3
$=\alpha(0+0)+-(\alpha 0) \quad$ by V 2
$=(\alpha \boldsymbol{0}+\alpha \boldsymbol{0})+-(\alpha \boldsymbol{0})$ by V5
$=\alpha 0+(\alpha 0+-(\alpha 0))$ by V1
$=\alpha \mathbf{0}+\mathbf{0}$ by $\mathrm{V} 3$
$=\alpha 0 \quad$ by $\mathrm{V} 2$

(d) Here we have to show that $(-\alpha) \mathbf{v}$ is the additive inverse of $\alpha \mathbf{v}$. To establish this we show that it has the property claimed for the additive inverse in $\mathrm{V} 3$; namely that when added to $\alpha \mathbf{v}$ we get 0 . Now
\begin{aligned} (-\alpha) \mathbf{v}+\alpha \mathbf{v} &=(-\alpha+\alpha) \mathbf{v} \quad \text { by } \mathrm{V} 6 \ &=\mathbf{0} \mathbf{v} \quad \text { because } F \text { is a field } \ &=\mathbf{0} \quad \text { by } \mathbf{v} 9 \ \text { Similarly, } \quad \quad \quad \alpha \mathbf{v}+(-\alpha) \mathbf{v} &=\mathbf{0} . \end{aligned}

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