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Please rate and comment. I want to improve; constructive criticism is highly appreciated. Please take style into account as well.

The following proof is based on vector space related axioms. Axiom names are italicised. They are defined in Wikipedia (see vector space article). Additionally, this trivial result is used.

Multiplying by $-1$ yields inverse element of vector addition.

Let $V$ be a vector space over a field $F$. \begin{array}{lrll} \text{By} & \dots & \text{we denote} & \dots \\ \hline & 1 && \text{a multiplicative identity element in $F$.} \\ & (-1) && \text{an additive inverse of $1$ in $F$.} \\ & 0 && \text{an additive identity element in $F$.} \\ & \mathbf{0} && \text{an additive identity element in $V$.} \\ \end{array} Let $\mathbf{v} \in V$. We want to prove that $$(-1)\mathbf{v} \text{ is an additive inverse of } \mathbf{v} \text{ in } V.$$ Proof. We prove that $\mathbf{v} + (-1)\mathbf{v} = \mathbf{0}$. \begin{align*} \mathbf{v} + (-1)\mathbf{v} &= 1\mathbf{v} + (-1)\mathbf{v} && \text{by }\textit{Identity element of scalar multiplication} \\ &= (1 + (-1))\mathbf{v} && \text{by }\textit{Distrib. of scalar mult. (field addition)} \\ &= 0\mathbf{v} && \text{by }\textit{Inverse elements of field addition} \\ &= \mathbf{0} && \text{by the result linked above} \end{align*} QED

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    $\begingroup$ Yeah, that's fine $\endgroup$ – Hagen von Eitzen Aug 20 '14 at 16:18
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    $\begingroup$ Your proof is good. I will just comment that in the right-hand column in your table, all four "a"/"an" should be "the", because the three identities and -1 are all unique. $\endgroup$ – Bungo Aug 20 '14 at 16:25
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    $\begingroup$ @Bungo I did not use "the" because uniqueness is not explicit in the axioms (in Wikipedia). $\endgroup$ – DracoMalfoy Aug 20 '14 at 16:28
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    $\begingroup$ @draco - OK, fair enough :-) $\endgroup$ – Bungo Aug 20 '14 at 17:06
  • $\begingroup$ i.e. scale $\,\bf v\,$ by $\ 0\, =\, -1 + 1\ \ $ $\endgroup$ – Bill Dubuque Aug 20 '14 at 17:50

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