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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 solely based on vector space axioms. Axiom names are italicised. They are defined in Wikipedia (vector space).

Another proof of uniqueness of identity element of addition of vector space

We prove the uniqueness of an identity element of addition of a vector space.

Let $V$ be a vector space. It remains to prove that an identity element of addition of $V$ is unique. We define a predicate $P$ over $V$ as follows. For every $v \in V$, $$P(v) \quad := \quad \big(\, u + v = u \; \text{ for all } u \in V \,\big).$$

By Identity element of addition, there exists an element of $V$ for which $P$ holds. Let $0 \in V$ such that $P(0 )$ holds; let $0' \in V$ such that $P(0')$ holds. It remains to prove that $0 = 0'$. \begin{align*} 0 & = 0 + 0' && \text{since } P(0') \text{ holds.} \\ & = 0' + 0 && \text{by }\textit{Commutativity of addition.}\ \\ & = 0' && \text{since } P(0) \text{ holds.} \end{align*} QED

P.S.: This is intended to improve my previous, roundabout proof. Of course, I could only incorporate a small subset of the helpful advice given there.

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DracoMalfoy
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    It's not wrong per se, but I found it odd that the phrase 'it remains to prove...' is at the beginning of the proof. It may also be a good idea to clearly state that your choice of $0,0'$ were arbitrary, but that is probably not necessary. – fixedp Aug 09 '14 at 10:03
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    This is not new. The equation $0=0+0'=0'$ is in every book on algebra. The same works for uniqueness of identity in groups. Moreover, the predicate-introduction is completely unnecessary. – Peter Franek Aug 09 '14 at 10:39
  • @fixedp: What wording would you use to state that they are arbitrary? – DracoMalfoy Aug 09 '14 at 16:48

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