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

With one exception, the following proof is solely based on vector space axioms. Axiom names are italicised. They are defined in Wikipedia (vector space).

Vector spaces - If an addend adds nothing, then the addend is the zero vector.

Let $V$ be a vector space. By this proof, we know that a zero vector of $V$ is unique; let $0$ be the zero vector of $V$. Let $v_1, v_2 \in V$.

If $v_1 + v_2 = v_1$, then $v_2 = 0$.

Proof. We assume that $v_1 + v_2 = v_1$. It remains to prove that $v_2 = 0$. Let $(-v_1)$ be an additive inverse of $v_1$. \begin{align*} 0 &= v_1 + (-v_1) && \text{by }\textit{Inverse elements of addition} \\ &= (v_1 + v_2) + (-v_1) && \text{by assumption} \\ &= (v_2 + v_1) + (-v_1) && \text{by }\textit{Commutativity of addition} \\ &= v_2 + (v_1 + (-v_1)) && \text{by }\textit{Associativity of addition} \\ &= v_2 + 0 && \text{by }\textit{Inverse elements of addition} \\ &= v_2 && \text{by }\textit{Identity element of addition} \\ \end{align*} QED

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DracoMalfoy
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    Looks like a great proof to me. I especially like that you justified each step. – layman Aug 10 '14 at 14:38
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    The proof is absolutely correct. To avoid the use of the axiom of Commutativity of addition, you can add $-v_1$ to the left. – Taladris Aug 11 '14 at 01:49
  • @Taladris In Wikipedia, the axiom of Inverse elements of addition only stated that $0 = v_1 + (-v_1)$, not that $0 = (-v_1) + v_1$. If that were stated, I could see how to improve the proof. Currently, I don't see how I could do it. – DracoMalfoy Aug 11 '14 at 06:48
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    @dracomalfoy: OK. A vector space is a commutative group by definition, so it follows directly that $(-v_1)+v_1=0$. Actually, many books would state the axiom of existence of an additive inverse as $\forall v\in V,\ \exists w\in V;\ v+w=w+v=0$. – Taladris Aug 11 '14 at 09:38

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The proof is correct. More intuitively, the proof is by cancelling $\,v_1$ from both sides, by adding its inverse, which makes it clear that the proof generalizes to the following cancellation law

$$ v+ u = v + w\ \Rightarrow\ u = w $$

This simple inference, that invertible elements are cancellable, is ubiquitous in algebra.

Bill Dubuque
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