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We are supposed to prove that the zero vector is unique with the additive identity $x+O=x$ and then by creating another additive identity with $O'$, then $x+O'=x$.

We set them equal $x+O=x+O'$ and use the vector cancelling theorem to get $O=O'$, but how does this prove that the zero vector is unique?

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If there are two distinct additive identities, $O$ and $O'$, then by your steps we have $O=O'$, contradiction.

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Well, you proved that two "zeroes" are actually the same, so you only have one. A cleaner proof is $$0 \stackrel{(1)}{=} 0+0' \stackrel{(2)}{=} 0',$$where in (1) use use that $0'$ is neutral for $+$ (on the right), and in (2) use that $0$ is neutral for $+$ (on the left).

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Let (V, +, ·) be a vector space over ℝ. According to the definition of zero vector, u + 0 = 0 + u = u, ∀ u ∈ V on the contrary, let 0' be another zero vector and 0'∈V such that u + 0' = 0' + u = u, ∀ u ∈ V

Consider 0+0' since 0 is zero vector of V, u + 0 = 0 + u = u, ∀ u ∈ V taking u = 0', 0+0' = 0' (i)

0' is also a zero vector of V according to our assumption. Therefore ∀ u ∈ V, u + 0' = 0' + u = u taking u = 0, 0+0' = 0 (ii)

from (i) and (ii), 0' = 0 as required

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  • $\begingroup$ This adds nothing new to the already existing answers. $\endgroup$ – José Carlos Santos Nov 9 '18 at 8:51

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