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Proof of uniqueness of identity element of addition of vector space
This proof is solely based on vector space axioms.
Axiom names are italicised.
They are defined in Wikipedia (vector space).
Let $V$ be a vector space.
We prove the uniqueness of an identity element of addition (IEOA).
By Identity element of addition,
there exists an IEOA.
Let this element be denoted by $0$.
For the sake of contradiction, we assume that the IEOA is not unique.
That is, there exists an IEOA $0'$ such that $0' \ne 0$.
Obviously, $V \ne \emptyset$. Let $v \in V$.
By Identity element of addition, $v + 0 = v$ and $v + 0' = v$.
Hence, $$v + 0 = v + 0'.$$
By Commutativity of addition, $$0 + v = 0' + v.$$
By Inverse elements of addition, there exists an additive inverse of $v$.
Let $-v$ denote the additive inverse of $v$.
Due to the foregoing equality, $$(0 + v) + (-v) = (0' + v) + (-v).$$
By Associativity of addition, $$0 + (v + (-v)) = 0' + (v + (-v)).$$
By Inverse elements of addition, $$0 + 0 = 0' + 0.$$
By Identity element of addition, $$0 = 0'.$$
The foregoing equality contradicts our assumption.
Thus, our assumption is false and its negation is true.
That is, the IEOA is unique. QED
P.S.: I wrote another version of the proof, which is less roundabout.
Best Answer
This is "too thorough", to put it nicely. One wonders why you just didn't start with $0' + 0 = 0' + 0.$ and conclude $0 = 0'$, using commutativity if you felt necessary.
As another general piece of advice, at the end of each proof by contradiction, look back and see if you can't write it as a straightforward "direct proof" that doesn't use contradiction. This particular proof is a good candidate for that. Unnecessarily using contradiction proofs can lead to proofs that are more complex than necessary.
That said, proofs by contradiction are sometimes the way to go, or else lead you to a method for a direct proof, so they have their uses.