# How to end the proof

Hi eveyone,

I was trying to prove that for the vector spaces, there is a unique vector that satisfy "u + 0 = u" and I used contradiction technique. The last point that I reached is $$u + 0_1 = u + 0_2$$. However, I don't know whether I can say $$0_1 = 0_2$$ after this statement or there are some other operations that I must do (like this statement needs a proof as well?).

Thank you.

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## Answers and Replies

Defennder
Homework Helper
??? Why do you need to prove that? Isn't that an axiom that every vector space has to satisfy? Namely that every vector space has a unique zero vector? What axioms do you start off with?

You can "prove" this by noting that along with your last step, -u also exists in the same vector space.

Sorry, I guess I wrote my question wrong. I was trying to prove there is a unique identitiy element in summation and what I did is to select two different vectors and at the end of it, to show they are the same. I used the axiom in the question.

HallsofIvy
Science Advisor
Homework Helper
Uniqueness of the identity is an axiom in groups but can be proved in a vector space.

In any vector space, there exist additive inverses and addition is commutative. Add the additive inverse of u to both sides of your equation.

Uniqueness of the identity is an axiom in groups but can be proved in a vector space.

In any vector space, there exist additive inverses and addition is commutative. Add the additive inverse of u to both sides of your equation.

Is this true? Almost every group theory book I have looked at proves the uniqueness of the identity as a theorem.

JasonRox
Homework Helper
Gold Member
Uniqueness of the identity is an axiom in groups but can be proved in a vector space.

In any vector space, there exist additive inverses and addition is commutative. Add the additive inverse of u to both sides of your equation.

The axioms say there must exist a zero vector. It does not say it is unique or must be unique. You prove that it is unique if there exists such a vector.

It is impossible to reply. You do not said definition of the vector space, and do not said about preceding procedure.