How to end the proof

  • Thread starter soul
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  • #1
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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 [tex]u + 0_1 = u + 0_2[/tex]. However, I don't know whether I can say [tex]0_1 = 0_2[/tex] 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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  • #2
Defennder
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??? 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.
 
  • #3
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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.
 
  • #4
HallsofIvy
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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.
 
  • #5
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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.
 
  • #6
JasonRox
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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.
 
  • #7
It is impossible to reply. You do not said definition of the vector space, and do not said about preceding procedure.
 

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