Prove that the additive identity in a vector space is unique

[zeion]

Homework Statement

Prove that the additive identity in a vector space is unique

Homework Equations

Additive identity

There is an element 0 in V such that v + 0 = v for all v in V

The Attempt at a Solution

Assume that the additive identity is NOT unique, then there exists y, z belong to V such that
A + y = A + z = A, then y = z = 0, which is a contradiction.

Is this enough to prove??

 

[tiny-tim]

… Assume that the additive identity is NOT unique, then there exists y, z belong to V such that
A + y = A + z = A, then y = z = 0, which is a contradiction.

Is this enough to prove??

Hi zeion!

hmm … you're assuming that A - A = 0, which is sort-of begging the question.

Hint: what is y + z ?

 

[zeion]

Since y, z belong to V, and y, z are the zero vectors in V, then
y + z = y = z, which is a contradiction.

 

[tiny-tim]

Yup!

(except it's not actually a contradiction … unless you state at the start that y and z are different, which you don't have to).

 

[zeion]

Ooh okay! Thanks ^_^

But why couldn't I say that A - A = 0?
Could I do that if I stated that I assumed A was in V?

..or is it because then I would be assuming that A was unique?

 

[tiny-tim]

… that 0 was unique? yes!