Prove that the additive identity in a vector space is unique

1. Jan 16, 2010

zeion

1. The problem statement, all variables and given/known data

Prove that the additive identity in a vector space is unique

2. Relevant equations

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

3. 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??

2. Jan 16, 2010

tiny-tim

Hi zeion!

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

Hint: what is y + z ?

3. Jan 16, 2010

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.

4. Jan 16, 2010

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).

5. Jan 16, 2010

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?

6. Jan 17, 2010

tiny-tim

… that 0 was unique? yes!