Uniqueness of identity element of addition

[supermiedos]

Homework Statement

Here, V is a vector space.
a) Show that identity element of addition is unique.
b) If v, w and 0 belong to V and v + w = 0, then w = -v

Homework Equations

The Attempt at a Solution

a)
If u, 0', 0* belong to V, then
u + 0' = u
u + 0* = u
Adding the additive inverse on both sides of the equations:

0' = u -u
0* = u - u

So, 0' = 0*. Therefore, the identity element of addition is unique.

b) If v, w and 0 belong to V and v + w = 0, then w = -v

v + w + (-v) = 0 + (-v)
w + 0 = 0 + (-v)
w = -v

Is that correct? Because I often see the procedure is more complex.

Thanks in advance

 

[andrewkirk]

That 's all OK. The only reservation I have is that you have used the associativity and commutativity of addition in a vector space to move things around and get them to cancel. For instance to get from #v+w+(-v)$to$w+0## requires one use of the commutative law and one of the associative law. Normally one would do this without bothering to point it out but, since the problem is concerned with the most basic, fundamental algebraic properties of a vector space, your lecturer may expect you to be explicit about things like that.

 

[HallsofIvy]

This:

a)
If u, 0', 0* belong to V, then
u + 0' = u
u + 0* = u
Adding the additive inverse on both sides of the equations:

0' = u -u
0* = u - u

So, 0' = 0*. Therefore, the identity element of addition is unique.

Assumes the additive inverse is also unique. If you are not assuming the additive identity is unique, why should you assume additive inverses are unique? Instead of u+ 0' and u+ 0*, look at 0'+ 0*, first using the fact that 0' is an additive identity, then using the fact that 0* is an additive identity.

 

[andrewkirk]

Assumes the additive inverse is also unique.

I think it's only the use of the word 'the' that implies uniqueness (probably inadvertently) and, as far as I can see, that is not used in the proof. ie if the wording is changed from 'Adding the additive inverse' to 'Adding an additive inverse -u of u' then I think the proof still works.

Also, I see (b) as essentially a proof of the uniqueness of the additive inverse.

 

[supermiedos]

That 's all OK. The only reservation I have is that you have used the associativity and commutativity of addition in a vector space to move things around and get them to cancel. For instance to get from #v+w+(-v)$to$w+0## requires one use of the commutative law and one of the associative law. Normally one would do this without bothering to point it out but, since the problem is concerned with the most basic, fundamental algebraic properties of a vector space, your lecturer may expect you to be explicit about things like that.

That makes sense. So I must be explicit in the use of the properties. Thank you

 

[supermiedos]

This:

Assumes the additive inverse is also unique. If you are not assuming the additive identity is unique, why should you assume additive inverses are unique? Instead of u+ 0' and u+ 0*, look at 0'+ 0*, first using the fact that 0' is an additive identity, then using the fact that 0* is an additive identity.

That's what confuses me. I tought since the problem states that we're going to find the unicity of the identity element of addition, we could use all the rest of the properties without questioning its unicity.

Since 0' + 0* = 0´ and 0* + 0' = 0* then 0* = 0'. Is that correct?

 

[vela]

That's what confuses me. I tought since the problem states that we're going to find the unicity of the identity element of addition, we could use all the rest of the properties without questioning its unicity.

You can't assume the identity is unique when trying to prove the identity is unique. That's begging the question.

 

[andrewkirk]

Since 0' + 0* = 0´ and 0* + 0' = 0* then 0* = 0'. Is that correct?

Your original proof was valid provided you replaces 'the additive inverse' by 'an additive inverse'. But what you've written here, based on HoI's suggestion, is more elegant.

 

[supermiedos]

Thank you so much for your replies. I understand now :)