Is the Additive Inverse in a Field Always Unique?

[JPanthon]

Homework Statement

Suppose F is a field and a is an element of F . Prove that the additive inverse of a is
unique (and so we may write it as -a). Justify each line of your proof
in terms of the eld axioms

Homework Equations

The field axioms.
http://mathworld.wolfram.com/FieldAxioms.html

The Attempt at a Solution

Given: (additive inverse) for every a in F, there exists a b in F, such that a + b = b + a = 0

Proof

a + b = b + a = 0
a + b = 0 (additive inverse)
b = -a (adding (-a) to both sides)
a - a = 0 (substitute -a in place of b)This proof seems incomplete to me. Have I made assumptions anywhere?
Anything anyone would change?

Also, this is for a first year uni maths course.

thank you in advance

 

[Robert1986]

Well, when you say you add "-a" you are sort of assuming that which you are trying to prove. Forget about "-a" for the proof; the book only mentioned "-a" to give the motivation for wanting to prove this is true in a field. What you need to do is start off by saying something like:

Let a be an element of F and let b and b' be additive inverses of a. So, a + b = b + a = 0 and a + b' = b' + a = 0.


Now, using those relations and the field axioms, you need to prove that b = b'.

 

[JPanthon]

Well, when you say you add "-a" you are sort of assuming that which you are trying to prove. Forget about "-a" for the proof; the book only mentioned "-a" to give the motivation for wanting to prove this is true in a field. What you need to do is start off by saying something like:

Let a be an element of F and let b and b' be additive inverses of a. So, a + b = b + a = 0 and a + b' = b' + a = 0.


Now, using those relations and the field axioms, you need to prove that b = b'.

Thank you for your reply!

Okay, your method makes much more sense, and would prove uniqueness. But, without subtracting a to both sides, how can I re-write b = b' as -a?

 

[Robert1986]

Thank you for your reply!

Okay, your method makes much more sense, and would prove uniqueness. But, without subtracting a to both sides, how can I re-write b = b' as -a?

Until you complete this proof, forget that "-a" is something that you are familiar with; just act as though "-a" does not exist. This uniquiness thing is essentially telling you "hey, you can write additive inverses as '-a' just as you did with addition of numbers."

So, I'd start by writting:

b = b + 0 (by the definition of 0)
= what does this equal?
...
= b'

Now, use what you know about b,b' and a to get the RHS of an equation to be b' as above.

EDIT: One last thing to point out, if you have this equation:

a+b = a+b'

you are NOT allowed to do something like this (though it may be tempting):

b + a + b = b + a + b' implies b = b'

Why can't you do this? This "cancelation" idea is not an axiom of a field. It can be deduced, but you haven't proven that yet.

 

[JPanthon]

Until you complete this proof, forget that "-a" is something that you are familiar with; just act as though "-a" does not exist. This uniquiness thing is essentially telling you "hey, you can write additive inverses as '-a' just as you did with addition of numbers."

So, I'd start by writting:

b = b + 0 (by the definition of 0)
= what does this equal?
...
= b'

Now, use what you know about b,b' and a to get the RHS of an equation to be b' as above.

EDIT: One last thing to point out, if you have this equation:

a+b = a+b'

you are NOT allowed to do something like this (though it may be tempting):

b + a + b = b + a + b' implies b = b'

Why can't you do this? This "cancelation" idea is not an axiom of a field. It can be deduced, but you haven't proven that yet.

Thank you again. Does this work?

(1) a + b = b + a = 0
(2) a + b' = b' + a = 0

Let a = 0

(1) b + (0) = 0
[Zero additive]
b = 0

(2) b' + (0) = 0
b' = 0

b = 0 = b'
Therefore, b = b'


Please reply.

 

[Robert1986]

Thank you again. Does this work?

(1) a + b = b + a = 0
(2) a + b' = b' + a = 0

Let a = 0

(1) b + (0) = 0
[Zero additive]
b = 0

(2) b' + (0) = 0
b' = 0

b = 0 = b'
Therefore, b = b'


Please reply.

No, this doesn't work. First of all, at best all this does is prove that 0 is the unique additive inverse of 0 (and whether you have proven that is VERY debateable.) You are proving that FOR ALL a in F, the additive inverse of a is unique. So, you are NOT allowed to say "a=0" (or a equals anything, for that matter.) ALL you can assume about a is that it is in F.

Let me give you another hint:

Let a be an element of F and let b and b' be additive inverses of a, that is a+b = b+a = 0 and a+b' = b'+a = 0. Then,
b = b + 0 = b + (a + b') = ...

Now, keep manipulating this until the RHS of the last "=" is " b' ". Then you have proven that b = b'. Now, once you do that, you are going to have to re-write it line-by-line and explain why you can get from one step to the other from the field axioms.